(Ebook) - Engineering - Basic Engineering Circuit Analysis, 7Ed(Solution Manual)

(Ebook) - Engineering - Basic Engineering Circuit Analysis, 7Ed(Solution Manual)

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Unformatted text preview: Problem 1.1 If 60 C of charge pass through an electric conductor in 30 seconds, determine the current in the conductor. Suggested Solution i= Q 60 = =2A t 30 i =2A Problem 1.2 If the current in an electric conductor is 2.4 A, how many coulombs of charge pass any point in a 30 second interval? Suggested Solution Q = i t = 2.4 30 = 72 C Q = 72 C Problem 1.3 Determine the time interval required for a 12-A battery charger to deliver 4800 C. Suggested Solution t = Q 4800 = = 400 s i 12 t = 400 s Problem 1.4 A lightning bolt carrying 30,000 A lasts for 50 microseconds. If the lightning strikes an airplane flying at 20,000 feet, what is the charge deposited on the plane? Suggested Solution Q = i t = 30, 000 50 10-6 = 1.5 C Q = 1.5 C ( ) Problem 1.5 Determine the energy required to move 240 C through 6 V. Suggested Solution w = V Q = 6 240 = 1440 J w = 1440 J Problem 1.6 Determine the amount of power absorbed or supplied by the element shown if (a) V1 = 4V , I = 2 A (b) V1 = -4V , I = -2 A I + V1 - Suggested Solution (a) P = 4 2 = 8W 8W absorbed (b) P = ( -4 )( -2 ) = 8W 8W absorbed Problem 1.7 Determine the amount of power absorbed or supplied by the element shown if (a) V1 = -6V , I = 3 A (b) V1 = 6V , I = -3 A I + V1 - Suggested Solution (a) P = ( -6 ) 3 = -18W 18W supplied (b) P = 6 ( -3) = -18W 18W supplied Problem 1.8 Determine the missing quantity in the circuits shown. I=? + V1 = 6 V - P = 24 W + V1 = 12 V - P=? I =2A (a) (b) Suggested Solution (a) P = 6 I = 24 I =4A I =4A (b) P = 12 ( -2 ) = -24 W P = 24W supplied Problem 1.9 Determine the missing quantity in the circuits shown. I=? V1 = -6 V + I = -2 A V1 = ? + P = 12 W P = -24 W (a) (b) Suggested Solution (a) P = ( -6 ) I = 12 I = -2 A I = -2 A (b) P = V1 - ( -2 ) = -24 V1 = -12 V V1 = -12 V Problem 1.10 Determine the missing quantity in the circuits shown. I = -3 A + V1 = 4 V - I = -4 A V1 = -6 V + P=? P=? (a) (b) Suggested Solution (a) P = 4 ( -3) = -12 W P = 12 W supplied (b) P = ( -6 ) ( -4 ) = 24 W P = 24W absorbed Problem 1.11 Determine the power supplied to the elements shown. 4A 1 + 2V + 8V - 2 4A Suggested Solution P = 2 4 = 8W 1 P = 8W 1 P2 = 8 4 = 32 W P2 = 32 W Problem 1.12 Determine the power supplied to the elements shown. 2A 1 + 4V 8V + 2 2A Suggested Solution P = 4 2 = 8W 1 P = 8W 1 P2 = ( -8 ) 2 = -16W P2 = -16 W Problem 1.13 Two elements are connected in series, as shown. Element 1 supplies 24W of power. Is element 2 absorbing or supplying power, and how much? 1 + 3V 6V + 2 Suggested Solution If element 1 is supplying power, the direction of the current must be as shown. I + 3V 6V + 1 2 P = 3 ( - I ) = -2 4 1 I =8A Then, P2 = 6 8 = 48W P2 = 48W absorbed Problem 1.14 Two elements are connected in series, as shown. Element 1 absorbs 36W of power. Is element 2 absorbing or supplying power, and how much? 1 12 V + + 4V - 2 Suggested Solution If element 1 is absorbing power, the direction of the current must be as shown. I 12 V + + 4V - 1 2 P = 12 I = 36 1 I =3A Then, P2 = 4 ( -3) = -12 W P2 = 12W supplied Problem 1.15 Determine the power that is absorbed or supplied by the circuit elements shown. + 10 V 1 2A 2A + 14 V - +6V1 1A 18 V 1A 24 V 2 12 V 2A 1A (a) (b) Suggested Solution (a) P24V = 24 ( -2 ) = -48W P24V = 48W supplied P = 10 2 = 20W 1 P = 20W absorbed 1 P2 = 14 2 = 28W P2 = 28W absorbed (b) P V = 18 ( -1) = -18W 18 P V = 18W supplied 18 P = 6 1 = 6W 1 P = 6 W absorbed 1 P2 = 12 1 = 12W P2 = 12W absorbed Problem 1.16 Find the power that is absorbed or supplied by the circuit elements shown. +6V1 2A Ix = 4 A +8V1 + 20 V - 2A 14 V 16 V 4A 2Ix 2A 4A (a) (b) Suggested Solution (a) P2 A = 20 ( -2 ) = -40 W P2 A = 40 W supplied P = 6 2 = 12W 1 P = 12 W absorbed 1 P V = 14 2 = 28W 14 P V = 28W absorbed 14 P4 A = 16 ( -4 ) = -64W P4 A = 64W supplied P = 8 4 = 32 W 1 P = 32W absorbed 1 P2 I x = ( 2 I x ) 4 = 8 4 = 32 W P2 I x = 32 W absorbed (b) Problem 1.17 Compute the power that is absorbed or supplied by the elements in the network shown. Ix = 4 A + 12 V 1 2A + 24 V + 28 V 1Ix 2A 36 V 2 3 Suggested Solution P36V = 36 ( -4 ) = -144W P36V = 144 W supplied P = 12 4 = 48W 1 P = 48W absorbed 1 P2 = 24 2 = 48W P2 = 48W absorbed P I x = (1I x ) ( -2 ) = 4 ( -2 ) = -8W 1 P I x = 8W supplied 1 P3 = 28 2 = 56 W P3 = 56W absorbed Problem 1.18 Find I x in the network shown. + 1A + 12 V 6A + 12 V 2A 1 + 12 V 8V 2 Ix 3 4V 4 + 12 V - 1A Suggested Solution P6 A = 12 ( -6 ) = -72W or 72W supplied P = 12 2 = 24W absorbed 1 P2 = 12 1 = 12W absorbed P3 = 4 I x = 4 I x W absorbed P8V = 8 I x = 8I x W absorbed P4 = 12 1 = 12W absorbed Since Power Supplied = Power Absorbed , then 72 = 24 + 12 + 4 I x + 8I x + 12 or 12 I x = 24 . Therefore, I x = 2 A . Ix = 2 A Problem 1.19 Is the source VS in the network shown absorbing or supplying power, and how much? 3A +6V2 6A 9A 10 V + 1 9A 16 V + 3 8V + VS 3A Suggested Solution P = 10 3 = 30W absorbed 1 P9 A = 16 ( -9 ) = -144 W or 144 W supplied P3 = 8 6 = 48W absorbed P2 = 6 3 = 18W absorbed In order to satisfy Conservation of Power, P + P2 + P9 A + P S + P3 = 0 1 V or 30 + 18 - 144 + P S + 48 = 0 V Therefore, P S = 48W . V P S = 48W absorbed V Problem 1.20 Find VS in the network shown. 2A 6V 2V + 1 2 + 4V 6A 4 3 - 8V + VS 6V + 4A Suggested Solution P6V = 6 ( -2 ) = -12W or 12W supplied P2 = 4 6 = 24 W absorbed P S = VS ( -4 ) = -4VS W or 4VS W supplied V P4 = 6 4 = 24 W absorbed P3 = 8 4 = 32W absorbed P = 2 2 = 4 W absorbed 1 In order to satisfy Conservation of Power, P + P6V + P2 + P3 + P S + P4 = 0 1 V or 4 - 12 + 24 + 32 - 4VS + 24 = 0 Therefore, 4VS = 72 or VS = 18V . VS = 18V Problem 1.21 Find I o in the network shown. 6A + 8V 1 4A 24 V 2 + 10 V + Io 4 16 V + 6V 1A 3A Ix = 2 A - 6V + 3 4Ix 3A 5 - 8V + 6 Suggested Solution P24V = 24 ( -6 ) = -144 W or 144 W supplied P4 I x = ( 4 I x ) ( -3) = 8 ( -3) = -24 W or 24W supplied P = 8 6 = 48W absorbed 1 P3 = 6 I o = 6 I o W absorbed P6 = 8 3 = 24W absorbed P2 = 10 4 = 40 W absorbed P5 = 6 1 = 6W absorbed P4 = 16 2 = 32W absorbed In order to satisfy Conservation of Power, P24V + P4 I x + P + P3 + P6 + P2 + P5 + P4 = 0 1 or -144 - 24 + 48 + 6 I o + 24 + 40 + 6 + 32 = 0 Therefore, 6 I o = 18 or I o = 3 A . Io = 3 A Problem 2.1 Find the current I and the power supplied by the source in the network shown. I 6V 20 k Suggested Solution I= 6 = 0.3 mA 20 103 P = VI = (6)(0.3 10-3 ) = 1.8 mW Problem 2.2 In the circuit shown, find the voltage across the current source and the power absorbed by the resistor. + 3 mA VCS _ 3 k Suggested Solution VCS = ( 3 10-3 )( 3 103 ) = 9 V P = VI = ( 3 10-3 ) ( 9 ) = 27 mW Problem 2.3 If the 10-k resistor in the network shown absorbs 2.5 mW, find VS . VS 10 k Suggested Solution P= VS2 10 k or VS = P R = ( 2.5 10 )(10 10 ) = 5 V -3 3 Problem 2.4 In the network shown, the power absorbed by Rx is 5 mW. Find Rx . 1 mA 2 Rx Suggested Solution Rx = P 5 10-3 = = 20 k I 2 ( 0.5 10-3 )2 Problem 2.5 In the network shown, the power absorbed by Rx is 20 mW. Find Rx . 2 mA Rx Suggested Solution R = P 20 10-3 = = 5 k I 2 ( 2 10-3 )2 Problem 2.6 A model for a standard two D-cell flashlight is shown. Find the power dissipated in the lamp. 1- lamp 1.5 V 1.5 V Suggested Solution Vlamp = 1.5 + 1.5 = 3 V Plamp = 2 Vlamp Rlamp = ( 3) 1 2 =9W Problem 2.7 An automobile uses two halogen headlights connected as shown. Determine the power supplied by the battery if each headlight draws 3 A of current. 12 V Suggested Solution PS = VI = (12 )( 3 + 3) = 72 W Problem 2.8 Many years ago a string of Christmas tree lights was manufactured in the form shown in (a). Today the lights are manufactured as shown in (b). Is there a good reason for this change? (a) (b) Suggested Solution First of all, one must recognize the fact that a failed lamp becomes an open circuit. Then: In (a), one faulty lamp would take out the whole string. There would be no way to identify the faulty lamp. In (b), if one lamp fails, the others stay lit. It is easy to identify the faulty lamp. Problem 2.9 Find I1 in the network shown. 4 mA 12 mA I1 2 mA Suggested Solution I2 4 mA 12 mA I1 2 mA I 2 = 0.004 + 0.002 = 0.006 A 0.012 = I1 + I 2 I1 = 6 mA Problem 2.10 Find I1 and I 2 in the circuit shown. I3 = 5 mA IS I1 I6 = 3 mA I2 I4 = 6 mA Suggested Solution By KCL, I1 = I 4 = 6 mA Also, I 6 + I 2 = I 4 or 0.003 + I 2 = 0.006 I 2 = 3 mA Problem 2.11 Find I x and I y in the network shown. 4 mA 12 mA Ix Iy 8 mA Suggested Solution 0.012 = 0.004 + I x I x = 8 mA 0.004 = I y + 0.008 I y = -4 mA Problem 2.12 Find I x , I y and I z in the circuit shown. Ix Iy 12 mA 4 mA 2 mA Iz Suggested Solution Ix Iy 4 mA 2 mA Iz 12 mA I x + 0.012 = 0.004 I x = -8 mA I y + 0.004 = 0.002 + 0.012 I y = 10 mA I z + 0.004 = 0.002 I z = -2 mA Problem 2.13 Find I x in the circuit shown. Ix 1 mA 2 Ix 4 mA Suggested Solution Ix 1 mA 2 Ix 4 mA I x + 0.001 = 0.004 + 2 I x I x = -3 mA Problem 2.14 Find I x , I y and I z in the network shown. 3 mA 12 mA Ix Iy 4 mA Iz 2 mA Suggested Solution 3 mA 12 mA Ix Iy 4 mA Iz 2 mA I x + 0.003 = 0.012 I x = 9 mA I y + 0.012 + 0.002 = 0.004 I y = -10 mA I z + 0.004 = 0.002 I z = -2 mA Problem 2.15 Find I x in the circuit shown. 2 mA 2Ix Ix 6 mA Suggested Solution 2 mA 2Ix Ix 6 mA 2 I x + 0.002 = I x + 0.006 I x = 4 mA Problem 2.16 Find I x in the network shown. 4 Ix 1 mA Ix 6 mA Suggested Solution 4 Ix 1 mA Ix 6 mA 4 I x + 0.001 = I x + 0.006 Ix = 5 mA 3 Problem 2.17 Find Vx in the circuit shown. + Vx - 9V 2 Vx Suggested Solution KVL: -9 + Vx + 2 V = 0 Vx = 3 V Problem 2.18 Find Vbd in the circuit shown. a + 12 V 4V b 6V c + 2V _ d Suggested Solution KVL: -12 + 4 + Vbd = 0 Vbd = 8 V Problem 2.19 Find Vad in the network shown. a 4V + e 3V d 3V + b 2V + 12 V c Suggested Solution KVL: -4 + Vad - 3 = 0 Vad = 7 V Problem 2.20 Find Vaf and Vec in the circuit shown. a _ 2V + g 2V + f 1V + e 12 V b + 1V c 3V d + 3V _ Suggested Solution KVL: 2 + Vaf + 2 = 0 Vaf = -4 V KVL: Vec + 3 + 3 = 0 Vec = -6 V Problem 2.21 Find Vac in the circuit shown. a + 12 V 4V b 3 Vx c + Vx = 2 V _ d Suggested Solution KVL: Vac + 2 - 12 = 0 Vac = 10 V Problem 2.22 Find Vad and Vce in the circuit shown. a 4 Vx + e 1V d 1V + b Vx = 2 V 12 V + c Suggested Solution KVL: Vad - 12 + 2 + 1 = 0 Vad = 9 V KVL: Vce + 1 - 12 = 0 Vce = 11 V Problem 2.23 Find Vab in the network shown. a 3 12 V 6 b c Suggested Solution Using voltage division, 3 Vab = 12 = 4 V 3+ 6 Problem 2.24 Find Vbd in the network shown. a 3 k 12 V b 1 k 4V c d Suggested Solution a 3 k 12 V I d b 1 k 4V c I= 12 - 4 = 2 mA 3000 + 1000 Vbd - 4 - 1000(0.002) = 0 KVL: Vbd = 6 V Problem 2.25 Find Vx in the circuit shown. 2 k 24 V 5 k + Vx _ 6V 2 k Suggested Solution Using voltage division, 5000 Vx = ( 24 - 6 ) = 10 V 2000 + 5000 + 2000 Problem 2.26 Find Vx in the circuit shown. 24 V _ Vx + 4 k 6 k 6V 8V Suggested Solution Using voltage division, 4000 Vx = ( 6 + 8 - 24 ) = -4 V 4000 + 6000 Problem 2.27 Find Vx in the network shown. 15 5V 3 Vx 5 + Vx _ Suggested Solution 15 5V I 3 Vx 5 + Vx _ KVL: -5 + 15 I - 3Vx + Vx = 0 , where Vx = 5 I Therefore, -5 + 15 I - 3 ( 5 I ) + 5 I = 0 I =1 A Vx = 5 V Problem 2.28 Find V1 in the network shown. + 20 k + V1 _ Vx 40 k 5V - Vx 2 Suggested Solution + I 20 k + V1 _ Vx 40 k 5V - Vx 2 KVL: - Vx + 20000 I + 40000 I + 5 = 0 , where Vx = 40000 I 2 40000 I + 20000 I + 40000 I + 5 = 0 2 Therefore, - I = -0.125 mA V1 = Vx + 5 = 40000 I + 5 = -5 + 5 = 0 V Problem 2.29 Find the power absorbed by the 30-k resistor in the circuit shown. 30 k 12 V 2 Vx 10 k + Vx _ Suggested Solution I 30 k 12 V 2 Vx 10 k + Vx _ KVL: -12 + 30000 I + 2Vx + Vx = 0 where Vx = 10000 I Therefore, -12 + 30000 I + 2 (10000 I ) + 10000 I = 0 I = 200 A P30 k = I 2 ( 30000) ) = ( 200 10-6 ) ( 30000 ) = 1.2 mW 2 Problem 2.30 Find I o in the network shown. 12 mA 2 k Io 6 k 3 k Suggested Solution 1 2000 Io = (12 mA ) = 6 mA 1 1 1 + + 2000 6000 3000 Problem 2.31 Find I o in the circuit shown. 8 k 120 mA 4 k Io 4 k Suggested Solution Using current division, 1 4000 Io = ( 0.120 ) = 90 mA 1 1 + 4000 8000 + 4000 Problem 2.32 Find I o in the network shown. 6 k 12 mA 12 k Io 12 k Suggested Solution Using current division, 1 12000 Io = - ( 0.012 ) = -3 mA 1 + 1 + 1 6000 12000 12000 Problem 2.33 Find Vo in the circuit shown. 6 k 24 mA 8 k + 6 k Vo _ 8 k Suggested Solution Combining the 8-k resistors in parallel yields the following circuit. I 6 k 24 mA 4 k + 6 k Vo _ Using current division, 1 6000 + 6000 I = ( 0.024 ) = 6 mA 1 1 + 4000 6000 + 6000 Then, Vo = 6000 I = 36 V Problem 2.34 Find I o in the network shown. 3 k Io 6 k 4 k 12 mA 2 k Suggested Solution Combining the 3-k and 6-k resistors in parallel yields the following equivalent circuit. I1 2 k 12 mA 4 k 2 k 4000 I1 = ( 0.012 ) = 6 mA 4000 + ( 2000 + 2000 ) Then, 6000 Io = I1 = 4 mA 6000 + 3000 Problem 2.35 Determine I L in the circuit shown. Ix 2 k 6 mA IL Ix 2 4 k Suggested Solution Ix 2 k 6 mA IL 4 k + Ix 2 V1 _ KCL: I x - 0.006 + Ix V V + I L = 0 , where I x = 1 and I L = 1 2000 4000 2 Therefore, V1 V V - 0.006 + 1 + 1 = 0 2000 4000 4000 V1 = 6 V Then, IL = 6 = 1.5 mA 4000 Problem 2.36 Determine I L in the circuit shown. Ix 2 k 3 Ix IL 3 k 6 mA 6 k 3 mA Suggested Solution Ix 2 k 3 Ix IL 3 k + V1 _ 6 k 6 mA 3 mA KCL: V1 V V - 0.006 + 3 I x + 0.003 + I x + I L = 0 , where I x = 1 and I L = 1 . 6000 2000 3000 Therefore, V1 V V V - 0.006 + 3 1 + 0.003 + 1 + 1 = 0 6000 2000 3000 2000 V1 = 6 = 1.2 V 5 and IL = 1.2 = 0.4 mA 3000 Problem 2.37 Find RAB in the circuit shown. 2 k A 2 k RAB B 3 k 6 k 12 k Suggested Solution The network can be redrawn as shown below. C A 2 k RAB B C B A 3 k 6 k 12 k B 2 k A Then, At A-A: 6000 12000 = 4 k At B-B: 2000 + 4000 = 6 k At C-C: 3000 6000 = 2 k and RAB = 2000 + 2000 = 4 k Problem 2.38 Find RAB in the circuit shown. 5 k A 2 k RAB B 2 k 2 k 2 k 2 k Suggested Solution D 5 k A 2 k RAB B D C B A 2 k 2 k 2 k 2 k C B A At A-A: 2000 + 2000 = 4 k At B-B: 2000 4000 = 4 k 3 At C-C: 2000 + 4000 10 = k 3 3 At D-D: 2000 Then, 10000 = 1250 3 RAB = 5000 + 1250 = 6250 Problem 2.39 Find RAB in the network shown. 5 k A 4 k RAB 6 k 3 k B 3 k 8 k 4 k Suggested Solution The network can be redrawn as shown below. C 5 k A RAB 6 k 3 k 3 k 4 k 8 k B A 4 k B C B A At A-A: 4000 + 8000 = 12 k At B-B: 3000 3000 4000 12000 = 1 k At C-C: 5000 + 1000 = 6 k Therefore, RAB = 6000 6000 = 3 k Problem 2.40 Find RAB in the circuit shown. A 2 k 4 k RAB 3 k 6 k 4 k B 12 k Suggested Solution The circuit can be redrawn as shown below. C A 2 k B 4 k RAB 12 k A 6 k 4 k B C 3 k At A: At B: 6000 3000 = 2 k 4000 + 2000 = 6 k At C-C: 6000 12000 = 4 k Then, RAB = 2000 + 4000 + 4000 = 10 k Problem 2.41 Find RAB in the network shown. 6 k A 6 k RAB B 6 k 2 k Suggested Solution The 2-k resistor is shorted. Therefore, the network reduces to: A RAB B 6 k 6 k 6 k Then, RAB = 6000 6000 6000 = 2 k Problem 2.42 Find RAB in the circuit shown. 12 k A 2 k RAB B 4 k 6 k 12 k Suggested Solution The circuit maybe redrawn as follows: A B 2 k RAB 4 k A 12 k 6 k 12 k B At A: At B: Then, 12000 6000 = 4 k 2000 + 4000 = 6 k RAB = 4000 6000 12000 = 2 k Problem 2.43 Find RAB in the circuit shown. 2 k 2 k 2 k 4 k A 2 k B 2 k 2 k RAB 4 k 2 k 2 k Suggested Solution Combining each series pair of 2-k resistors, the circuit can be redrawn as follows: 4 k 4 k 2 k 4 k A 4 k B RAB 4 k 4 k A 2 k 4 k B RAB 4 k or A RAB B 6 k 6 k Then, RAB = 6000 6000 = 3 k Problem 2.44 Find the range of resistance for the following resistors: a) 1 k with a tolerance of 5%. b) 470 with a tolerance of 2%. c) 22 k with a tolerance of 10%. Suggested Solution a) Minimum value = (1 - 0.05 )(1 k ) = 950 Maximum value = (1 + 0.05 )(1 k ) = 1050 b) Minimum value = (1 - 0.02 )( 470 ) = 460.6 Maximum value = (1 + 0.02 )( 470 ) = 479.4 c) Minimum value = (1 - 0.1)( 22 k ) = 19.8 k Maximum value = (1 + 0.1)( 22 k ) = 24.2 k Problem 2.45 Given the network shown, find the possible range of values for the current and power dissipated by the following resistors: a) 390 with a tolerance of 1%. b) 560 with a tolerance of 2%. I 10 V R Suggested Solution (10 V ) = 100 . 10 V Note that I = and P = R R R 2 a) Minimum resistor value = (1 - 0.01)( 390 ) = 386.1 Maximum resistor value = (1 + 0.01)( 390 ) = 393.9 Minimum current value = 10 = 25.39 mA 393.9 10 = 25.90 mA Maximum current value = 386.1 100 = 253.9 mW 393.9 100 = 259 mW Maximum power value = 386.1 Minimum power value = b) Minimum resistor value = (1 - 0.02 )( 560 ) = 548.8 Maximum resistor value = (1 + 0.02 )( 560 ) = 571.2 Minimum current value = 10 = 17.51 mA 571.2 10 = 18.22 mA Maximum current value = 548.8 100 = 175.1 mW 571.2 100 = 182.2 mW Maximum power value = 548.8 Minimum power value = Problem 2.46 Given the circuit shown: a) Find the required value of R . b) Use Table 2.1 to select a standard 10% tolerance resistor for R . c) Calculate the actual value of I . d) Determine the percent error between the actual value of I and that shown in the circuit. e) Determine the power rating for the resistor R . I = 40 mA 510 100 mA R Suggested Solution a) From KCL, the current in the 510- resistor is 100 mA - 40 mA = 60 mA . From Ohm's Law, the voltage across the circuit is ( 510 )( 60 mA ) = 30.6 V . Therefore, the required value of R is ( 30.6 V ) ( 40 mA ) = 765 . b) From Table 2.1, the nearest 10% resistor value is 820 . c) To find the actual value of I, first find the actual voltage across the circuit, which is 31.44 V = 38.35 mA . (100 mA ) ( 510 820 ) = 31.44 V . Then, I = 820 ( 38.35 - 40 ) d) The percent error is (100 ) = -4.14 % . 40 e) Since the power consumption in R is ( 31.44 V ) 820 2 = 1.206 W , a 2-W resistor should be used. Problem 2.47 The resistors R1 and R2 shown in the circuit are 1 with a tolerance of 5% and 2 with a tolerance of 10%, respectively. a) What is the nominal value of the equivalent resistance? b) Determine the positive and negative tolerance for the equivalent resistance. R1 Req R2 Suggested Solution a) The nominal value is Req = R1 + R2 = 3 b) The minimum and maximum values of R1 are 0.95 and 1.05 , and the minimum and maximum values of R2 are 1.8 and 2.2 . Thus, the minimum and maximum values of Req are 0.95 + 1.8 = 2.75 and 1.05 + 2.2 = 3.25 . The positive and negative tolerances are Positive Tolerance = ( 3.25 - 3) 3 (100 ) = 8.33 % (100 ) = -8.33 % Negative Tolerance = ( 2.75 - 3) 3 Problem 2.48 Find I1 and Vo in the circuit shown. I1 6V 1 k 3 k + Vo _ 12 k Suggested Solution From Ohm's Law: I1 = -6 V = -0.5 mA 12 k 3 k By application of voltage division: Vo = ( -6 V ) = -4.5 V 3 k + 1 k Problem 2.49 Find I1 and Vo in the circuit shown. 2 k 12 V 8 k 6 k I1 4 k + Vo _ Suggested Solution Combining resistors 6 k ( 8 k+4 k ) = 4 k reduces the network to the following: 2 k + 12 V V1 _ 4 k Using voltage division, then 4000 V1 = (12 V ) = 8 V . 2000 + 4000 Looking back at the original circuit, 2 k 12 V + 8V _ 8 k 6 k I1 4 k + Vo _ Ohm's Law: I1 = 8V 4 = mA 6 k 3 4000 8 Vo = (8) = V 3 8000 + 4000 Voltage division: Problem 2.50 Find I1 in the circuit shown. I1 2 k 2 k 12 mA 10 k 2 k 2 k Suggested Solution The circuit can be simplified as follows: I2 2 k 12 mA 10 k 2 k ( 2 k+2 k) = 4 k 3 10000 ( 0.012 ) = 9 mA Then, I 2 = 10000 + 2000 + 4000 3 2000 and I1 = ( 9 mA ) = 3 mA 2000 + ( 2000 + 2000 ) Problem 2.51 Determine Vo in the network shown. 4 k 14 k 9 k 2 k 12 k 2 mA + Vo _ Suggested Solution The network can be redrawn as: A 4 k 14 k 9 k 2 k 12 k 2 mA + Vo _ A The equivalent resistance to the left of A-A is (14 k + 4 k ) 9 k 12 k = 4 k . 2 k 4 k 2 mA + Vo _ Then, Vo = - ( 2 k + 4 k )( 2 mA ) = -12 V . Problem 2.52 Find Vo in the network shown. 2 8 2 + 4 24 V 4 Vo _ Suggested Solution The network can be redrawn as: 8 2 2 4 24 V 4 + Vo _ Combining the four resistors on the right-hand side ( 4 + 2 ) ( 8 + 4 ) = 4 yields: 2 I1 24 V 4 and I1 = (2 + 4 ) 24 V = 4 A. Then, reconsidering the original circuit, 2 8 2 + 4 24 V 4A 4 Vo _ I2 ( 4 + 2) 4 I2 = (4 A) = A 4 + 2 ) + (8 + 4 ) 3 ( and 16 4 V Vo = - ( 4 A ) A = - 3 3 Problem 2.53 Find I o in the circuit shown. 9 k 24 V 6 k 4 k 3 k 6 k Io Suggested Solution The circuit can be redrawn as: B A I1 24 V 9 k 6 k I2 4 k 6 k Io 3 k B A At A-A: 6000 3000 = 2 k At B-B: 6000 ( 4000 + 2000 ) = 3 k The circuit simplifies to: I1 24 V 9 k 3 k Then, I1 = 24 6000 = 2 mA I 2 = I1 = 1 mA 9000 + 3000 6000 + ( 4000 + 2000 ) 3000 1 and I o = I 2 = mA . 3000 + 6000 3 Problem 2.54 Find Vo in the circuit shown. 4 k 2 k 2 k 12 k 3 k 6 mA + Vo _ Suggested Solution A B 4 k 2 k 2 k 12 k 3 k 6 mA + Vo _ A B At A-A: At B-B: ( 2000 + 4000 ) 12000 = 4 k ( 4000 + 2000 ) 3000 = 2 k Then, Vo = ( 2 k )( 6 mA ) = 12 V . Problem 2.55 Find I o in the network shown. 1 k 10 k 2 k 6 k 12 V 4 k 4 k 6 k 3 k Io Suggested Solution Redrawing the network: D I1 1 k C 10 k B 2 k A 12 V 4 k Io 6 k 4 k 3 k 6 k D C B A At A-A: 3000 6000 = 2 k At B-B: 4000 ( 2000 + 2000 ) = 2 k At C-C: 6000 (10000 + 2000 ) = 4 k At D-D: 4000 4000 = 2 k I1 = 12 = 4 mA 1000 + 2000 4000 I1 = 2 mA Using current division, I o = 4000 + 4000 Problem 2.56 Find Vo in the network shown. 9 k 6 k 12 mA 8 k 4 k 4 k + Vo _ Suggested Solution A I1 9 k 6 k 12 mA 8 k 4 k 4 k I2 + Vo _ A At A-A: 4000 ( 8000 + 4000 ) = 3 k 6000 Using current division, I1 = 0.012 = 4 mA 6000 + ( 9000 + 3000 ) 4000 Again using current division, I 2 = I1 = 1 mA 4000 + ( 8000 + 4000 ) Then, Vo = 4000 I 2 = 4 V . Problem 2.57 Find I o in the circuit shown. 6 k 6 k 3 k 12 V 2 k 4 k Io Suggested Solution The circuit can be redrawn as: A 6 k 6 k 12 V 3 k 6 k I1 2 k 12 V 2 k 4 k Io 2 k 4 k Io A At A-A: I1 = ( 2000 + 2000 ) 4000 = 2 k 12 = 1.5 mA 6000 + 2000 ( 2000 + 2000 ) Using current division, I o = I1 = 0.75 mA ( 2000 + 2000 ) + 4000 Problem 2.58 If Vo = 4 V in the network shown, find VS . 8 k VS 4 k + Vo = 4 V _ Suggested Solution 8 k VS 4 k I + Vo = 4 V _ I= Vo 4 = = 1 mA 4000 4000 Then, VS = 8000 I + Vo = 8 + 4 = 12 V . Problem 2.59 If the power absorbed by the 4-k resistor in the network shown is 36 mW, find I o . 9 k 12 k Io 4 k Suggested Solution 9 k 12 k Io + V1 _ 4 k P4 k = Io = V12 = 36 mW 4 k V1 = 12 V 12 V = 1 mA 12 k Problem 2.60 If the power absorbed by the 4-k resistor in the circuit shown is 36 mW, find VS . 9 k VS 12 k 4 k Suggested Solution I2 9 k VS 12 k Io I1 + V1 _ 4 k P4 k = 36 mW = V12 4000 V1 = 12 V Io = I1 = V1 = 1 mA 12 k V1 = 3 mA 4 k I 2 = I o + I1 = 1 mA + 3 mA = 4 mA VS = ( 9 k ) I 2 + V1 = 48 V Problem 2.61 In the network shown, the power absorbed by the 4- resistor is 100 W. Find VS . 3 VS 7 4 3 Suggested Solution I3 VS I2 3 3 + V1 _ 7 4 I1 P4 = I12 ( 4 ) = 100 W V1 = 4 I1 = 20 V I1 = 5 A I2 = V1 =2A 7 +3 I 3 = I1 + I 2 = 7 A VS = 3 I S + V1 = 21 + 20 = 41 V Problem 2.62 In the network shown, Vo = 6 V . Find I S . 8 k 4 k IS 4 k 2 k + Vo _ Suggested Solution I2 8 k 4 k IS _ + 4 k VS 2 k I1 + Vo _ I1 = Vo 6 = = 3 mA 2 k 2000 VS = ( 4 k ) I1 + Vo = 12 + 6 = 18 V I2 = VS 18 = = 1.5 mA 8 k + 4 k 12000 I S = I1 + I 2 = 3 mA + 1.5 mA = 4.5 mA Problem 2.63 In the circuit shown, I = 4 mA . Find VS . 2 k VS 6 k I 5 k 3 k Suggested Solution I2 2 k VS 6 k _ I + V1 5 k 3 k I1 V1 = ( 6 k ) I = 24 V I1 = V1 24 = = 3 mA 5 k + 3 k 8000 I 2 = I + I1 = 0.004 + 0.003 = 7 mA VS = ( 2 k ) I 2 + V1 = 14 + 24 = 38 V Problem 2.64 In the circuit shown, I o = 2 mA . Find I S . 4 k 4 k IS 2 k 6 k Io 1 k 2 k Suggested Solution I3 4 k + 4 k V3 _ I2 IS + V2 2 k + V1 _ I1 1 k 6 k 2 k Io V1 = ( 6 k ) I o = 12 V I1 = V1 12 = = 4 mA 1 k + 2 k 3000 I 2 = I o + I1 = 0.002 + 0.004 = 6 mA V2 = ( 2 k ) I 2 = 12 V V3 = V2 + V1 = 12 + 12 = 24 V I3 = V3 24 = = 3 mA 4 k + 4 k 8000 I S = I 3 + I 2 = 0.003 + 0.006 = 9 mA Problem 2.65 In the network shown, V1 = 12 V . Find VS . 4 k 2 k VS 6 k + V1 1 k 3 k 4 k Suggested Solution A I3 2 k + VS V2 _ I1 I2 + 4 k V1 1 k 3 k 6 k 4 k A At A-A: 4000 (1000 + 3000 ) = 2 k I1 = V1 12 = = 3 mA 4 k 4000 V2 = V1 + ( 2 k ) I1 = 12 + 6 = 18 V I2 = V2 18 = = 3 mA 6 k 6000 I 3 = I1 + I 2 = 0.003 + 0.003 = 6 mA VS = ( 2 k ) I 3 + V2 = 12 + 18 = 30 V Problem 2.66 In the circuit shown, Vo = 2 V . Find I S . 12 10 IS 2 8 3 4 + Vo _ Suggested Solution I4 12 10 + V2 _ I3 2 + V1 _ I2 8 4 I1 + Vo _ 3 IS I1 = Vo 2 = = 0.5 A 4 4 V1 = ( 8 ) I1 + Vo = 6 V I2 = V1 =2A 3 I 3 = I 2 + I1 = 2.5 A V2 = ( 2 ) I 3 + V1 = 11 V I4 = V2 = 0.5 A 10 + 12 I S = I 3 + I 4 = 2.5 + 0.5 = 3 A Problem 2.67 In the network shown, Vo = 6 V . Find I S . 3 k IS 9 k + Vo _ 1 k 2 k 5 k Suggested Solution + + IS V2 _ I3 V1 3 k + Vo _ I2 I1 1 k 5 k 9 k I4 2 k I1 = Vo 6 = = 3 mA 2 k 2000 Vo 6 = = 1 mA 1 k + 5 k 6000 I2 = I 3 = I1 + I 2 = 0.003 + 0.001 = 4 mA V1 = ( 3 k ) I 3 = 12 V V2 = V1 + Vo = 12 + 6 = 18 V I4 = V2 18 = = 2 mA 9 k 9000 I S = I 3 + I 4 = 0.004 + 0.003 = 6 mA Problem 2.68 In the circuit shown, I o = 2 A . Find I S . 4 8 IS 6V 3 Io 4 Suggested Solution I3 4 8 + V2 _ IS I2 6V + V1 _ 3 Io 4 I1 V1 = ( 3 ) I o = 6 V I1 = V1 6 = = 1.5 A 4 4 I 2 = I o + I1 = 2 + 1.5 = 3.5 A V2 = 6 V + V1 = 6 + 6 = 12 V I3 = V2 12 = =1 A 8 + 4 12 I S = I 2 + I 3 = 3.5 + 1 = 4.5 A Problem 2.69 If I o = 4 mA in the circuit shown, find I S . 1 k Io 4 k IS 10 k 12 V 2 k Suggested Solution + + I2 I3 V2 Io 4 k IS V3 _ 10 k V1 _ 12 V 2 k I1 1 k + V1 = ( 4 k ) I o - 12 V = 16 - 12 = 4 V I1 = V1 4 = = 2 mA 2 k 2000 I 2 = I o + I1 = 0.004 + 0.002 = 6 mA V2 = (1 k ) I 2 = 6 V V3 = V2 + V1 = 6 + 4 = 10 V I3 = V3 10 = = 1 mA 10 k 10000 I S = I 2 + I 3 = 0.006 + 0.001 = 7 mA Problem 2.70 Find I o in the circuit shown. I1 = 5 mA 12 k 12 k 12 k 9 mA 2 k Io 8 k Suggested Solution I1 = 5 mA _ V1 + + 2 k Io 12 k 12 k V3 _ V2 + I3 8 k I2 12 k 9 mA I1 + I 2 = 9 mA I 2 = 4 mA V1 = (12 k ) I1 = 60 V V2 = (12 k ) I 2 = 48 V V3 = V1 - V2 = 60 - 48 = 12 V I3 = V3 12 = = 1 mA 12 k 12000 I o = I1 + I 3 = 0.005 + 0.001 = 6 mA Problem 2.71 Find Vo in the circuit shown. 12 k + 4 k 6 k 18 k 6 k I1 = 4 mA Vo 42 V _ Suggested Solution 12 k I 3 + V1 _ 4 k _ V2 + I2 I1 = 4 mA 42 V _ 6 k Vo + 6 k V3 18 k + V1 = ( 4 k ) I1 = 16 V V2 = 42 V - V1 = 42 - 16 = 26 V I2 = V2 26 13 = = mA 6 k 6000 3 13 1 mA - 4 mA = mA 3 3 I 3 = I 2 - I1 = V3 = (12 k ) I 3 = 4 V Vo = V1 - V3 = 16 - 4 = 12 V Problem 2.72 Find the power absorbed by the network shown. 6 k 21 V 12 k 6 k 2 k 18 k Suggested Solution Redrawing the network: 2 k 2 k 21 V 12 k 6 k 6 k Y 6 k 21 V 6 k 2 k 18 k 3 k The equivalent resistance seen by the source is: Req = ( 2 k + 6 k ) ( 6 k + 2 k ) + 3 k = 4 k + 3 k = 7 k Then, P= ( 21 V ) Req 2 = 441 = 63 mA 7000 Problem 2.73 Find I o in the circuit shown. 2 9 36 V 12 12 Io 3 4 5 18 Suggested Solution Note that the four right-most resistors can be combined as ( 3 + 4 + 5 ) 12 = 6 . Then the circuit can be redrawn as: I 9 2 6 36 V 18 12 Io 9 2 Io Y 3 36 V 2 6 I= 36 V ( 9 + 3 ) ( 2 + 2 ) + 6 =4A ( 9 + 3) Io = I =3A + 3) + ( 2 + 2 ) (9 Problem 2.74 Find I o in the circuit shown. 12 12 12 V 12 14 5 Io Suggested Solution Applying the Y transformation to the three 12- resistors yields: Ix 4 4 4 12 V 5 Io 14 ( 4 + 14 ) Io = - Ix 4 + 5 ) + ( 4 + 14 ) ( where I x = 12 V 4 + ( 4 + 5 ) ( 4 + 14 ) = 12 = 1.2 A 10 Io = - 18 12 4 i = - -0.80 A 27 10 5 Problem 2.75 Find I o in the circuit shown. 12 Io 6 4 12 A 18 6 Suggested Solution Converting the to a Y yields the circuit: Io 2 3 4 12 A 6 6 ( 2 + 4) Io = (12 A ) = 4 A ( 2 + 4) + ( 6 + 6) Problem 2.76 Find I o in the circuit shown. 12 A 4 12 4 12 4 Io Suggested Solution Applying the Y transformation: I 12 A 12 12 12 12 Io 12 12 Io 12 12 A 12 12 12 The circuit can be further simplified to: I 12 A 12 12 I= 12 A =6A 2 I =3A 2 Io = Problem 2.77 Find I o in the circuit shown. 6 12 V 12 6 Io 6 12 6 Suggested Solution Combining the two 12- resistors yields: 6 12 V Io 6 6 6 6 12 V 2 Io 2 Y 6 6 2 The circuit can be further simplified to the following: 12 V 8 Io 2 12 V 4 Io 2 8 Io = 12 V =2A 6 Problem 2.78 Find Vo in the circuit shown. IS 3 k 12 V 2000 IS 5 k + Vo _ Suggested Solution KVL: -12 + 3000 I S - 2000 I S + 5000 I S = 0 I S = 2 mA Vo = 5000 I S = 10 V Problem 2.79 Find Vo in the network shown. 2 k 12 V 2 Vo 2 k + Vo _ Suggested Solution I 2 k 12 V 2 Vo 2 k + Vo _ KVL: -12 + 2000 I + 2Vo + Vo = 0 where Vo = 2000 I Therefore, -12 + 2000 I + 4000 I + 2000 I = 0 I = 1.5 mA Vo = 2000 (1.5 10-3 ) = 3 V Problem 2.80 Find Vo in the network shown. Ix 5 mA 2 k 2 Ix 1 k + Vo _ Suggested Solution KCL: -0.005 + Vo V V + 2 I x + o = 0 where I x = o 2000 1000 2 k Therefore, -0.005 + Vo V V + 2 o + o = 0 Vo = 2 V 2000 2000 1000 Problem 2.81 Find I o in the network shown. 8 6 + Io 4 Vx _ 5A 3 Vx Suggested Solution + 8 6 5A + 4 _ Vx _ 3 Vx V1 Io KCL: V1 V V1 4 + 1 - 5 + 3Vx = 0 where Vx = V1 = 6 8+4 3 8+4 Therefore, V1 V1 V + -5+3 1 = 0 6 12 3 V1 = 4 V Io = V1 4 2 = = A 6 6 3 Problem 2.82 Find Vo in the circuit shown. 2 k 6 k Vo 2000 4 mA + 1 k Vo _ Suggested Solution + 2 k 6 k Vo 2000 V1 4 mA + 1 k Vo _ _ KCL: V V1 V1 V1 1000 - o - 0.004 + = 0 where Vo = V1 = 6000 2000 2000 + 1000 2000 + 1000 3 Therefore, 3Vo V 3Vo - o - 0.004 + =0 6000 2000 3000 Vo = 4 V Problem 2.83 A single-stage transistor amplifier is modeled as shown. Find the current in the load RL . + RS = 1 k Rb = 250 110 Ib Ro = 4 k RL = 400 VS = 250 mV _ Ib Io Suggested Solution Ib = VS 0.25 = = 0.2 mA RS + Rb 1000 + 250 4000 Io = - (110 I b ) = -100 I b = -20 mA 4000 + 400 Problem 2.84 A typical transistor amplifier is shown. Find the amplifier gain G (i.e., the ratio of the output voltage to the input voltage). 100 500 5 k + 300 Vo _ VS = 250 mV 5 k Ib 3.5 105 Ib Suggested Solution I1 100 500 5 k + 300 Vo _ VS = 250 mV 5 k Ib 3.5 105 Ib I1 = VS 0.25 = = 451 A 100 + ( 5 k 500 ) 554.5 5000 Ib = I1 = 409.8 A 5000 + 500 300 5 Vo = ( -3.5 10 ) I b = -8.12 V 300 + 5000 G Vo -8.12 = -32.5 VS 0.250 Problem 2.85 For the network shown, choose the values of Rin and Ro such that Vo is maximized. What is the resulting ratio, Vo VS ? RS + VS Rin Vin _ Ro + Vin RL Vo _ Suggested Solution Rin Vin = VS RS + Rin RL Rin RL Vo = ( Vin ) = Ro + RL RS + Rin Ro + RL VS Therefore, Rin RL Vo = VS RS + Rin Ro + RL Clearly, to maximize Vo Rin , we must maximize Rin . This is because is always less than 1, but gets VS RS + Rin closer and closer to 1 as Rin is made larger and larger. On the other hand, to maximize Vo RL , we must minimize Ro . This is because is always less than VS Ro + RL 1, but gets closer and closer to 1 as Ro is made smaller and smaller. In the limit, V lim lim o Ro 0 V S Rin RL lim = Rlim R 0 in R + R o Ro + RL S in = 1 1 = Rin Problem 2.86 In many amplifier applications we are concerned not only with voltage gain, but also with power gain. Power gain = Ap = ( power delivered to the load ) ( power delivered by the input ) Find the power gain for the circuit shown, where RL = 50 k . RS = 2 k + VS Ro = 100 k V/100 + Rin = 8 k V _ RL = 50 k Vo _ Suggested Solution RS = 2 k IS VS + Ro = 100 k V/100 Io + Vo _ Rin = 8 k V _ RL = 50 k 8000 V = VS = 0.8VS 8000 + 2000 100, 000 V -3 Io = - = -5.33 10 VS 100, 000 + 50, 000 100 Pload = I o2 RL = ( 5.33 10-3 VS ) ( 50, 000) ) = 1.422 VS2 2 VS VS2 Pin = VS I S = VS = 8000 + 2000 10, 000 Ap = Pload 1.422 VS2 = = 14.22 103 Pin VS2 10, 000 Problem 2.87 Find the power absorbed by the 10-k resistor in the circuit shown. + 4 k 11 mA 2 k 4 k 10 k Vx 2000 Vo + Vx _ 3 k _ Suggested Solution Combining the two 4-k resistors in parallel yields the following simplified circuit: + 2 k 11 mA 2 k Vx 2000 Vo + Vx _ 3 k _ 10 k -0.011 + Vo V Vo Vo 3000 + x + + = 0 where Vx = Vo = 0.6Vo 2000 + 3000 2 k 2000 2 k + 3 k 10 k Therefore, -0.011 + Vo 3Vo V Vo + + o + = 0 Vo = 10 V 2000 10, 000 5000 10, 000 2 P k = 10 (10 ) Vo2 = = 10 mW 10 k 10, 000 Problem 2.88 Find the power absorbed by the 12-k resistor in the circuit shown. + 4 k 3 Io 6 mA 6 k Vo 6 k 3 k _ Io 12 k Suggested Solution + 4 k 3 Io 6 mA 6 k I1 Vo 6 k 3 k _ Io 12 k -0.006 + Vo Vo Vo 6000 2 + 3 Io + + = 0 where I o = I1 = I1 6 k 12 k 6000 + 3000 3 4 k + ( 6 k 3 k ) I1 = Vo V V 2 V = o Io = o = o 3 6000 9000 4 k + ( 6 k 3 k ) 6000 Then, substituting: -0.006 + Vo Vo V V + 3 o + o + = 0 Vo = 8 V 6000 9000 6000 12, 000 P k = 12 V02 64 = = 5.33 mW 12 k 12, 000 Problem 2FE-1 Find the power generated by the source in the network shown. 5 k 6 k 12 k 120 V 18 k 4 k 6 k Suggested Solution Applying the Y transformation yields: 5 k 5 k 3 k 3 k 120 V 2 k 6 k 120 V 4 k I 4 k 6 k I= 120 V = 10 mA 12 k P = (120 V ) I = 120 0.010 = 1.2 W Problem 2FE-2 Find the equivalent resistance of the circuit shown at the terminals A-B . A 12 k 6 k RAB 6 k 12 k 12 k 12 k 4 k B Suggested Solution Combining resistors, 12 k 6 k = 4 k 12 k (12 k + 12 k ) = 8 k The circuit is now reduced to: A 4 k RAB B 6 k 4 k 8 k and further to: A 4 k RAB B 6 k ( 8 k+ 4 k) = 4 k Therefore, RAB = 4 k + 4 k = 8 k Problem 2FE-3 Find the voltage Vo in the network shown. 1 k 2 k + 3 k 24 mA 6 k Vo _ 6 k 12 k Suggested Solution Combining resistors: 1 k + 3 k = 4 k 6 k 12 k = 4 k The network is now reduced to: 2 k I + 4 k 24 mA 6 k 4 k Vo _ Using current division, 4000 I = ( 24 mA ) = 6 mA 4000 + ( 2000 + 6000 + 4000 ) Vo = ( 6 k ) I = 36 V Problem 2FE-4 Find the current I o in the circuit shown. 3 k 4 k 12 V 12 k 6 k 6 k 3 k Io 6 k Suggested Solution Combining resistors: 3 k + 6 k = 9 k 3 k 6 k = 2 k Thus, the circuit simplifies to: I 12 V 12 V 9 k 18 k 6 k Ix 6 k I 12 V 9 k 6 k 4 k 12 k Ix 2 k I 6 k I =- 12V = -1 mA 6 k + 6 k 9000 1 Ix = I = - mA 3 9000 + 18, 000 6000 2 Io = I x = - mA 3000 + 6000 9 Problem 3.1 Find I0 in the circuit using nodal analysis. 10k 12k 6 mA 6k Io 3k Suggested Solution V1 V2 12k 6 mA 6k Io 3k V1 V -V 6 + 1 2 = 12k 10k k V2 - V1 V2 V2 + + =0 10k 6k 3k 1 1 12k + 10k V1 V = -1 2 10k -1 10k 1 1 1 + + 10k 6k 3k -1 1mA 6 k = 36 I = 6 = 1mA 6 0 6 0 Problem 3.2 Find I0 in the circuit using nodal analysis. Io 5k 10 mA 2k 2k Suggested Solution V1 3k V2 Io 5k 10 mA 2k 2k V1 V1 - V2 10 + = 5k 3k k V2 - V1 V2 V2 + + =0 3k 2k 2k 1 1 + V1 5k 3k V = -1 2 3k -1 3k 1 1 1 + + 3k 2k 2k -1 10 k = 22.2 I = 5.56 = 2.78mA 5.56 0 2k 0 2.78mA Problem 3.3 Find V2 in the circuit shown using nodal analysis. 4 k 2 k 4 mA 6 k + V2 - 8 k 6 mA Suggested Solution 8 k + 2 k V1 - 4 mA 4 k 6 k + V2 - + V3 - 6 mA V1 V -V + 1 2 = 4 mA 2 k 4 k V -V V2 - V1 V + 2 + 2 3 =0 4 k 6 k 8 k V3 - V2 = 6 mA 8 k In matrix form: 1 1 2000 + 4000 - 1 4000 0 1 4000 1 1 1 + + 4000 6000 8000 1 - 8000 - V1 0.004 1 - V = 0 2 8000 V3 0.006 1 8000 0 V2 = 22 V Alternately, since we know the current through the 8 k resistor is 6 mA, we know that V3 = V2 + 48 V . Therefore, we really need only 2 equations to solve this problem. Those are: V1 V -V + 1 2 = 4 mA 2 k 4 k Problem 3.4 Use nodal analysis to find Vo in the circuit shown. 4 k 6 k - 3 k 4 mA 2 mA Vo + 4 k Suggested Solution 6 k + V3 - 4 mA + V1 - - 3 k 2 mA 4 k Vo + + V2 - 4 k V3 - V1 = 4 mA 6 k V1 - V3 V V -V + 1 + 1 2 =0 6 k 3 k 4 k V2 - V1 V + 2 = -2 mA 4 k 4 k In matrix form: 1 - 6000 1 + 1 + 1 6000 3000 4000 1 - 4000 0 1 - 4000 1 1 + 4000 4000 1 6000 V 0.004 1 1 - V2 = 0 6000 V3 -0.002 0 V1 6.5455 V = -0.7273 2 V3 30.5455 Vo = V2 - V1 = -0.73 - 6.55 = -7.28 V Alternately, since we know the current through the 6 k resistor is 4 mA, we know that V3 = V1 + 24 V . Therefore, we really need only 2 equations to solve this problem. Those are: Problem 3.5 Find I0 in the circuit using nodal analysis 3k 9V 2k 6k 2k Suggested Solution V1 3k 9V V2 2k V3 6k 2k V1 = 9V 9 - V2 3k = V2 V2 - V3 + 6k 2k V3 = 1.2V I 0 = 0.6mA 1.2 V3 - V2 V3 V3 + + = 0 V3 = 1.2V , I 0 = = 0.6mA 2k 2k 2k 2k Problem 3.6 Use nodal analysis to find both V1 and Vo in the circuit shown. 2 mA V1 6 k 4 mA 3 k 2 k 12 k 2 k + Vo - Suggested Solution 2 mA V1 6 k 4 mA 3 k V2 2 k 12 k 2 k + Vo - V1 V -V + 1 2 = -4 mA - 2 mA 3 k 6 k V2 - V1 V2 V2 + + = 2 mA 6 k 12 k 2 k + 2 k In matrix form: 1 1 3000 + 6000 - 1 6000 1 V1 -0.006 6000 = 1 1 1 V2 0.002 + + 6000 12000 4000 - 2000 V2 = 0 V 2000 + 2000 V1 -12 V = 2 0 V1 = -12 V and Vo = Problem 3.7 Find V1 and V2 in the circuit using nodal analysis Then solve the problem using Matlab and compare your answers 4mA + V1 3k 6k + V2 12k 10k 6mA Suggested Solution V1 V1 - V2 4 6 + = - k k 6k 4k V2 - V1 V2 V2 -4 + + = k 4k 5k 10k 1 -1 1 6k + 4k V 4k 1 = 1 1 1 V2 -1 + + 4k 4k 5k 10k V1 = -12.6V ;V2 = -13V Nodal - Equations 6m + V1 / 6k + (V1 - V2 ) / 4k - 4m = 0 4m + (V2 - V1 ) / 4k + V2 / 5k + V2 /10k = 0 or -24 = 5V1 - 3V2 -80 = -5V1 + 11V2 -1 -2 k -4 k In Matlab >> g=[5 -3;-5 11] g= 5 -3 -5 11 >> i=[-24;-80] i= -24 -80 >> v=inv(g)*i v= -12.6000 -13.0000 V1 = -12.6V ;V2 = -13V V1 = -12.6V ;V2 = -13V Problem 3.8 Find I0 in the network using nodal analysis I0 12k 4k 6V 6k 12V Suggested Solution V1 12V 12k V2 4k 6V 6k I0 V3 V1 = 12V ;V3 = -6V V2 - 12 V2 V2 + 6 + + = 0 V2 = -1V 12 K 6k 4k I0 = V2 - V3 -1 + 6 = = 1.25mA 4k 4k V2 = -1V ; I 0 = 1.25mA Problem 3.9 Find V0 in the network using nodal analysis V0 + 6k 12V 6k 12k Suggested Solution V1 6k 12V 6k V2 12k 6V V0 + V3 V1 = 12V ;V3 = 6V V2 - 12 V2 V2 - 6 + + = 0 V2 = -6V 6K 6k 12k V2 = 6v V0 = V2 - V3 = 0V V2 = 6v V0 = V2 - V3 = 0V Problem 3.10 Use nodal analysis to find I o and I1 in the circuit shown. 6 k 3 k 12 mA 4 k Io 12 V 2 k I1 Suggested Solution 6 k + 12 mA V1 - 3 k 4 k Io 12 V + V2 - 2 k I1 V2 = -12 V I1 = V2 -12 = = -6 mA 2 k 2000 V1 = 8 V Io = V1 8 = = 2 mA 4 k 4000 -12 mA + V - ( -12 V ) V1 - ( -12 V ) V1 + 1 + =0 4 k 6 k 3 k I o = 2 mA , I1 = -6 mA Problem 3.11 Find I0 in the circuit using nodal analysis I0 12k 6V 12k 12k 12k Suggested Solution V1 12k 6V 12k 12k 12k I0 V2 V1 + 6 V1 V1 - V2 + + =0 12k 12k 12k V2 - V1 V2 2 + = 12k 12k K -1 3V1 V2 - = 12k 12k 12k -V1 2V2 2 + = 12k 12k 12k 3V1 - V2 = -6 -V1 + 2V2 = 24 12 V1 3 -1 -6 5 V = -1 2 24 = 66 2 5 12 66 - I 0 = 5 5 = -0.9mA 12k I 0 = -0.9mA Problem 3.12 Use nodal analysis to find V0 in the network + 4k 6V 2k 2k 2k V0 - Suggested Solution V1 4k 6V 2k 2k 2k V1 - (-6) V1 2 V1 + + + =0 4k 2k k 2 k + 2k 2k -7 -7 ( ) V1 = V ;V0 = 2 2 2k + 2k -7 V0 = V 4 -7 V 2 -7 V0 = V 4 V1 = Problem 3.13 Find I0 in the network 6V 3mA 3mA 6k 4k 3k 6k I0 Suggested Solution 3mA 3/k 3k I0 I0 = -3 3k ( ) = -1mA k 3k + 6k I 0 = -1mA Problem 3.14 Find V0 in the network 2k 6V 4k 2mA 12k 8k 12V 4k Suggested Solution + 8k 12V 4k V0 - V0 = 12( 4k ) = 4V 4k + 8k V0 = 4V Problem 3.15 Use nodal analysis to find V0. 8k 8k 4k 6V + 4k V0 - 3V Suggested Solution -6V 8k V1 8k 4k 6V + 4k V0 3V -3V V1 + 6 V1 V +3 -5 + + 1 = 0 V1 = V 6k 2k + 2 k 2k 2 -5 1k -5 ( )= V V0 = 2 1k + 2k 6 V0 = -5 V 6 Problem 3.16 Write the node equations for the circuit in matrix form and find the node voltages using Matlab 3mA 4k 3k 6mA 2k Suggested Solution 3 V1 V1 - V2 + = 2k 2k + 2k k V2 - V1 V2 - V3 6 + = 1k 3k k V3 - V2 V3 -3 + = 3k 4k k 3 1 3 V1 ( ) - V2 ( )0 = 2k k k 1 4 1 6 -V1 ( ) + V2 ( ) - V3 ( ) = 3k 3k k k 1 7 -3 0 - V2 ( ) + V3 ( )= 3k 12k k 3 V1 2k -1 V2 = k V3 0 -1 3 0 k k 4 -1 6 3k 3k k -1 7 - 3 3k 12k k Nodal - Equations V1 / 2k (V1 - V2 ) / k - 3m = 0 (V3 - V2 ) / 3k + V3 / 4k + 3m = 0 V1 / 2k + V3 / 4k - 6m = 0 or 6 = 3V1 - 2V2 -36 = -4V1 + 7V2 24 = 2V1 + V3 Using Matlab EDU g=[3 -2 0;0 -4 7;2 0 1] g= 3 0 2 -2 -4 0 0 7 1 EDU i=[6;-36;24] i= 6 -36 24 EDU v=inv(g)*i v= 10.8000 13.2000 2.4000 Problem 3.17 Find I1 in the network shown. 3V 2mA 3k I1 6k Suggested Solution V1 3V V1-3 2mA 3k I1 6k V1 V1 - 3 2 + = V1 = 5V ; I1 = 1.67mA k 3k 6k I1 = 1.67 mA Problem 3.18 Find I0 in the network shown. 4k 2mA 12k 6V 6k 3V 2k Suggested Solution I0 6k 6V 3V I0 = -6 - 3 = -1.5mA 6k I 0 = -1.5mA Problem 3.19 Find I0 in the circuit shown. 4k 6V 4k 2k I0 12V 6k Suggested Solution V1 4k 6V V2 4k 2k V3 I0 12V 6k V1 = 6V , V3 = 12V V2 - V1 V2 - V3 V2 + + =0 3k 2k 10k 150 204 V2 = V , V3 = 12 = V 17 17 150 204 -27 I0 = ( ) / 2k = mA = -1.59mA - 17 17 17 I 0 = -1.59mA Problem 3.20 Find I0 in the network using nodal analysis 4k 2mA 12V 3k 6k I0 2k Suggested Solution 4k V2 3k 6k I0 2mA V3 12V 2k V2 - 12 V2 2 + = V2 = 12VandI 0 = 2mA 3k 6k k I 0 = 2mA Problem 3.21 Use nodal analysis to find V0 in the network shown. Then solve this problem using Matlab and compare your results. 6k 12V 6k 2mA 6k 6k Suggested Solution 12V V2-12 6k V2 6k V3 2mA 6k 6k V2 - 12 V2 - 12 - V3 V2 V2 - V3 + + + =0 1k 1k 1k 1k V3 - V2 V3 - (V2 - 12) 2 + = 1k 1k k 4 2 24 -2V2 2V3 -10 V2 - V3 = ; + = k k k k k k 4V2 - 2V3 = 24; -2V2 + 2V3 = -10 2V2 = 14 V2 = 7VandV3 = 2V Nodal - Equations V2 - V1 = 12 V1 /1k + V2 /1K = 2m (V0 - V2 ) /1k + (V0 - V1 ) /1k = 2m or 12 = -V1 + V2 2 = V1 + V2 2 = -V1 - V2 + 2V0 EDU g=[-1 1 0;1 1 0;-1 -1 2] g= -1 1 -1 1 1 -1 0 0 2 EDU i=[12;2;2] i= 12 2 2 EDU v=inv(g)*i v= -5 7 2 V0 = 2V Problem 3.22 Find V0 in the circuit shown using Nodal Analysis 12V 2k 6V 1k 1k + V0 1k - Suggested Solution 12V -6V V0+12 2k 6V 1k 1k V0 + V0 1k - V0 + 12 V0 + 12 + 6 V0 + 6 V0 + + + =0 1k 2k 1k 1k 7 -27 V0 ( ) = 2k 1k V0 = -7.17V V0 = -7.17V Problem 3.23 Use nodal analysis to find I0 in the network shown. 12V 3k 2k I0 2k 3k 4mA 1k Suggested Solution V1-12 12V V1 V2 3k 2k I0 2k 3k 4mA 1k V1 - 12 V1 V1 - V2 + + =0 3k 2k 3k V2 - V1 4 V2 + + =0 3k k 3k 7V1 V2 4 - = 6k 3k k -1V1 2V2 -4 + = 3k 3k k 7 V1 6k V = -1 2 3k -1 3k 2 3k -1 I 0 = 1mA 4 k 2 = 1mA V1 = 2VandI 0 = -4 2k k Problem 3.24 Find I o in the network shown. 1 k 1 k 12 V 1 k Io 1 k 6 mA Suggested Solution 1 k 1 k + V1 - 12 V + 1 k Io V2 - + 1 k V3 - 6 mA V2 = V3 + 12 V V1 V -V V -V + 1 2 + 1 3 =0 1 k 1 k 1 k V -V V2 - V1 V + 2 + 3 1 - 6 mA = 0 1 k 1 k 1 k V2 - V3 = 12 3V1 - V2 - V3 = 0 -2V1 + 2V2 + V3 = 6 In matrix form: 0 1 -1 V1 12 3 -1 -1 V = 0 2 -2 2 1 V3 6 Io = V1 =0A 1 k V1 = 0 V Problem 3.25 Use nodal analysis to find V0 in the circuit shown. 9mA 12K + VO 6K 6V 4K 6mA Suggested Solution 9mA 12K VO 6V 6/K Note the position of the reference node KCL 9 6 V0 + + =0 K K 6K V0 = -90V V0 = -90V Problem 3.26 Find the VO in the circuit shown using nodal analysis. 12k 12k 18k 12V 4k 9V 6V Suggested Solution 12k VO 12k 18k 12V 4k 9V 6V V0 - 12 V0 + 9 V0 + 6 + + =0 12 K 18K 16 K 1 1 1 V0 ( + + ) = 0.621V 12 18 16 V0 = 0.621V Problem 3.27 Find Vo in the network shown using nodal analysis. + 2 k 4 k 4 k Vo - 12 V 2 k Suggested Solution + 2 k + V1 4 k 12 V + V2 4 k - - Vo - 2 k V1 - V2 = 12 V Vo - V1 V + o =0 2 k 4 k V1 - Vo V V + 1 + 2 =0 2 k 4 k 2 k 3Vo - 2V1 = 0 -2Vo + 3V1 + 2V2 = 0 In matrix form: 0 1 -1 Vo 12 3 -2 0 V = 0 1 -2 3 2 V2 0 Vo = 4.36 V Problem 3.28 Find Vo in the network shown using nodal analysis. + 2 k 2 mA 4 k 2 k Vo - 12 V 2 k Suggested Solution + 2 k 2 mA + V1 4 k 12 V + V 2 k - - Vo - 2 2 k V1 - V2 = 12 V Vo - V1 V + o - 2 mA = 0 2 k 2 k V1 - Vo V V + 1 + 2 + 2 mA = 0 2 k 4 k 2 k 2Vo - V1 = 4 -2Vo + 3V1 + 2V2 = -8 In matrix form: 0 1 -1 Vo 12 2 -1 0 V = 4 1 -2 3 2 V2 -8 Vo = 4.5 V Problem 3.29 Suggested Solution 1 1 1 6 2 V1 - 6 2 V1 - V2 ) - V2 ( )= + + = 0 V1 ( + - 4K 4K 4K 4K 4K 4K K K 1 1 1 4 V2 - V1 4 V2 ) + V2 ( )= - + = 0 -V1 ( + 4K 4K 8K 4 K K 2K + 2K K V2 = 10V V0 = 10( V0 = 5V V2 = 10V 2K ) = 5V 2K + 2K V0 = 5V Problem 3.30 Find Io in the circuit shown using nodal analysis. 4K 6K 8K 4K 2mA 6K Suggested Solution 4K V1 6K 8K 4K 2mA 6K V1 V 2 V -6 + 1 - + 1 = 0 V1 = 6V 4 K + 8K 4 K K 6K 6 IL = = 1.5mA 4K V1 = 6V I L = 1.5mA Problem 3.31 Find Vo in the circuit shown using nodal analysis. 1K + V0/2 2K 1K VO - Suggested Solution 1K + V0/2 2K 1K VO - V0 2 + V0 - 12 + V0 = 0 1K 2K 1K V0 = 9V V0 - 12 - V0 = 9V Problem 3.32 Use nodal analysis to find Vo in the network shown. 12V 1K VO/2K 1K 1K + VO - Suggested Solution V1-12 + 1K 1K VO - 12V 1K VO/2K V1 - 12 V0 V1 V1 - + + =0 1K 2 K 1K 1K + 1K 1K V , V1 = 1 V0 = 1K + 1K 2 V 5 12 )- 0 = V1 ( 2K 2K K 5 12 V 2V0 ( - 0 )= 2K 2K K V0 = 2.667V V0 = 2.667V Problem 3.33 Find the Vo in the circuit shown using nodal analysis. Then solve the problem using matlab and compare your results. 1K 12V 1K 2IX 1K IO 1K + VO - Suggested Solution V1-12 12V 1K 2IX 1K IO 1K V1 1K + VO - V1 - 12 V1 V1 + + =0 1K 1K 2 K Where 2I X + IX = V1 1K 2V1 V1 12 V1 V1 + - + + = 0 V1 = 24 / 9V 1K 1K K 1K 2 K 4 V0 = 1/ 2V1 = V 3 4 V0 = 1/ 2V1 = V 3 Nodal _ Equations 2 I X + V1 /1K + V2 /1K + V0 /1K = 0 (V2 - V0 ) /1K = V0 /1K V2 - V1 = 12 I X = V2 /1K OR V1 + 3V2 + V 0 = 0 V2 - 2V0 = 0 -V1 + V2 = 12 EDU G=[1 3 1;0 1 -2;-1 1 0] G= 1 0 -1 3 1 1 -2 1 0 EDU I=[0;0;12] I= 0 0 12 EDU V=inv(G)*I V= -9.3333 2.6667 1.3333 Problem 3.34 Find Io in the network shown. 1K 4000IX 1K 2K IX 4mA 4K IO Suggested Solution 1K v1+ 4000IX 1K 2K IX 4mA 4K IO V1 + 4 KI X V1 + 4 KI X V1 4 + + = 2K 2K 4K K V1 + 4 KI X -V = IX IX = 1 2K 2K 5 4V 4 V1 - 1 = 4K 2K K V1 = -16 V 3 -4 I0 = mA 3 V1 = -16 V 3 -4 I0 = mA 3 Problem 3.35 Find Vo in the circuit shown using nodal analysis. 2kIx 1 k 1 k + 1 k Ix 12 V 1 k Vo - Suggested Solution 2kIx 1 k + V2 - 1 k Ix + V1 - 1 k + 12 V 1 k Vo - Vo - V2 = 2kI x V1 = 12 V Vo - V2 - 2kI x = 0 V -V V V2 - V1 V + 2 + o 1 + o =0 1 k 1 k 1 k 1 k Ix = V2 1 k Vo - V1 + V2 = 0 V2 - 1000 I x = 0 In matrix form: Problem 3.36 Find Io in the circuit shown using nodal analysis. 10k 10k 1k IX 2mA 2IX 10k Io Suggested Solution 10k 10k 1k IX 2mA 2IX 10k Io V1 2 V -V - + 1 2 =0 10 K K 10 K V2 - V1 V + 2I X + 2 = 0 10 K 10 K V2 = -4V I 0 = -0.4mA V2 = -4V I 0 = -0.4mA Problem 3.37 Find Vo in the network shown using nodal analysis. Ix 10 k 4000 Ix 10 k 4 mA 10 k + Vo - Suggested Solution Ix + 4000 Ix V1 - 10 k 10 k 4 mA 10 k + Vo - V1 = -4000 I x V1 + 4000 I x = 0 Vo - V1 Vo - 4 mA + =0 10 k 10 k Ix = Vo - V1 10 k 2Vo - V1 = 40 Vo - V1 - 10000 I x = 0 In matrix form: 4000 Vo 0 0 1 2 -1 0 V1 = 40 1 -1 -10000 I x 0 Vo = 15 V Problem 3.38 Find Vo in the network shown. In addition, determine all branch currents and check KCL at every node. 10K IX 12V 10K + 10K 6V IX 10K VO - Suggested Solution 10K IX 12V V1 V2 IX 10K V3 + 10K 6V IX I1 10K I3 VO - IX V1 - 6 V1 V + + IX + 2 = 0 10 K 10 K 20 K 6 - V1 IX = 10 K V0 = V2 / 2 V1 = -4V , V2 = 8V , V0 = 4V V0 = 0.4mA 10 K 6 - V1 IX = = 1mA 10 K I 2 = I 3 + I X = 1.4mA I3 = I1 = V1 = -0.4mA 10 K V0 = 0.4mA 10 K 6 - V1 IX = = 1mA 10 K I 2 = I 3 + I X = 1.4mA I3 = I1 = V1 = -0.4mA 10 K Problem 3.39 Find Vo in the circuit shown. 1000IX 5K IX + 6K 12mA VO 6K 2K Suggested Solution 1000IX 5K 2K VO-1000IX IX 6K V1 + VO 12mA 6K V0 - KI X V0 - KI X - V1 V0 - V1 V0 + + + =0 6K 2K 5K 6K V - KI X - V1 IX = 0 2K V0 = 32.25 V0 = 32.25 Problem 3.40 Find Vo in the circuit shown. 1K + VX 1K 2mA 2K + 1K VO - 2VX/1000 Suggested Solution VX + VX 1K 2mA 1K V1 2K VO 2VX/1000 1K + VO - 2 V -V V1 - + 1 0 =0 2K K 2K VX = 1/ 2V1 V0 - V1 2VX V0 - + =0 2K 1K 1K V0 = 4V V0 = 4V Problem 3.41 Use matlab to find the node voltages in the network shown. 1K 2mA 2K V2 1K 12V 4mA 2K V4 1K Suggested Solution Applying _ kcl V1 - V2 2 V1 - V4 - + =0 K 1K 2K V2 = 12 V3 - V2 2 4 V3 + + + =0 K K 2K 1K V4 - V1 4 V4 - + =0 K 1K 2K MATRIX _ FORM -0.0005 V1 0.002 0 0.0015 -0.001 0 1 0 0 V2 12 = 0 -0.001 0.0015 0 V3 -0.006 0 0 0.0015 VX 0.004 -0.0005 EDU g=[0.0015 -0.001 0 -0.0005;0 1 0 0;0 -0.001 0.0015 0;-0.0005 0 0 0.0015] g= 0.0015 -0.0010 0 -0.0005 0 1.0000 0 0 0 -0.0010 0.0015 0 -0.0005 0 0 0.0015 EDU i=[0.002;12;-0.006;0.004] i= 0.0020 12.0000 -0.0060 0.0040 EDU v=inv(g)*i v= 11.5000 12.0000 4.0000 6.5000 V= 11.5000 12.0000 4.0000 6.5000 Problem 3.42 Use mesh equations to find Vo in the circuit shown. 2K 2K + 12V 3K 4K VO - Suggested Solution 2K 2K + 12V I1 3K I2 4K VO - 5K -3K -3K I1 12 = 4 K I 2 0 -3K 12 4 K 0 -1 I1 5K I = -3K 2 1 A K V0 = 4V I2 = V0 = 4V Problem 3.43 Find Vo in the network shown using mesh equation. 12V 4V + 2K 2K 2K VO - Suggested Solution 12V 4V + 2K I1 2K I2 2K VO - 4K -2 K -2 K I1 12 = 4 K I 2 -4 -2 K I 2 4 K 0 -1 I1 4 K I = -2 K 2 2 A 3K 4 V0 = V 3 I2 = 4 V0 = V 3 Problem 3.44 Use mesh analysis to find Vo in the circuit shown. 4K 4K + 6K 2K 6K VO 12V - Suggested Solution 4K 4K c 2K 6K 6K + VO I1 12V I2 - 12 K -6 K -6 K I1 -12 = 12 K I 2 12 -6 K -12 12 K 12 -1 I1 12 K I = -6 K 2 2 mA 3 4 V0 = V 3 I2 = 4 V0 = V 3 Problem 3.45 Use mesh analysis to find Vo in the network shown. 3K 2K + 6K 12V VO 2mA 12V - Suggested Solution 3K 2K c 6K 12V I1 12V I2 + VO 2mA - -12 + 3KI1 + K ( I1 - I 2 ) = 0 2 K 8 I1 = 3K I2 = -12 + 3K ( 8 2 ) + 2 K ( ) = V0 = 0V 3K K V0 = 0V Problem 3.46 Use loop analysis to find Vo in the circuit shown. 2K 4K + 2K 12V VO 2mA - Suggested Solution 2K 4K + 2K I1 12V 2mA I2 VO - 12 = 2 KI1 + 6 KI 2 I 2 - I1 = I2 = 2 K 2 K V0 = 4V 2 V0 = (2 K )( ) = 4V K V0 = 4V Problem 3.47 Use loop analysis to find Io in the circuit shown. 2K 2mA 4mA 12V IO 2K 2K Suggested Solution 2K 2mA I1 4mA 12V I2 2K 2K I3 12 = 2 KI 2 + 2 KI 3 2 K 4 I 2 - I3 = K I1 - I 2 = I1 = I 0 = 7 mA I1 = I 0 = 7 mA Problem 3.48 Use both nodal analysis and mesh analysis to find Io in the circuit shown. C 2K 12V 2K 1K 2mA Suggested Solution 2 4 V1 - V2 + =0 K K 2K 4 V2 - V1 V2 V3 + + - =0 2K 1K 2 K K Where V3 - V2 = 12 4mA V2 = -16 V 3 I 0 = -5.33mA I1 = 2K I1 12V -2 4 ,I2 = K K 1K ( I1 - I 3 ) - 12 + 2 KI 3 = 0 2K 1K 2mA I2 I3 10 3K I 0 = I1 - I 3 = -5.33mA I3 = I 0 = I1 - I 3 = -5.33mA Problem 3.49 Find Io in the network shown using mesh analysis. 4K IO 2K 2mA 6K 4mA 12V Suggested Solution 4K IO 2K 2mA 6K I1 12V I2 I3 -2 K 4 I3 = K -12 + 2 K ( I 2 - I1 ) + 4 KI 2 + 6 K ( I 2 - I 3 ) = 0 I1 = 16 mA 6 I 0 = I 2 = 2.67mA I2 = I 0 = I 2 = 2.67 mA Problem 3.50 Find Io in the network shown using mesh analysis. 2K 1K 4mA 2K IO 2mA 12V Suggested Solution 2mA 2K I2 12V 2K I1 IO I3 1K 4mA -12 + 2 K ( I1 - I 3 ) + 2 KI 2 + 1K ( I 2 - I 3 ) = 0 I1 - I 2 = I3 = 2 K I 0 = I1 - I 3 = 5.2mA -4 K 6 I1 = A 5K I 0 = I1 - I 3 = 5.2mA Problem 3.51 Find Vo in the circuit shown using mesh analysis. 1K 12V 1K 2mA + VO - 1K 1K Suggested Solution 12V I2 1K 2mA + VO I3 - 1K I1 1K 2 KI1 - 1KI 3 = -12 2 KI 2 - 1KI 3 = 12 I3 = -2 K 2 KI1 + 2 = -12 2 KI 2 + 2 = 12 I1 = -7 5 , I2 = K K V0 = 2V 7 5 (1K ) - (1K ) = 2V K K V0 = 2V VO = Problem 3.52 Use loop analysis to find Vo in the network shown. 1K 1K 12V 1K 6mA 1K VO - Suggested Solution 1K 1K 12V + 1K I1 6mA I2 1K VO - -6 K 1K ( I1 + I 2 ) + 1K ( I1 + I 2 ) + 1KI 2 + 12 + 1KI 2 = 0 I1 = I2 = 0 V0 = 0 V0 = 0 Problem 3.53 Find Io in the network shown using loop analysis. Then solve the problem using matlab and compare your answers. 6K 6K 6K 6K 6V 5mA IO Suggested Solution I3 6K 6K 6K 6V I1 5mA I2 6K IO 6 + 6 K ( I1 - I 3 ) + +6 K ( I 2 - I 3 ) + 6 KI 2 = 0 6 KI 3 + 6 K ( I 3 - I 2 ) + 6 K ( I 3 - I1 ) = 0 I 2 - I1 = 6K -6 K -1K 5 K 12 K -6 K 1K -12 K I1 -6 18K I 2 = 0 0 I 3 5 Using Matlab EDU Z=[6000 12000 -12000;-6000 -6000 18000;-1000 1000 0] Z= 6000 -6000 -1000 12000 -6000 1000 -12000 18000 0 EDU V=[-6;0;5] V= -6 0 5 EDU I=inv(Z)*V I= -0.0046 0.0004 -0.0014 Problem 3.54 Use loop analysis to find Io in the circuit shown. 4K 3K 2mA 12V IO 6K 2K Suggested Solution I2 3K 4K 2mA 12V I1 IO 6K I3 2K -12 + 3K ( I1 - I 2 ) + 6 KI 2 = 0 2 K 9 KI1 = 18 I2 = I1 = I 0 = 2mA I1 = I 0 = 2mA Problem 3.55 Find Vo in the network shown using both mesh and nodal analysis. 6V 4K 2K 6K VO 2K 2mA - Suggested Solution 6V I1 2K 4K I2 6K VO + 2K 2mA I3 - 2 K 6 = 6 KI1 - 4 KI 2 - 2 KI 3 I3 = 0 = -4 KI1 + 12 KI 2 - 2 KI 3 I 2 = 1.14mA V0 = 6 KI 2 = 6.86V V0 = 6.86V V1 - V2 V3 - V2 V2 + = 2V1 + V3 - 5V2 = 0 2K 4K 2K V 2 V2 + 3 = 3V2 + V3 = 12 2K 6K K V3 - V1 = 6 V3 = V0 = 6.86V Problem 3.56 Use loop analysis to find Vo in the circuit shown. 1K 6V 1K + 2K 1K 1K VO 2mA 12V - Suggested Solution 1K 6V 1K I2 + 2K 1K 1K VO I3 - I1 2mA 12V 2 K 1KI 2 - 6 + 2 K ( I 2 - I 3 ) + 1K ( I 2 - I1 ) = 0 I1 = -12 + 1K ( I 3 - I1 ) + 2 K ( I 3 - I 2 ) + 1K ( I 3 ) = 0 4 K -2 K I 2 8 -2 K 4 K I 3 = 14 6 KI 3 = 36 I 3 = 6mA V0 = 6V I 3 = 6mA V0 = 6V Problem 3.57 Use loop analysis to find Io in the network shown. 1K 1K 12V 1K 1K 2mA 1K 4mA 1K Suggested Solution 1K I1 12V 1K 1K 1K I2 2mA 1K I3 4mA I4 1K -12 + 1K ( I1 - I 2 ) + 1K ( I1 - I 3 - I 4 ) + 1K ( I1 ) = 0 2 K -4 I3 = K 1K ( I 3 + I 4 ) + 1K ( I 3 + I 4 - I1 ) + 1K ( I 4 - I 2 ) + 1KI 4 = 0 I2 = 3KI1 - 1KI 4 = 10 -1KI1 + 4 KI 4 = 10 40 A 11K I 0 = I 4 - I 2 = 1.64mA I4 = I 0 = I 4 - I 2 = 1.64mA Problem 3.58 Use loop analysis to find Vo in the network shown. 3mA 2K 4K 12K 4K + VO - 2K 6V 1mA Suggested Solution 3mA I1 4K 2K I2 4K + VO - 12K 6V I4 1mA I3 2K I1 = 3 K 1 K -6 + 4 K ( I 4 - I1 ) + 12 K ( I 3 - I 2 ) + 2 KI 3 I3 - I 4 = 0 = -2 KI1 - 12 KI 3 + 18 KI 2 31 A 15K V0 = 4 I 2 = 8.27V I2 = V0 = 4 I 2 = 8.27V Problem 3.59 Find Vo in the network shown. 4mA 2K 1K 1K + 1K 1mA 2K VO 2mA Suggested Solution 4mA I1 1K 2K 1K + 1K I3 1mA I4 2K VO I2 2mA 4 -2 -1 , I2 = , I3 = K K K 1K ( I 3 + I 4 ) + 1K ( I 3 + I 4 - I1 ) + 1K ( I 4 - I 2) + 2 KI 4 = 0 I1 = 5KI 4 = 4 4 5K 8 V0 = V 5 I4 = 8 V0 = V 5 Problem 3.60 Find Vo in the circuit shown. 2mA 1K 2mA 2K 2K 1K 4mA 1K + VO - Suggested Solution 2mA 1K I1 I2 2mA 2K 2K I3 1K I4 1K + VO - 4mA 0 = 2 KI 2 + 1KI 4 + 1K ( I 4 - I 3 ) + 2 K ( I1 - I 3 ) + 1KI1 = 0 4 2 2 , I 4 - I 2 = , I 2 - I1 = K K K 4 I4 = K V0 = 1KI 4 = 4V I3 = V0 = 1KI 4 = 4V Problem 3.61 Find Io in the circuit shown. 2K 6V 4K 2K 2mA 2K 6K 1mA Suggested Solution 6V I2 2K I3 4K 2K 2mA 6K I1 1mA I4 6 = 4 KI 3 + 2 KI 4 + 6 KI1 0 = -2 KI 2 - 2 KI 4 + 8 KI 3 2 1 , I 4 - I1 = K K I 2 = -1.5mA I1 - I 2 = I3 = 0 I 0 = I 2 - I 3 = -1.5mA I 0 = I 2 - I 3 = -1.5mA Problem 3.62 Use loop analysis to find Vo in the network shown. 2K Ix + 1K 2Ix 1K VO 6V Suggested Solution 2K 6V + 1K I1 2Ix Ix 1K VO - I1 = -2 I X 1K ( I1 + I X ) + 2 KI X - 6 + 1KI X = 0 3 A K V0 = 3V IX = V0 = 3V Problem 3.63 Use mesh analysis to find Vo in the circuit shown. 2K 4VA + 4K 12V 6K VO Suggested Solution 2K 4VA + 4K 12V I1 6K I2 VO 12 = 6 KI1 - 4 KI 2 0 = -4 KI1 + 10 KI 2 + 4VA VA = 2 KI1 I 2 = -0.63mA VO = 6 KI 2 = -3.79V VO = 6 KI 2 = -3.79V Problem 3.64 Use loop analysis to find Vo in the circuit shown. 4K 4K 2K + VO 6V 2VO/1000 - Suggested Solution 4K 4K 2K I2 2VO/1000 + VO 6V I1 2V0 ,V0 = 2 KI 2 K 6 = 4 KI1 + 6 KI 2 I 2 - I1 = I 2 = -1mA V0 = 2 KI 2 = -2V V0 = 2 KI 2 = -2V Problem 3.65 Find Vo in the circuit shown using mesh analysis. 2K + 6V 4K 2000Ix Ix 2K VO Suggested Solution 2K + 6V 4K 2000Ix I1 I2 2K VO I2 = I X 2 KI X = 6 KI1 - 4 KI 2 6 = -4 KI1 + 6 KI 2 I 2 = 3mA V0 = 2 KI 2 = 6V V0 = 2 KI 2 = 6V Problem 3.66 Use both nodal analysis and mesh analysis to find Vo in the circuit shown. 10VX 12mA 6K + 12K 4K VX + 8K VO - Suggested Solution 10VX 12mA 6K I2 + + I3 8K VO - 12K I1 4K - VX 22 KI1 - 6 KI 2 - 4 KI 3 = 0 12 KI1 + 10VX + 8 KI 3 = 0 VX = 4 K ( I1 - I 3 ) 12 K I 3 = -20.35mA I3 - I 2 = V0 = 8 KI 3 = -162.78V Nodal _ Analysis V2 - V1 V2 12 + + =0 6K 4K K V1 V V + 2 + 3 =0 12 K 4 K 8 K V1 - V3 = 10VX VX = V2 V3 = V0 = -162.78V V3 = V0 = -162.78V Problem 3.67 Using mesh analysis finds Vo in the circuit shown . VX/2000 2K 4K 6K VO 2mA 6K + Suggested Solution + I1 4K 2K 6K VO 2mA I3 6K I2 VX 2 , R3 = , VX = ( I 3 - I 2 )6 K 2K K ( I 2 - I 3 )6 K + ( I 2 - I1 )2 K + 6 KI 2 = 0 I1 = 6 I2 = A 5K V0 = 6 KI 2 = 36 V 5 V0 = 6 KI 2 = 36 V 5 Problem 3.68 Find Vo in the network shown. + VX 1K 2K 1mA 1K + VO 1K 1K 2K 2VX/1000 Suggested Solution + VX 1K 2K I1 1K 1mA I4 1K 2K 2VX/1000 I2 I3 1K + VO - 1 2 -4 , I 2 = , VX = -1KI 4 , I 3 = K K 2K 2 K ( I1 + I 4 ) + 1KI 4 + 1KI 4 + 1K ( I 4 - I 3 ) + 1K ( I 4 + I1 - I 2 ) = 0 I1 = I 2 = -2 I 4 -7 8K -7 V0 = V 8 I4 = V0 = -7 V 8 Problem 3.69 Use matlab to find the mesh currents in the network shown. 2K 2mA 1K I1 I2 1K 2K I3 2K 1K I5 6V 1K I4 Suggested Solution 2 K - I1 (1K ) + I 2 (5 K ) - I 4 (2 K ) = -12 I1 = - I1 (2 K ) - I 4 (1K ) + I 3 (3K ) = 6 - I 3 (1K ) - I 2 (2 K ) + I 4 (4 K ) - I 5 (1K ) = 0 - I 4 (1K ) + I 5 (2 K ) = 12 0 0 0 0 I1 0.002 1 -1000 5000 0 0 I 2 -12 -2000 -2000 0 3000 -1000 0 I 3 = 6 -2000 -1000 4000 -1000 I 4 0 0 0 0 0 -1000 2000 I 5 12 EDU r=[1 0 0 0 0;-1000 5000 0 -2000 0;-2000 0 3000 -1000 0;0 -2000 -1000 4000 -1000;0 0 0 -1000 2000] Using Matlab r= 1 0 -1000 5000 -2000 0 0 -2000 0 0 0 0 0 -2000 0 3000 -1000 0 -1000 4000 -1000 0 -1000 2000 0 EDU v=[0.002;-12;6;0;12] v= 0.0020 -12.0000 6.0000 0 12.0000 EDU i=inv(r)*v i= 0.0020 -0.0011 0.0041 0.0023 0.0071 Problem 3.70 Find Vo in the circuit shown. 5K 1K + 1V VO - Suggested Solution KCL at the intersecting input 1-0 0 -V 0 = 0.1K 5K V 0 == -5V V 0 == -5V Problem 3.71 Find Vo in the network shown and explain what effect R1 has on the output. 10 2 + 2V 10 VO - Suggested Solution No current in R1. It has no effect. KCL at the inverting input. 2-0 0 - V0 = 2 10 V0 = -10V V0 = -10V Problem 3.72 Find Vo in the network shown. 4 R1 1 + 5V VO 4V - Suggested Solution KCL at the inverting input is 5 - 4 4 - V0 = 1 4 V0 = 0V Problem 3.73 The network shown is a current-to-voltage converter or transresistance amplifier. Find Vo/Is for this network. 1 is VO - Suggested Solution KCL at the inverting input is 5 - 4 0 - V0 = 1 1 V0 = -1 is 0 V0 = -1 is 0 Problem 3.74 Find Vo in the circuit shown. 40 5K 5K 5V 4V 20K VO - Suggested Solution 40 5K 5K 5V 20K VO - Applying Voltage Division 20 VX =4( )=3.2V 20+5 KCL at the inverting input 5 - VX VX - V0 = 5K 40 K V0 = -11.2V V0 = -11.2V Problem 3.75 Find Vo in the circuit shown. 80K 40 10K 5V + VO - 20 40 Suggested Solution 80K 40 10K 5V + VO - 20 40 KCL at the inverting input 6 - V0 9 - 6 10 - 6 -12 - 6 6 + + = + 10 K 20 K 30 K 40 K 100 K V0 = 31V VX = 0, VY = V0 Node - Z 40 K + 20 3 VZ = ( )V0 = V0 40 K 2 Node - X 5 3V0 V + + 0 =0 10 K 160 K 40 K V0 = -11.4V V0 = -11.4V Problem 3FE-1 Find Vo in the circuit shown. 2 VX 6 1 12V 6V + 2 VO - Suggested Solution KCLat node Vx VX - 12 VX VX + 6 + + =0 2 3 6 VX = 5V V0 = VX ( 2 10 )= V 2 +1 3 V0 = VX ( 2 10 )= V 2 +1 3 Problem 3FE-2 Determine the power dissipated in the 6-ohm resistor in the network shown. 4 VX I1 6 12V 2I1 12 Suggested Solution KCL at node Vx VX - 12 VX 12 - VX + - 2 4 6 4 VX = 9V V 2 92 P6 = = = 13.5W R 6 VX =0 + 12 P6 = 13.5W Problem 3FE-3 Find the current Ix in the 4-ohm resistor in the circuit shown. 12V + 6 2A VX 4 Ix 2VX 3 Suggested Solution KCL at VX - 12 V V - 2VX -2+ X + X =0 6 4 3 VX = 48V I X = 12 A I X = 12 A Problem 3FE-4 Determine the voltage Vo in the circuit shown. 12V VX 4 + 4 4 2 Ix 2IX VO - Suggested Solution KCL at Non Reference nodes VX - 12 VX VX - V0 + + =0 4 2 4 V0 - VX V + 2I X + 0 = 0 4 4 V IX = X 2 1 VX - V0 = 3 4 3 1 VX + V0 = 0 4 2 1 VX V = 3 0 4 V0 = -3.27V -1 4 3 1 0 2 -1 V0 = -3.27V Problem 3FE-5 Given the summing amplifier shown, select the values of R2 that will produce an output voltage of 3v. R2 12K 24K 2V VO - 1V Suggested Solution V0 = - R2 R (2) - 2 (-1) 12 K 24 K -2 R2 R -3 = + 2 =0 12 K 24 K R2 = 24 K R2 = 24 K Problem 3FE-6 Determine the output voltage Vo of the summing op-amp circuit shown. 36K 12K 12K 24K 24K 2V 1V 3V 24K + VO Suggested Solution At the output of 1st op-amp -36(2) 36(-1) -9 - = V 12 24 2 24 9 24 V0 = -( )(- ) - ( )(-3) = 6V 12 2 24 V0 = 6V V1 = V0 = 6V Problem 4.1 Find Io in the circuit shown using linearity and the assumption that Io = 1mA. I2 12K 4K I3 4K 4mA 4K 4K I1 2K + - V1 - Suggested Solution + Io = 1 A k I2 12K 4K V2 I3 IT 4K 4K 4K I1 2K If Io = 1 k A 1 Then V 2 = k (4 K + 2 K ) = 6V , I 1 = Then I 2 = I 1 + Io = 5 mA and V 2 = V 1 + 4 KI 2 = 16V 2 Then I 3 = So 7/2mA 1mA V2 4K + 12K 4mA x = 1mA IT = I 2 + I 3 = 7 mA 2 + - V1 - + 4/K A Io = 1 A k 6 4k = 3 mA 2 = 8 x = Io = 7 mA Problem 4.2 Find Io in the network shown using linearity and the assumption that Io=1mA. 2K 64V 6K 3K 2K 6K Io Suggested Solution I2 I4 64V Vs 2K I3 4K Io 6 2K 3K V1 6K + 2K I1 V2 If Io = 1mA, V 1 = 6V . I 1 = V 2 = V 1 + 3KI 2 = 18V , I 3 = Vs = V 2 + 2 KI 4 = 32V 32 1mA + = 3mA. I 2 = Io + I 1 = 4mA Then I4 = I 2 + I 3 = 7mA x = Io = 2mA V2 6K = 3mA 64 x = Problem 4.3 Find Vo in the network shown using linearity and the assumption that Vo=1mV 2K 6V 2K I2 2K 2K 2K + Vo - Suggested Solution + R1 V's R2 I2 V1 + V2 V3 R2 I3 + V4 I4 + R4 R5 V'o=1mV I5 - All R = 2 K , o I 5 = VR'5 = 1 A 2 Vs = 6V I 4 = V 'o = 1 A RT 2 V 2 = V 3 + V ' o = 3mV V 1 = I 1R1 = 5mV I 3 = I 4 + I 5 = 1 A I 2 = V 2 = 1.5 A R2 V ' s = V 1 + V 2 = 8mV Vo = 0.75V V 3 = I 3 R 3 = 2mV I 1 = I 2 + I 3 = 2.5 A Vo Vs Vs = V 's Vo = ( 86 )(1m) = 0.75V m Problem 4.4 Find Vo in the circuit shown using linearity and the assumption that Vo=1V 4K 6K 8K 12V 2K Vo 6K + Figure P4.4 Suggested Solution Vx I1 4K 6K + 8K Vs Is 12V 2K Vo 6K I2 Assume Vo = 1V I2 = Vo 2K = 1 mA 2 Vx = I 2(8 K ) = 4V Vx I 1 = 12 K = 0.333mA Is = I 1 + I 2 = 0.8333mA Vs = Is (6 K ) + Vx = 9V 9V 1V = 12V Vo Vo = 1.33V Problem 4.5 In the network shown find Io using superposition 6K 12V 6K 6mA Io 6K 6K Suggested Solution 6K 6K 6K 6mA I'o Zero the indep. voltage source K I ' o = -0.006( 9 K9+ 6 K ) = - 18 mA 5 6K 9K 6mA 6K I'o I 6K 12V 6K 6K 6K I''o Zero the indep. current souce I= 12 6 K + 6 K (6 K + 6 K ) = 6 mA 5 2 I '' o = I ( 6 K6+K K ) = 5 mA 12 2 Io = I ' o + I '' o = ( - 18 + 5 )mA = - 16 mA 5 5 Problem 4.6 Find Io in the circuit shown using superposition 6K 30V 2K 12K Io 2K 30mA 2K Suggested Solution Zero the indep. current source I 30V 6K 2K 12K I'o 2K I= 30 6 K +12 K ||6 K = 3mA, 6 I ' o = I 18K = 1mA K Zero the indep. voltage source 6K 2K 12K I''o 30mA I1 2K 2K I 1 = 0.03( 4 K + 2 K4+K6 K ||12 K ) = 12mA 6 I '' o = -0.012( 18K ) = -4mA K Io = I ' o + I '' o = 3mA Problem 4.7 In the network shown find Vo using superposition 1K 1K 1K 12V 6mA 1K + Vo Suggested Solution Zero the indep. voltage source 1K 1K 1K 6mA 1K K V ' o = 0.006( 2 K2+ 2 K )(1k ) = 3V + V'o Zero the indep. current source + 12V V''o 1K - 1K 1K 1K V '' o = -12 4K (1K ) = -3V Vo = V ' o + V '' o = 0V Problem 4.8 Find Vo in the network shown using superposition 3K 6K 3K + 9V 3K Vo 3K Suggested Solution R1 R3 + 9V R2 Vo R3=6K R1 R2 9V R5 R4 12V All other R=3K A + VAB _ B R3 + Vo1 R4 R5 RAB = R 2 || [ R 3 + ( R 4 || R 5)] = 2.14 K R VAB = 9( RABABR1 ) = 3.75V + 4 5 V 01 = VAB ( ( R 4RR||5R+ R 3 ) = 0.75V || ) R3 + R1 R2 Vo2 - C R5 12V R4 D RCD = R 4 || [ R 3 + ( R1 || R 2)] = 2.14 K R V 02 = -12( RCDCDR 5 ) = -5V + Vo = V 01 + V 02 = -4.25V Problem 4.9 Find Io in the network shown using superposition. 6K 3K 6V 2K Io 2mA 3K Suggested Solution C 3K Ic 6K + 6K Io2 D B 2K 2mA 3K 6V VAB 2K Io1 3K A 3K RCD = 3K + (6 K || 2 K ) = 4.5 K Ic = 2m( 3 K3+K CD ) = 0.8mA R K I 02 = Ic( 6 K6+ 2 K ) = 0.6mA RAB = 2 K || (3K + 3K ) = 1.5 K RAB VAB = 6( RAB + 6 K ) = 1.2V I 01 = VAB / 2 K = 0.6mA Io = I 01 + I 02 = 1.2mA Problem 4.10 Find Io in the network shown using superposition 3K 6V Io 1K 9mA 2K 2K Suggested Solution A IA 3K Io1 1K 9mA 2K 2K B RAB = 3K + (2 K || 2 K ) = 4 K IA = I 01 = 9m( 1K1+KRAB ) = 1.8mA C I02 + 1K 2K VCD 2K 3K 6V D RCD = (1K + 3K ) || 2 K = 1.33K R VCD = -6[ RCDCD2 K ] = -2.4V + VCD I 02 = - 1K +3 K = 0.6mA I 0 = I 01 + I 02 = 2.4mA, I 0 = 2.4mA Problem 4.11 Find Io in the network shown using superposition. 2K 6K Io 4mA 12V 3K 4K Suggested Solution 6K I02 4mA 3K 4K 2K Req1 Re q1 = 6 K + (2 K || 4 K ) = 7.33K 3 I 02 = 4m[ 3 K +K q1 ] = 1.16mA Re 2K + 6K 12V 4mA I01 3K Req2 4K Re q 2 = 9 K || 4 K = 2.77 k , Vx I 01 = - 9 K = -0.77 mA Vx = 12[ ReRe2q 2 K ] = 6.97V q +2 I 0 = I 01 + I 02 = 0.39mA Problem 4.12 Find Io in the network shown using superposition 12K 12K 4mA Io 4mA 12K 6V Suggested Solution 12K 12K 4mA 12K 4mA 12K 6V 12K 12K 12K 6V 12K I'o Io -6 -3 I ' o = 12 K +12 K ||(12 K +12 K ) = 10 mA 12K 12K 4mA I1 12K 4mA 12K 6V 12K 12K 12K 4mA I''o 12K I 1 = 4m( 12 K +12 K ||12 K ) = 1.6mA 12 K I '' o = 1 I 1 = 0.8mA 2 Io = I ' o + I '' o = -0.003 10 + 0.008 = 1 mA 10 2 Problem 4.13 Find Io in the circuit shown using superposition Io 12V 4K 4K 6K 6K 6mA Io Suggested Solution Io 12V 4K 4K 6K 6K I'o 0.012 12 I 'o = = 1mA 4K 4K 6mA 6K 4K 4K 6K 6K I''o 6K 6mA I''o Current splits equally: I '' o = -3mA Io = I ' o + I '' o = -2mA Problem 4.14 Use superposition to find Io in the circuit shown 2K 12V 2K 6V Io 2mA 2K 2K Suggested Solution 2K 12V 2K 12V 2K I'o 2K I''o 2K 2K 2mA 2K 2K I'o = 12/4K = 3mA Current Splits equally I''o=1mA 2K 2K 6V I'''o 2K 2K I'''o = -6/4K = -3/2 mA Then Io=I'o + I''o + I'''o = 3mA + 1mA -3/2 mA = 5/2 mA Problem 4.15 Find Io in the network shown using superposition 6K 6K 6K 6V 5mA 6K Io Suggested Solution 6K 6K 6K 6V I'o I 'o = -6 6 K + 6 K ||(6 K + 6 K ) - = 106 A K 6K 6K 6K 6K 6K 6K I''o 5mA 6K I''o 6K 5mA 6K I '' o = 5m( 6 K + 6 K + 6 K ||6 K )( 1 ) = 1mA 2 3 2 Io = I ' o + I '' o = 1mA + 5 mA = 5 mA Problem 4.16 Find Io in the network shown using superposition. 4mA 2K 12V 1mA 1K Io 2K Suggested Solution 12V 2K 2K 1K I01 Io due to 12V source 2K 1K I02 Io due to 2mA source 2K 1K I03 Io due to 4mA source 2K I01= -12/(1K+12K) = -4mA Io = I01 + I02 + I03 = -5.33mA I02 = 2m[2K/(2K+3K)] = -1.33mA I03=0A Problem 4.17 Find Io in the network shown using superposition 6V 4K 3K 9V 2K Io 2mA 4K Suggested Solution Io due to 9V source Io due to 6V source (redrawn) 4K 3K 9V 2K I01 4K 6V 4K 2K I02 3K I 01 = 9 4K = 2.25mA I 02 = -6 4K = -1.5mA Io due to 2mA source 4K 3K 2mA 4K 2K I03 (4 K ) I 03 = 0 I 03 = 0 Io = I 01 + I 02 + I 03 Io = 0.75mA Problem 4.18 Find Vo in the network shown using superposition 12V 6K 2mA + 6K 6K 6K 6K Vo - Suggested Solution 12V Vo due to 2mA source R2 2mA + R1 R3 R4 R5 All R=6K 6K||6K=3K Vo 6K 6K 6K 6K 6K 2mA + Vo due to 12V sosurce R2 12V A 2mA + 6K 2mA + R1 R3 R4 R5 B V01 6K 3K 6K - V02 R2 R1||R5=3K A R2 RAB R1 R3 R4 3K R5 + V02 - R1 B + RAB = R1||[R2+(R3||R4)] RAB = 3.6K ohms V01 = 12[R5/(R5+RAB)] V01 = 7.5V 2mA V02 I02 3K R2 3K I02 = 2m[R2/(R2+3K+3K)] = 1mA V02 = I02(3K)=3V Vo = V01+V02=10.5V Vo=10.5 Problem 4.19 Use source transformation to find Io in the circuit shown 6K 3K 2K Io 2mA 3K Suggested Solution 6K 3K 6K 3K+3K=6K 2K Io 2mA 3K 6V Io 2mA I=6/6K + 6/6K = 2mA 3K = 6K || 6K I 3K 2K Io Io = I(3K/(2K+3K)) = 1.2mA Problem 4.20 Find Vo in the network shown using source transformation 2K 2K + 6V 2K Vo 2K 4mA Suggested Solution 2K + 2K 3mA 2K Vo 2K 4mA 3mA 2K + 1K Vo 2K 4mA 1K 2K + 2K 4mA 1mA 3K Vo 2K 4mA 3K 3V + Vo - 3K||2K=1.2K + 5mA 1.2K Vo _ Vo = (5m)(1.2K) = 6V Vo = 6V Problem 4.21 Use source transformation to find Vo in the network shown. 6V + 2mA 6K 4K Vo 12K 24V Suggested Solution 6V 12K 6V 12K + 2mA 6K 4K Vo 24V 6K 4K 12V + Vo 24V + 3mA 6K Vo 4K 12K 2mA + Vo 1mA 2K Vo = 2V Problem 4.22 Find Vo in the network shown using source transformation 3K 1K 5V 6V 2K 2K 2mA 1K Vo + Suggested Solution 1K 5/2mA 2mA 3K 2mA 2K 1K Io 3/4K 3/2mA 1K + Vo - Io = 3 2K 3/ K ( 3/ 4 K4+ 2 K ) = 0.41mA Vo = 1K ( Io) = 0.41V Problem 4.23 Find Io in the circuit shown using source transformation 12V 3K 6K 12K 12K 3K 2mA Suggested Solution 12V 3K 12V 6K 12K 12K 6K Io Figure P4.23 12V 12V 2mA 12K 3K Io 6K 12K 3K 3K 6V 12K 6K Io 12K 6K 1mA 12K Io 4K 4V 4K 4K 6K Io 12K 8V 6K 12K 4K 2mA Io = 2m(3K/(3K+6K)) = 0.67mA 6K Io 3K 2mA Io = 0.67mA Problem 4.24 Find Io in the network shown using source transformation. 6mA 4K Io 2mA 6K 18K 1mA Suggested Solution 6K 5K 18K 4K Io 12V 18K 12V 1mA 9K 4K Io 18K 6V 1mA 4K Io 2/3mA 9K 18K 1mA 4V Io 1/3mA 6K 12V 2 12 Io = -14/10K=-7/5mA 6K 4K Io Problem 4.25 Use source transformation to find Io in the circuit shown. 6K 3K 12V 12K 6K Io 6K 3K 2mA Suggested Solution 2mA 6K 12K 6K Io 3K 6 3K 6K 4K 2m 6K Io 12K 1/2m 3K 3/2m 6K Io Io = -3/2m(3K/9K) = -1mA Problem 4.26 Find Vo in the network shown using source transformation. + Vo 2K 2K 2mA 3K 12K 4K 12K 12V Suggested Solution + Vo 2K 2K 2mA 3K 12K 4K 12K 12V 6K + Vo - 2mA 6K 12K 12K 4K 3mA 4K 3K 4K 8V 3K I 3K 9V - 8 + I (4K + 3K + 3K) + 9 = 0 I = - 0.1mA Vo = I (3K) = - 0.3V Vo = -0.3V Problem 4.27 Find Vo in the circuit shown using source transformation 8K 4K 4K 12V 3K 9V 1mA 3K 4K Io 3K Suggested Solution 3K 8K 4K 4K 12V 12K 4K 3m 3K 3K 3K 3m 6K 6K 9V 3K 9V 2m 6K 1/2m 3V 3K Io 3K 3K 6K 2m 3K Io 3K 2m 3K Io 3K Io = - 1mA Problem 4.28 Find Io in the network shown using source transformation. 6V 3K 2mA 2K 6K 4K 4K 1mA Io 2mA Suggested Solution 4K 3K 6V 6 6K 4m 8V 3K 6K 1m 4K 1m 4K 4K Io 2m 8K 4V 4K Io 2m Io = -3/2m(8K/12K) 1/2m 8K 4K Io 2m 3/2m 8K 4K Io Io = -1mA Problem 4.29 Find Io in the network shown using source transformation. 3K 4mA Io 6K 9K 6V 12K 4K 3K Suggested Solution 3K 4mA 9K 12V 6K Io 6K 3K 6V 12K 4K 3K 6V 12K 4K || 12K = 3K Io 6K Io 6K 4K 6V 12K 6K 2m 6V 8V - 6 + Io ( 6K + 4K) + 8 = 0 4K Io = - 0.2 mA Problem 4.30 Find Io in the network shown using source transformation. 12K 6V 4K 2K 18K 2mA Io 8K 2mA Suggested Solution 18K 1/2mA 12K 12V 8K 2mA 6K Io Io 12K || 24K 8K 8K 1/2mA 12K 1/2mA 24K 8K Io 2mA 2mA Io = 2m [ 8K / (8K + 8K) ] Io = 1 mA Problem 4.31 Find Io in the network shown using source transformation. + 6V 6K V1 12K Io 3K 6K 2K 2mA Suggested Solution + 6V 6K V1 2K 2mA 6V 12K 2K 3K || 6K 2K 12K 6K Io 2mA 3K 6K Io = 2m ( 6 / (6 + 12 ) = 0.67mA Io = 0.67 mA Problem 4.32 Use Thevenin's Theorem to find Vo in the network shown. + Vo - 3K 12V 4K 6K 2K 2mA Suggested Solution + 3K 12V Voc 4K 6K 2K 2mA Voc = 12 ( 6K / 9K ) + 2m 2K = 12V Rth = 3K || 6K + 12K = 4K Rth 3K 6K 2K 12V 4K 4K + Vo = 6V Problem 4.33 Use Thevenin's Theorem to find Vo in the network shown. 6V 12V + 2K 4K 2K Vo - Suggested Solution 6V 12V + + 2K 4K 2K Find Voc Rth + 2K 4K Rth = 2K || 4K = 1.33K Voc 2K Vo Vo = 4.8V Vo 2K 4K V1 Voc 6V 12V + V1 = - 6 [ 4K / (4K + 2K )] V1 = - 4V Voc = 12 + V1 Problem 4.34 Use Thevenin's Theorem to find Io in the network shown. 2mA 6K 2K 12V 2K Io 1K Suggested Solution 2mA 6K I2 12V + Voc Then Voc = 12 - 6K( I2 -2m ) = 16 / 3V 1K 2K I1 = 2mA -12 + 6K ( I2 - 2m) + 2K (I2 -2m) + 1K I2 = 0 I2 = 28 / 9 mA I1 6K 2K 2K Rth 1K 16/3 V Voc Io = 4/3 mA 2K Problem 4.35 Find Io in the network shown using Thevenin's Theorem. 4K 3K 12V 6K Io 2K 2mA Suggested Solution 2mA Voc = 12 + 3K (2m) = 18V 3K 12V Rth 2K 4K 4K Rth = 3K 3K 2K Rth 18 V Voc 6K Io = 2 mA 3K Problem 4.36 Find Vo in the circuit shown using Thevenin's Theorem. 2mA 1K 2K + 12V Vo 2K Suggested Solution 2mA Voc = 12 + 2K (2m) = 16V 2K + 12V Voc 1K 1K 2K 1K Rth = 2K 3K 1K 16V + Vo 2K Vo = 16 (2K / (2K + 2K)) = 8V Rth Problem 4.37 Find Io in the circuit shown using Thevenin's Theorem. 12V 4K 4K 6K 6K 6mA Io Suggested Solution + 12V 4K Voc 4K -24 V 6K 6mA 6K 4K 4K 6K 6K Io = -2 mA Rth = 6K Voc = 12 - 6K (6m) = -24V Problem 4.38 Find Io in the network shown using Thevenin's Theorem. 2mA 2K 4mA 12V Io 2K 2K Suggested Solution 2mA 2K 4mA Voc = - 2K (2m) - 6m (2K) = - 16V 2m 4m 6m 6m + Voc 2K 2K 2K 2K 16 4K Io 12 Io = (16 + 12) / ($K) Io = 7mA Rth Rth = 4K Problem 4.39 Find Vo in the circuit shown using Thevenin's Theorem. 4mA 2K 1K 1K 1mA 2mA + 2K Vo - 1K Suggested Solution KCL 4mA 2K 1K 1K 1mA 1K Voc 2mA 4m 6m 2m 5m 1m Voc = 1K(-2m)+1K(5m)+1K(1m) Voc = 4V 1K 1K 1K 3K 1K Rth = 3K 4 2K + Vo = 4 / (3K + 2K) *2K Vo = 8/5 V Problem 4.40 Find Vo in the circuit shown using Thevenin's Theorem. + 6V 2K 2K 2K Vo 2mA 2K - Suggested Solution + 6V I1 2K 2K 2K I2 = 2m - 6 + 4K I1 + 2K(I1 - 2m) = 0 I1 = 10/6K = 5/3 mA Voc = 4K I1 + 2K I2 = 4K (5/3m) + 2K 2m = 32/3V Voc 2mA I2 2K - 4K 2K Rth = 2K + 4K || 2K = 10K/3 2K 10K/3 + 32/3 Vo 6K Vo = 32/3 [6K/(6K + 10/3 K)] = 48/7 V Problem 4.41 Find Io in the network shown using Thevenin's Theorem. 1K 1mA 2mA 6V Io 1K 1K Suggested Solution Find Voc 1K 1mA 1K I1 2mA 1mA 2mA + 6V Io 1K 1K Voc 1K I2 1K Find Rth 1K Rth = 1K + 1K I1 = - 1 mA I1 - I2 = 2mA , I2 = - 3 mA Voc = (1K)I1 + (1K)I2 = - 4V - 6 + Io (Rth) + Voc = 0 Rth = 2K Rth Voc Io 6V Io = (6 - Voc) / Rth Io = 5mA 1K Rth 1K Problem 4.42 Find Io in the network shown using Thevenin's Theorem. 2K 12V 2K 2K 2K Io 2mA Suggested Solution 2K 12V Find Voc I2 = - 2mA Voc = 12 - (4K) I2 Voc = 20V I1 2K + 2K I2 Voc 2mA Rth 2K 2K 4K Rth Both 2K resistors are shorted! Rth = 4K Voc 6K Io = 2mA Io Io = Voc / (Rth + 6K) Problem 4.43 Find Io in the network shown using Thevenin's Theorem. 2K 4mA Io 2K 2K 12V 2K Suggested Solution Find Voc 2K 4mA I2 2K I1 - I2 = 4mA 12 = (2K) I2 + 4K (I2) so, I2 = 2mA, I1 = 6mA Voc = 3K(I1 - I2) - 4K (I2) Voc = 4V 12V I1 2K 2K Find Rth 2K Rth = 3K + (2K || 4K) Rth = 4.33K Vo 3K 4K 6K Rth Io Rth Io = Voc/(Rth + 6K) Io = 0.39mA Problem 4.44 2K 12V 2mA 4K 4K Io Suggested Solution Find Voc I1 12V 2K 2mA I2 4K Voc 2K I2 - I1 = 2mA (4K)I2 + (2K)I1 + (2K)I2 = 0 yields: I2 = 0.5mA Voc = 12 -(4K)I2 = 10V Rth 2K 4K Rth 2K Voc Rth = 4K || (2K + 2K) Rth = 2K Io 6K Io = 1.25 mA Io = Voc / (Rth + 6K) Problem 4.45 Find Vo in the network shown using Thevenin's Theorem. 2mA 1K 1mA 1K 1K 1mA 1K 1K + Vo - Suggested Solution Find Voc I1 = 2mA, I2 = - 1mA, I3 = 1mA 2mA I1 1K I2 + Voc 1mA Find Rth Rth 1K 1K 1K + Rth = 3K Voc 1K Vo _ Vo = Voc [ 1K / (1K + Rth)] Vo = 0.75 V 1mA (1K)I3 + 1K(I3 - I1) - (1K)I2 + Voc = 0 Voc = 3V 1K 1K I2 1K Problem 4.46 Find Vo in the network shown using Thevenin's Theorem. + 6mA 4K 4K 4K 8K Vo - 12V 2mA 4K Suggested Solution + 6mA 4K I1 4K Voc 4K I3 4K -12 + 8K(I2 - I1) + (28K) I3 = 0 Voc = (4K)I1 + (24K) I3 Yields Voc = 74.67V I1 = 6mA I3 - I2 = 2mA 12V I2 2mA 4K Rth 24K Rth = 4K + [(8K + 4K) || 24K ] Rth = 12K Rth + 8K Voc 4K - Vo Vo = Voc[ 8K / (8K + Rth) ] Vo = 29.87V Problem 4.47 Use a combination of Thevenin's Theorem and superposition to find Vo in the circuit shown. 4K 8K 3K 12V 6K + 4mA 2K 4K _ Vo Suggested Solution Find Voc1 and Rth1 4K 8K 3K 12V 6K + 4mA 2K 4K _ Vo 12V 6K voc1 Rth1 = 3K || 6K = 2K New circuit: Find Voc2 due to 8V + 8K V'oc2 2K Find Voc2 due 4mA souce 4K 8K V'oc2 2K I2 + I1 = -4m(8+2)/(2+2+8+4) I1 = - 1mA Rth2 Rth2 = 6K||10K = 3.75K + Vo Voc Voc = Voc2 + V'oc2 Voc = 3V 8K 2K 2K Rth2 Voc2 = 8[(8+2)/(8+2+4+2)] Voc2 = 5V 4K Find Rth2 3K + Voc1 = 12[6/(6+3)] Voc1 = 8V 2K 4K I1 I2 = 4m [ (4 + 8) / ( 4 + 8 + 2 + 20] = 3mA V'oc2 = 8K I1 + 2K I2 = - 2V Vo = Voc[4K/(Rth + 4K)] Vo = 1.55V Problem 4.48 In the network shown find Vo using Thevenin's Theorem. 12V + 2K 1K 1/2 Vo 1K Vo - Suggested Solution 12V + 2K Voc = 12 + Voc/2 [ 2K / (1K 5K) ] = 12 + Voc/3 Voc = 18V Voc - 1K 1/2 Vo 12 V'o = 0V + 12V 1K 1/2 V'o 2K V'o - 1K 2K Isc Isc = 12 / (1K || 2K) = 18mA , Rth = Voc/Isc = 1K 1K 18 1K Vo = 9V + Problem 4.49 Find the Thevenin equivalent of the network shown at the terminals A-B. 1000 Ix A 1K 1K 2K Ix 2K B Suggested Solution 1000 Ix 1K 1K 2K Ix Figure P4.49 B 2K A ITest VTest Apply a test source, VTest RAB = VTest / ITest Ix = Vx 2K Vx - VTest = 1000 Ix = Vx 2 So, Vx = 2VTest At the reference node, Vx ITest = V2Test + 2 K + 1KVx1K = + K Test RAB = VTest = I 1 2K [VTest + 2Vx] = 1 2K [5VTest ] 2K 5 = 400 RAB = 400 Problem 4.50 Find the Thevenin equivalent of the network shown at the terminals A-B. + Vx - 1K 1K 1K Vx / 1000 1K A B Suggested Solution ITest VTest B Apply a test source, VTest, RAB = VTest / ITest + Vx - 1K 1K 1K Vx / 1000 1K A K Vx = Vy[ 1K1+1K ] Vx = Vy 2 At Vy: Vx 1K = Vy 2K - + VTestKVy yields Vy = VTest 2 VTest ITest - At VTest: ITest = V1Test + VTestKVy K 2 = RAB = 1K Problem 4.51 Find Vo in the network shown using Thevenin's Theorem. 1K + VA VA/2 + A 6V 2K 3K - B Suggested Solution 1K + VA + 2K 3K 6V + VA/2 Find Voc 1K VA 2K VA/2 + Voc VA = 6 [1K / (2K+1K) ] VA = 2V 6V 1K + VA - I1 Isc Voc = - VA/2 -VA + 6 = 3V Rth Rth = Voc / Isc + 6V 1K I2 VA/2 Voc 3K Vo - Rth = 1K VA = 6 - VA/2 => VA = 4V I1 = VA / 1K = 4mA, I2 = (VA/2)/2K = 1mA Isc = I1 - I2 = 3mA Vo = Voc[3K/(3K + Rth)] = 2.25V Vo = 2.25V Problem 4.52 Find Vo in the network shown using Thevenin's Theorem. 4K + VA 2VA 2K 4K Vo + 12V Suggested Solution 4K + V'A 2V'A 2K Voc Figure P4.52 + Using voltage division V'A = 12(4K/6K) = 8V 12V 4K + V''A 2V''A 2K Isc KVL outer loop -12 + V''A + 2V''A = 0 V''A = 4V 12V If V''A = 4V 1/K 4/K 3/K Isc = - 3mA and Rth = Voc/Isc = (-12) / (-3m) = 4K 4K 12V 4K + Vo Vo = - 6V Problem 4.53 Find Vo in the circuit shown using Thevenin's Theorem. + Vo - 4K 6V + 2K 2K 4K 12V Vx/1000 Vx - Suggested Solution + 4K 6V + Voc 2K 2K Vy Vx/1000 Vx = 6 [ 2K / (4K + 2K) ] = 2V 4K Vy = 12 - 4K ( Vx/1000 ) = 4V 12V Voc = Vx + Vy = - 2V Vx - Find Isc 4K + 6V 2K 12V Vx/1K Isc 4K Isc = Vx/1K + (Vx - 12) / 4K (6 - Vx)/4K = Vx / 2K + Isc Rth + Voc 2K Vo Vx = 9/4 V Isc = - 0.19 mA Rth = Voc / Isc = 10.67 K Vo = Voc[ 2K / ( 2K + Rth) ] Vo = - 0.32 V Problem 4.54 Use Thevenin's Theorem to find Vo in the circuit shown. Ix 2K 1K 2 Ix _ + 6V 1K Vo Suggested Solution I'x 2K 1K 2 I'x _ 6V + Voc I'x = 0 , Voc = 6V I''x 2K 1K I1 6V Isc 2 I''x I''x = Isc, I1 = - 2 Isc 2K Isc + 1K Isc +1K I1 - 6 = 0 Isc = 6mA Rth = Voc / Isc = 1K 6V 1K 1K + Vo _ Vo = 3V Problem 4.55 Use Thevenin's Theorem to Find Io in the circuit shown. + Vx 2K 20V 2K Io 4K 2Vx Suggested Solution + V'x 2K 20V Voc 4K 2V'x - 20 + 2K I + 4K I + 2 Vx = 0 V'x = 2K (I) I = 2mA Voc = 20 -2m (2K) = 16V + V''x 2K 4K 2V''x Isc V''x = 20 Isc = 20 / 2K + 40 / 4K = 20 mA Rth = Voc / Isc = 4/5 K 20V 4/5 K 16V 2K Io = 16 / (2 + 4/5)K = 40 / 7K = 5.71 mA Io = 5.71 mA Io Problem 4.56 Find Vo in the network shown using Thevenin's Theorem. 2K 2K 2Vx + 12V 1K + Vx 2K 2K Vo - Suggested Solution Find Voc 2K 2K 2Vx + 12V 1K + Vx 2K 2K Vo 2K 1K 2K V1 + (Vx - V1)/2K = V1/1K + Vx/2K yields, Vx = Voc = 18 V Now, Vx = 0! and V2 = - 12V V2/2K + V2/1K + Isc = 0 , Isc = 18 mA Find Isc V2 + 2Vx 1K 2K 0V 12V + Vx=0V 2K 2K Isc Isc 2K Rth = Voc/Isc = 1K Rth Voc 2K + Vo Vo = Voc [ 2K / ( 2K + 2K + Rth)] = 7.2V Vo = 7.2V Problem 4.57 Use Thevenin's Theorem to find Vo in the network shown. 12V 2K + 2K 2KIo 2K 1K Io 1K Vo - Suggested Solution 12V Voc-12 2K 2KI'o 2K 1K I'x 12V Voc + Supernode - KCL (Voc - 12 - 2KIx) / 2K + (Voc - 12) / 2K + Voc / 1K = 0 I'x = Voc / 1K Then Voc = 12 V 1K Resistor is shorted I''x = 0 2K 2K Isc = 12mA and Rth = Voc / Isc = 1K 1K 12 2K 1K + Vo = 12(1K/4K) = 3V - Problem 4.58 Find Vo in the network shown using Thevenin's Theorem. 6K 100Ix 1K 3V 2K Ix 1mA 2K + Vo - Suggested Solution 6K 100Ix 1K 3V 2K Ix 1mA 2K 6K + Vo 3V 2K Ix Ix = - 3/8K = - 0.38 mA 100Ix 3 + (6K)Ix - (1K)Ix + Voc = 0 Vx = 2000Ix Vx = - 1000 Ix Vx = 0 and Ix = 0A Now, -3 / 6K = Isc => Isc = -0.5mA Voc = - 1.13V Rth = Voc / Isc = 2.25 K Voc 100Ix + - Find Isc 6K 3V 2K Ix Isc Rth 100Ix Find Voc 1K Rth + + 1mA V'oc - Voc Ix Voc 1mA 2K _ Vo Voc = -1.13V, Rth = 2.25K V'oc = (1m)Rth + Voc V'oc = 1.13V + 2K Vo - Rth Rth R'th = Rth 1K V'oc = V'oc[2K / (2K+1K+R'th)] Vo = 0.43V Problem 4.59 Use Thevenin's Theorem to Find Vo in the circuit shown. + 1K Vo - 12V 2Ix 1K 1K Ix 1K Suggested Solution + 12V 2I'x 1K 1K I'x Voc - (Voc - 12) / 1K + 2I'x + Voc/1K = 0 I'x = Voc / 1K Voc = 3V I''x = 0, The network is reduced to 12V 2I''x 1K 1K I''x Isc 1K 12 Isc = 12mA Rth = Voc / Isc = 3 / 12m = 0.25 K + 1K Vo Vo = 4/3 V Vo = [ 3 / (2 + 1/4)K ] 1K = 4/3 V 1/4 K 1K Problem 4.60 + 1K 2K Vx 1K 1K 1mA 1K 2Vx / 1000 2K 1K 4mA + Vo Suggested Solution + 1K 2K Vx 1K 1K 1mA 1K 2K 2Vx / 1000 Find Isc 1K 2K I1 Vx 1K 1mA Isc Voc 1K I2 2Vx / 1000 2K 4mA Vo = Voc [ 1K / (1K + Rth ) ] 1K I3 Rth = Voc / Isc Rth 1K + Vo Rth = 8K 1K 4mA 1K I2 2Vx / 1000 + (3K)I1 + Voc - (1K) I3 + 1K (I1 - I2) = 0 Voc = -8V 2K 1K 4mA I3 Find Voc + Vo 2K 1K I1 + Vx 1K 1mA + Voc Vx = 0 I1 = 1mA I3 = - 4mA I2 = 0 I1 - Isc = 1mA, I3 = -4 mA Vx = - Isc (1K), I2 = 2Vx / 1K 3KI1 + (1K) Isc + 1K (Isc -I3) + 1K (I1 - I2) = 0 yields, Isc = -1mA Vo = - 0.89 V Problem 4.61 Use Norton's Theorem to find Vo in the network shown. 2000Ix 4K 2K Ix 2mA 4K + Vo - 6K 6V Suggested Solution 2000Ix 4K 2K Ix 2mA 4K + Vo Find Isc I1 6V 4K Isc 2mA 6K 6V 6K 2K Ix 2000Ix I2 I1 6V 6K 2K Ix 2KIx I2 3mA 4K + Voc - 6 + 6K I1 + 2K(I1 -I2) = 0 2K (I2 - I1) - 2K Ix + 4K Isc = 0 Ix = I1 - I2 Isc - I2 = 3mA yields: Isc = 0.86mA Same equations as those used to find Isc except Isc -> 0. Also, Voc = 2K Ix + 2K Ix yields: Voc = 6V Isc Rth 4K + Vo - Rth = Voc / Isc = 7K Vo = Isc[ Rth || 4K ] Vo = 2.18V Problem 4.62 Find Io in the network shown using Norton's Theorem 2K 12V 2K Io 4K 4V Suggested Solution Find Isc 2K 12V 2K Io 4K 4V I1 2K 12V Isc I2 Io 2K Isc Req Io = 2mA 4K 4V I1 = 12 / 2K = 6mA I2 = 4 / 4K = 1mA Isc = I1 - I2 = 5mA Io = Isc[ Req/( Req + 2K) ] Req = 2K || 4K 2K Req 4K Req = 1.33 K Problem 4.63 Use Norton's Theorem to find Io in the circuit shown. 3K 12V 3K 2K Io 2K 4mA Suggested Solution Isc = 12 / 3K - 4m[3K / (3K + 3K)] 3K 12V Isc 3K 2K 4mA 1K Isc = 2mA 3K 2K Rth = 3K || 6K = 2K 3K Rth 1K 2mA 2K 2K Io = 1mA Problem 4.64 Find Io in the network shown using Norton's Theorem. 6K 3K 6V 2K Io 3K 2mA Suggested Solution 6K 3K Find Isc 6K Isc 6V 2K Io Find req 1K Req = 6K || (3K + 3K) 1K Req 1K Req = 3K Isc Req Io 2K 3K 2mA I1 6V I2 6V I1 = 6 / 6K = 1mA I2 = 6/ 6K = 1mA Isc = I1 + I2 = 2mA 3K Io = Isc [ Req / (Req + 2K)] Io = 1.2 mA Problem 4.65 Find Io in the network shown using Norton's Theorem. 2K Io 6K 3K 12V 2K 4K Suggested Solution Find Isc Isc V1 = 12V 6K V1 12V 2K V2 3K (V1 - V2) / 6K + (V1 - V2) / 3K = V2 / 2K yields, V2 = 6V 4K Isc = (V1 - V2) / 3K + V1 / 4K = 5mA Req 6K 3K Isc 2K 4K 1K Req 2K Io Io = Isc[ Req / (Req + 2K)] Io = 2.57mA Req = 4K || [3K + (6K || 2K)] Req = 2.12 K Problem 4.66 Use Norton's Theorem to find Vo in the network shown. 1K 2K + 12V Vo _ 2K 1K 2mA Suggested Solution 1K 2K 2mA Isc = 12 / 2K + 2 / 1K = 8mA 12V 1K Rth = 2K 1K + 1K Rth 8m 2K 2K Vo Vo = 8V 2K Problem 4.67 4K 3K 4mA 2K 8K + Vo 4K - 6V 6K Suggested Solution 4K 3K 4mA 2K 8K + 6V 6K 4K Vo 4K Find Isc 2K I2 4mA 2K 8K Isc 3K 6K 2K 8K 2mA Find Req 4K 3K 6K 4m 2K 8K Isc Find Isc 4K 4V I1 I1 - I2 = 4mA, 2K(Isc - I1) + 8K (Isc - I2) = 0 4 = (2K) I1 + (4K) I2 yields: Isc = 0.1333 mA Req = (8K + 2K) || [4K + (6K || 3K)] Req = 3.75 K + Isc Req 4K Vo = Isc [ Req || 4K] = 258mV Vo = 258 mV Vo Problem 4.68 Given the linear circuit shown, it is known that when a 2-K load is connected to the terminals A-B, the load current is 10mA. If a 10-K load is connected to the terminals the load current is 6mA. Find the current in a 20K load. A Rth Voc B Suggested Solution Rth A + Voc RL VAB (Rth + RL) I = Voc if RL = 2K , I = 10mA => Voc = 20 + 0.01Rth if RL = 10K , I = 6mA => Voc = 60 + 0.006Rth yield Voc = 120V and Rth = 10K If RL = 20K , I = Voc / (RL + Rth) I = 4 mA - B I Problem 4.69 If an 8-K load is connected to the terminals of the network shown, VAB = 16V. If a 2K load is connected to the terminals VAB = 8V. Find VAB if a 20K load is connected to the terminals. A Linear circuit B Suggested Solution Rth A + Voc VAB B Thevenin eq. for linear circuit VAB = Voc[ RL / (RL + Rth) ] => Voc = VAB [ 1 + Rth / RL ] If RL = 8K , VAB = 16V => Voc = 16 [ 1 + Rth / 8K ] If RL = 2K , VAB = 8V => Voc = 8 [ 1 + Rth / 2K ] yield: Rth = 4K and Voc = 24V If RL = 20K , VAB = 24 [ 20 / (20 + 4) ] VAB = 20 V Problem 4.70 Find RL for maximum power transfer and the maximum power that can be transferred in the network shown. 12V 6K 6K RL 6K 3V Suggested Solution A Find RAB All 3 Resistors are attached to both node A and node B, so, 12V 6K 6K 6K RL RAB RAB = 2K 6K RAB = 6K || 6K || 6K 6K 3V 6K B Find maximum load tranfer V3 12V 6K 6K RL Yields: V3 - V1 = 12V V2 = 3V At supernode: V1 V2 3V (V3 - V2) / 6K + (V1 - V2) / 6K + V3 / 2K + V1 / 6K = 0 V3 = 5V PL = V23 / RAB PL = 12.5 mW 6K Problem 4.71 Find RL for maximum power transfer and the maximum power that can be transferred in the network shown. 3K 2mA 6K RL 5K Suggested Solution A Find RAB A 3K 2mA 6K RL 2K 3K RAB = 2K + 3K + 5K = 10K 5K 5K B B Find maximum load tranfer 3K I1 6K 2mA I3 RL I3 - I1 = 2mA I2 = 1mA (2K) I1 + (10K) I3 + 5K (I3 - I2) + 3K (I1 - I2) = 0 yields: I3 = 0.9mA PL = I23 RAB I2 5K PL = 8.1 mW Problem 4.72 In the network shown find RL for maximum power transfer and the maximum power that can be transferred to this load. RL 4K 3K 2mA 6K 3V Suggested Solution Rth = 4K + 3K || 6K = 6K 4K 3K 6K + Voc I1 = 2m 3K (I2 - I1) + 6K I2 + 3 = 0 6K I2 = 1/3 mA 3V I2 4K I1 2mA 3K Voc = 4K (2m) + 6K (1/3 m) = 10V 6K + 10 Vo RL = 6K Vo = 5V PL = V2o / RL PL = 25/6 mW Problem 4.73 Find RL for maximum power transfer and the maximum power that can be transferred in the network shown. 2K 1mA 4K RL + Vx / 2000 2K 2K Vx - Suggested Solution 2K To find Req, replace RL with a 1mA current source, Req = VTest / 1mA RL + Vx / 2000 2K 2K Vx 4K I1 2K I4 4K RL I3 2K + Vx Vx / 2000 2K 2K I3 VTest 1mA 4K + I2 2K Vx - Find max load power 1mA I1 Vx / 2000 I2 2K I1 = Vx / 2K , I2 = Vx / 2K I3 - I2 = 1mA (2K)I3 + (2K)I2 + 2K(I2 -I1) + 4K(I3 -I1) = 0 VTest = 4K(I3 - I1) + (2K)I3 yields: VTest = 3V Req = VTest / 1mA = 3K I1 = 1mA, I2 - I1 = Vx / 2K, I3 = Vx / 2K 4K (I4 - I2) + 2K I4 + 3K (I4 - I3) = 0 2K I3 + 2K (I3 - I2) + 3K ( I3 - I4) = 0 yields: I3 = 1.25mA and I4 = 1.42mA => PL = 3K (I4 - I3)2 PL = 83.3 W Problem 4.74 In the network shown find Io using PSPICE. 6K 12V 6K 6mA Io 6K 6K Suggested Solution I o = 3.2 mA. R1 + 12V V1 R2 6K I1 6mA IDC 6K R5 6K R3 6K 3.200mA Problem 4.75 In the network shown determine Vo using PSPICE. 6V + 2mA 6K 4K Vo 12K 24V Suggested Solution Vo = 2V V1 + IDC 2mA I1 R1 6K 6V R2 4K 2.000V + V2 24V R3 6K Problem 4.76 Find Vo in the network shown using PSPICE. + Vo - 3K 12V 4K 6K 2K 2mA Suggested Solution Vo = 5 - ( -1) = 6 R1 + 12V V1 R2 6K V. 6K R3 5.000V 6K -1.000V R4 2K I1 2mA IDC Problem 4.77 Find Vo in the network shown using PSPICE. 2K 2K 2Vx + 12V 1K + Vx 2K 2K Vo - Suggested Solution Vo = 8V R3 1K I1 2mA R1 2K + V1 12V R2 2K IDC 8.000V R4 2K Problem 4FE-1 Determine the maximum power that can be delivered to the load RL in the network shown. 1K 12V 4mA 1K 1K 2K RL Suggested Solution + Voc (Vx - 12) / 1K - 4m + Vx / 3K = 0 Vx = 12V Voc = Vx [ 2K / (2K + 1K)] = 8V 1K 12V 1K 2K 4mA 1K 1K Rth = 2K || 2K = 1K 1K 1K RL RL = 2K PRL = (8 / 4K)2 (2K) PRL = 8mW Problem 4FE-2 Find the value of the load RL in the network shown that will achieve maximum power transfer, and determine the value of the maximum power. 1K + Vx RL 2Vx 1K 12V Suggested Solution 1K + V'x + Voc 2Vx - 12 + 2K I + 1K I + 2V'x = 0 V'x = 2K I Voc = 12 - 2K I = 66/7 V 1K + V''x Isc 2V''x 1K Vx = 12 Isc = 12 / 2K + 24 / 1K = 30mA 1K I 12V 12V Rth = Voc / Isc = 2/7 K P = [(60/7)/(4/7 K)]2 (2/7 K) RL = 2/7 K P = 64.3 mW 2/7 K 66/7 V Problem 4FE-3 Find the value of RL in the network shown for maximum power transfer to this load. Ix 3 12V 12 12 2Ix RL Suggested Solution I'x 3 12V 12 2I'x + Voc (Voc - 12) / 3 + Voc / 12 - 2 I'x = 0 I'x = (12 - Voc) / 3 Voc = 144/13 V I''x I''x = 12/3 = 4mA 3 12V 12 2I''x Isc Then Isc = I''x + 2I''x = 12A Rth = Voc / Isc = (144/13)/12 = 12/13 Then RL = 12 + 12/13 RL = 12.92 for MPT Problem 5.1 A 12F capacitor has an accumulated charge of 480C. Determine the voltage across the capacitor after 4 s. Suggested Solution t2 v(t 2) - v(t1) = 1 C t1 i(t )dt C = 100 F i (t ) = 1mA = I v(t1) = 0. v(t 2) = 40V so, I v(t 2) = C (t 2 - t1) where t 2 - t1 = 4sec Problem 5.2 A 12F capacitor has an accumulated charge of 480C. Determine the voltage across the capacitor. Suggested Solution C=Q C Q = 480 C C = 12 F V = 40V Problem 5.3 A capacitor has an accumulated charge of 600C with 5V across it. What is the value of capacitance? Suggested Solution Q C=V Q = 600 C V = 5V C = 120 F Problem 5.4 A 25- F capacitor initially charged to 10V is charged by a constant current of 2.5 A. Find the voltage across the capacitor after 2 minutes. Suggested Solution V= = 1 C 0 t idt + v(0) 150 1 25 x10-6 150 0 2.5 x10-6 dt - 10 1 = 10 0 1 dt - 10 = 10 t |150 -10 0 V = 15 - 10 = 5V Problem 5.5 The energy that is stored in a 25-F capacitor is w(t)=12sin2377t J. Find the current in the capacitor. Suggested Solution w ( t ) = 1 Cv 2 (t ) = 12sin 2 377t 2 C = 25 F v 2 (t ) = 2 (12 ) sin 2 377t = 9.6 105 sin 2 377t -6 25 10 v(t ) = 979.8sin 377t V i (t ) = C dv ( t ) = 25 10-6 (979.8)(377) cos 377t = 9.23cos 377t A dt Problem 5.6 An uncharged 10-uF capacitor is charged by the current I(t)=10cos377t mA. Find (a) the expression for the voltage across the capacitor and (b) the expression for the power. Suggested Solution i (t ) = 10 cos(377t )mA a) 1 v(t ) = C i (t )dt = 0 t 0.01 1 C 377 C = 10 F v(0) = 0V sin(377t ) |t0 v(t ) = 2.65sin(377t ) V b) P (t ) = v(t )i (t ) = (2.65)(0.01) sin(377t ) cos(377t ) but cos( x) sin( x) = 1 sin(2 x) 2 p(t ) = 13.3sin(754t ) mW so, Problem 5.7 The voltage across a 100-uF capacitor is given by the expression v(t)=120sin(377t) V. Find (a)the current in the capacitor and (b) the expression for the energy stored in the element. Suggested Solution v(t ) = 120sin(377t ) V, C = 100 F dv a) i (t ) = C = (100 10-6 )(120)(377) cos(377t ) = 4.52 cos(377t ) A dt 1 2 1 2 b) w ( t ) = Cv ( t ) = (10-4 ) (120 ) sin 2 ( 377t ) 2 2 2 = 720sin ( 377t ) mJ = 360 - 360 cos 754t mJ Problem 5.8 A capacitor is charged by a constant current of 2mA and results in a voltage increase of 12V in a 10-s interval. What is the value of the capacitance? Suggested Solution t2 v(t 2) - v(t1) = 1 C t1 I I v(t 2 ) - v(t 1 ) = 12 = C (t2 - t1 ) = C (10) i (t )dt i (t ) = 2mA = I C = (2m)(10) /12 = 1.67mF C = 1.67mF Problem 5.9 The current in a 100F capacitor is shown in Figure P5.9. Determine the waveform for the voltage across the capacitor if it is initially uncharged. i(t)(mA) 10 0 1 2 Figure P5.9 t(ms) Suggested Solution C = 100 F Time(ms) 0t 2 t>0 v(t ) = 10 0 1 C i(t )dt v(t)(V) 100t 0 + v(2m) = 0.2V i(t)(mA) i(t)(mA) 10 0 1 2 t(ms) Problem 5.10 The voltage across a 100-F capacitor is shown in Figure P5.10. Compute the waveform for the current in the capacitor v(t)(V) 6 0 1 2 3 t(ms) Figure P5.10 Suggested Solution C = 100 F Time(ms) i (t ) = Cc dv(V/ms) dt i(t)(mA) 0 t 1 1 t 2 2t 3 v(t)(V) 6 6 0 -6 600 0 - 600 0 1 2 3 t(ms) Figure P5.10 Problem 5.11 The voltage across a 6F capacitor is shown in Figure P5.11. Compute the waveform for the current in the capacitor v(t)(V) 2 0 1 2 3 4 t(ms) Figure P5.11 Suggested Solution C = 6 F Time(ms) ( i (t ) = C dvdtt ) dv(V/ms) dt i(t)(mA) 0t 2 2t 4 61 -1 6 -6 v(t)(V) 2 0 1 2 3 4 t(ms) Problem 5.12 The voltage across a 50-F capacitor is shown in Figure P5.12. Determine the current waveform. v(t)(V) 10 0 2 4 6 8 10 12 t(ms) Figure P5.12 Suggested Solution C = 50 F Time(ms) i (t ) = C dv dt dv(V/ms) dt i(t)(mA) 0t 2 2t 4 4t 8 8 t 10 10 t 12 t > 12 v(t)(V) 10 5 0 -5 0 5 0 250 0 - 250 0 250 0 0 2 4 6 8 10 12 t(ms) Problem 5.13 The voltage across a 2-F capacitor is given by the waveform in Figure P5.13. Compute the current waveform. v(t)(V) 0 2 3 6 t(ms) -12 Figure P5.13 Suggested Solution C = 2 F Time(ms) i (t ) = C dv dt dv(V/ms) dt i(t)(mA) 0t 2 2t 3 3t 6 t >6 -6 0 4 0 - 12 0 8 0 v(t)(V) 0 2 3 6 t(ms) -12 Problem 5.14 The voltage across a 0.1-F capacitor is given by the waveform in Figure P5.14. Find the waveform for the current in the capacitor vc(t)(V) -12 0 -12 1 2 3 4 5 t(s) Figure P5.14 Suggested Solution C = 0.1F Time(s) i (t ) = C dv dt dv(V/ms dt i(t)(A) 0t 2 2t 3 3t 5 t >5 vc(t)(V) 6 - 24 6 0 0.6 - 2.4 0.6 0 -12 0 -12 1 2 3 4 5 t(s) Problem 5.15 The waveform for the current in a 200-F capacitor is shown in Figure P5.15. Determine the waveform for the capacitor voltage. i(t)(mA) 5 0 t(ms) 1 2 Figure P5.15 3 4 Suggested Solution C = 200 F 0 t 4ms i (t ) = 1.25t A v(t ) = 3125t 2 +v(0) assuming v(0) = 0V , v(t ) = 3125t 2 t > 4ms i (t ) = 0 A v(t ) = 0 + v(4m) v(t ) = 0 + 3125(4m) 2 = 50mV 3125t 2V v(t ) = 50mV 0 t 4ms t > 4ms v(t ) = 1 C i(t )dt Problem 5.16 Draw the waveform for the current in a 12-F capacitor when the capacitor voltage is as described in Figure 5.16 v(t)(V) 12 10 0 -8 6 16 t(s) Figure P5.16 Suggested Solution C = 12 F Time(s) i (t ) = C dv dt dv(V/s) dt i(t)(A) 0t 6 t 10 10 t 16 t > 16 2 -5 1.33 0 24 - 60 16 0 v(t)(V) 10 5 0 -5 -10 t(s) 0 2 4 6 8 10 12 14 16 i(t)(mA) 30 10 -10 -30 -50 -70 0 v(t)(V) 2 4 6 8 10 12 14 16 t(s) 10 5 0 -5 -10 t(s) 0 2 4 6 8 10 12 14 16 i(t)(mA) 30 10 -10 -30 -50 -70 0 2 4 6 8 10 12 14 16 t(s) Problem 5.17 Draw the waveform for the current in a 3-F capacitor when the voltage across the capacitor is given in Figure P5.17 v(t)(V) 4 2 9 0 -1 2 4 7 10 t(ms) Figure P5.17 Suggested Solution C = 3 F Time(ms) i (t ) = C dv dt dv(V/s) dt i(t)(A) 0t 2 2t 4 4t 7 7t 9 9 t 10 t>0 2 -1 0 -3 2 6 -3 0 - 4.5 3 0 1 0 v(t)(V) 4 2 0 -2 0 2 4 6 8 10 12 t(s) i(t)(mA) 6 3 0 -3 -6 0 2 4 6 8 10 12 t(s) Problem 5.18 The waveform for the current in a 100-F initially uncharged capacitor is shown in Figure P5.18. Determine the waveform for the capacitor's voltage. v(t)(mA) 5 0 -5 1 2 3 4 5 t(ms) Figure P5.18 Suggested Solution v(t)(mA) 5 0 -5 1 2 3 4 5 t(ms) C = 100 F v (0)=0V v(t ) = 1 C i(t )dt 0t 1ms 1mst 2ms i (t )=-5mA v (t ) =50t V i (t )=-5mA v(t ) = -50(t - 1m) + v(1m) v(t ) = -50t + 0.1 V 2ms t 3ms i (t ) = 5mA v(t ) = 50(t + 2m) + v(2m) v(t ) = 50t - 0.1 V i (t ) = -5mA v(t ) = -50(t - 3m) + v(3m) v(t ) = -50t + 0.2 V i (t ) = 5mA v(t ) = 50(t - 4m) + v(4m) v(t ) = 50t - 0.2 V i (t ) = 0 v(t ) = 50mV 0 t 1ms 1ms t 2ms 2ms t 3ms 3ms t 4ms 4ms t 5ms t > 5ms 3ms t 4ms 4ms t 5ms t > 5ms V 50t -50t + 0.1 V 50t - 0.1 V v(t ) = -50t + 0.2 V 50t - 0.2 V mV 50 Problem 5.19 The voltage across a 6-F capacitor is given by the waveform in Figure P5.19. Plot the waveform for the capacitor current. v(t)(mA) 5 0.5 0 -5 Figure P5.19 1 t(ms) Suggested Solution C = 6 F i (t ) = C dv dt if v(t ) = 10sin( t )V = 2 f = 2 / T where T = 1m sec, v(t ) = 10sin(2000 t )V i (t ) = (6 )(10)(2000 ) cos(2000 f ) i (t ) = 377 cos(2000 t )mA v(t)(V) 5 0 -5 0 i(t)(V) 0.5 1 t(ms) 5 0 -5 0 0.5 1 t(s) Problem 5.20 The current in an inductor changes from 0 to 200mA in 4ms in 4 ms and induces a voltage of 100mV. What is the value of the inductor? Suggested Solution di v(t ) = L dt i = 200mA vind = L ti t = 4ms v ind = 100mV Assuming the current change is linear, L = vind ( ti ) = 2mH L = 2mH Problem 5.21 The current in a 100-mH inductor is i(t)=2sin(377t)A. Find (a) the voltage across the inductor and (b) the expression for the energy stored in the element Suggested Solution L = 100mH a) b) w(t ) = L i 2 (t ) = 2 but 0.1 2 i (t ) = 2sin(377t ) A di v(t ) = L dt = (0.1)(2)(377) cos(377t ) = 75.4 cos(377t )V (2)2sin 2 (377t ) = 0.2sin 2 (377t ) so sin 2 x = 1 - 1 cos(2 x) 2 2 w(t ) = 0.1 - 0.1cos(754t ) J Problem 5.22 A 10-mH inductor has a sudden current change from 200mA to 100mA in 1ms. Find the induced voltage. Suggested Solution L = 10mH v(t ) = L di dt i = 100mA - 200mA = -100mA t = 1ms assuming the current changed linearly vIND = (0.01) ti = -1V vIND = -1V Problem 5.23 The induced voltage across a 10-mH inductor is v(t)=120cos(377t) V. Find (a) the expression for the inductor current and (b) the expression for the power. Suggested Solution L = 10mH a) i (t ) = b) 1 L v(t ) = 120 cos(377t )V 120 1 0.01 377 Assume i (0) = 0. v(t )dt = sin(377t ) = 31.83sin(377t ) A p(t ) = v(t )i (t ) = (120)(31.83) cos(377t ) sin(377t ) but sin( x) cos( x) = 1 sin( x) 2 so, p (t ) = 1910sin(754t )W Problem 5.24 The current in a 25-mH inductor is given by the expressions i(t)=0 t<0 -4t i(t)=10(1-te )mA t>0 Find (a) the voltage across the inductor, (b) the expression for the energy stored in it. Suggested Solution L = 25mH a) 0 i (t ) = -t 10(1 - e )mA di dt t<0 t >0 v(t ) = L t < 0: t > 0: v(t ) = 0 v(t ) = (0.025)(0.01)e - t = 250e - t V t<0 t>0 0 v(t ) = -t 250e V b) w(t ) = L i 2 (t ) = 1.25[1 - 2e - t + e -2t ] J 2 Problem 5.25 Given the data in the previous problem, find the voltage across the inductor and the energy stored in it after 1 s. Suggested Solution v(t ) = 250e - t V at t = 1sec. w(t ) = 1.25[1 - 2e -t + e -2t ] J v(1) = 91.97 V w(1) = 0.5 J Problem 5.26 The current in a 50-mH inductor is given by the expressions i(t)=0 t<0 -4t i(t)=2te A t>0 Find (a) the voltage across the inductor, (b) the time at which the current is a maximum, and (c) the time at which the voltage is a minimum Suggested Solution L = 50mH 0 i (t ) = - t 2te A t t<0 t >0 for t>0 ( a) v(t ) = L didtt ) = (0.05)(2e -4 t -8te -4 ) 0 v(t ) = -4 t 0.1e (1 - 4t )V b) t<0 t>0 The current will be at its maximum when di/dt = 0, or, v(t)= 0 v(tmax ) = 0 = 0.1e -4tmx (1 - 4tmax ) tmax = 0.25sec c) The voltage will be at its max. or min. when dv/dt = 0 di dt t =tmin | = [0.1e -4t (-4) + 0.1e -4t (4)(4t - 1)] |t = min = 0 tmin = 0.5 sec. yields : Problem 5.27 The current i (t) = 0 i (t) = 100te-t/10 A t<0 t>0 flows through a 150-mH inductor. Find both the voltage across the inductor and the energy stored in it after 5 seconds. Suggested Solution di v(t ) = L dt v(t ) = 100 x10-3 v(t ) = -1.5e -t /10 d dt (100e - t /10 ) 100 v(t ) = 1000 ( 100 )(-e -t /10 ) 10 v( s ) = -1.5e -1/ 2 = -0.91V w(t ) = 1 Li 2 2 w(t ) = 1 (50 x10-3 )(104 e 2 w(t ) = 1 (50)(10)e 2 w(t ) = 250e - -t 5 - -t 5 - -2 t 10 ) ) w(5) = 91.97 J Problem 5.28 The current in a 10-mH inductor is shown in Figure P5.28. Find the voltage across the inductor. i(t)(mA) +5 2 0 4 6 7 8 t(ms) -10 Figure P5.28 Suggested Solution L = 10mH Time(ms) di v(t ) = C dt di(A/ms) dt v(t)(mV) 0t 2 2t 4 4t 6 6t 7 t>7 -5 7.5 0 -5 0 - 50 75 0 - 50 0 Problem 5.29 The current in a 50-mH inductor is given in Figure P5.29. Sketch the inductor voltage. i(t)(mA) 100 4 0 -100 Figure P5.29 2 6 8 10 t(ms) Suggested Solution L = 50mH Time(ms) di v(t ) = L dt di(A/s) dt v(t)(V) 0t 2 2t 4 4t 8 8 t 10 t > 10 0 - 50 50 - 50 0 0 - 2.5 2.5 - 2.5 0 120 80 40 i(t)(mA) 0 -40 -80 -120 2 4 6 time (ms) 8 10 3 2 1 v(t)(V) 0 -1 -2 -3 2 4 6 8 10 time (ms) Problem 5.30 The current in a 16-mH inductor is given by the waveform in Figure P5.30. Find the waveform of the voltage across the inductor. i(t)(A) 12 -12 -24 2 4 5 9 11 12 t(ms) Figure P5.30 Suggested Solution L = 16mH Time(ms) di v(t ) = L dt di(A/ms) dt v(t)(V) 0t 2 2t 5 5t 9 9 t 11 11 t 12 t > 12 - 12 + 12 0 -12 +12 0 -192 192 0 - 192 192 0 i(t)(A) 12 0 -12 -24 0 5 time (ms) 10 Problem 5.31 Draw the waveform for the voltage across a 10-mH inductor when the inductor current is given by the waveform shown in Figure P5.31 i(t)(A) 4 2 6 -1 3 Figure P5.31 9 11 t(s) Suggested Solution L = 10mH Time(s) di v(t ) = L dt di(A/s) dt v(t)(mV) 0t 3 3t 6 6t 9 9 t 11 t > 11 2/3 -1 0 2.5 0 6.67 - 10 0 25 0 4 2 i(t)(A) 0 -2 0 3 6 time(s) 30 v(t)(mV) 15 0 9 11 12 -15 0 3 6 time(s) 9 11 12 Problem 5.32 The voltage across a 10-mH inductor is shown in Figure P5.32. Determine the waveform for the inductor current v(t)(mV) 10 0 1 Figure P5.32 2 t(ms) Suggested Solution L = 10mH 0 t 1ms assuming 1 t 2ms i (t ) = 1 L v(t )dt i (t ) = 500t 2 A v(t ) = 10t mV i (0) = 0 A, v(t ) = 10-2 - 10(t - 10-3 )V i (t ) = (2 - 1000t )dt + K where K = integration constant i (t ) = 2t - 500t 2 + K A Both equations for i(t) must be equal at t = 1ms. 500(10-3 ) 2 = 2(10-3 ) - 500(10-3 ) 2 + K So i (t ) = 2t - 500t 2 - 10-3 A 0 t 1ms 1 t 2ms t > 2ms 500t 2 A i (t ) = 2t - 500t 2 - 10-3 A 1 mA K = -1mA Problem 5.33 The waveform for the voltage across a 20-mH inductor is shown in Figure P5.33. Compute the waveform for the inductor current. v(t)(mV) 10 0 1 2 3 t(ms) -20 Figure P5.33 Suggested Solution L = 20mH i (t ) = 1 L v(t )dt 0 t 2ms v(t ) = 10mV if i (0) = 0 A, i (t ) = 0.5t A 2 t 3ms v(t ) = -20mV i (t ) = -(t - 2m) + i (2m) = -t + 2m + 1m i (t ) = (3 x10-3 - t ) A t > 3ms v(t ) = 0V i (t ) = 0 + i (3m) = 0 A 0.5t i (t ) = 3 x10-3 - t 0 A A A 0 t 2ms 2 t 3ms t > 3ms 10 v(t)(mV) 0 -10 -20 0 1 2 time (ms) 3 4 Problem 5.34 The voltage across a 2-H inductor is given by the waveform shown in Figure P5.34. Find the waveform for the current in the inductor. v(t)(mV) 1 0 1 2 3 4 5 t(ms) Figure P5.34 Suggested Solution L = 2H 0 t 1ms i (t ) = 1 L v(t )dt i (0) = 0 A assume v(t ) = 1mV i (t ) = 500t A 1 t 2mA 2ms t 3ms v(t ) = 0V i (t ) = i (1ms ) = 0.5 A v(t ) = 1mV i (t ) = 500(t - 2m) + K1 A K1 is an integrator constant, i (2m) = 0.5 A = 500(2m - 2m) + K1 K1 = 0.5 A so, i (t ) = 500t - 0.5 A 3ms t 4ms 4ms t 5ms v(t ) = 0 i (t ) = 0 + i (3m) = 1 A v(t ) = 1mV i (t ) = 500(t - 4m) + K 2 A i (4m) = 1 A = 500(4m - 4m) + K 2 K 2 = 1 A i (t ) = 500t - 1 A t > 5ms i (t ) = 1.5 A 500t 0.5 500t - 0.5 i (t ) = 1.0 500t - 1 1.5 A A A A A A 0 t 1ms 1 t 2ms 2 t 3ms 3 t 4ms 4 t 5ms t > 5ms Problem 5.35 Find the possible capacitance range of the following capacitors. a) 0.068 F with a tolerance of 10% b) 120pF with a tolerance of 20% c) 39F with a tolerance of 20% Suggested Solution a) Minimum capacitor value = 0.9C = 61.2nF Maximum capacitor value = 1.1C = 74.8nF b) Minimum capacitor value = 0.8C = 96pF Maximum capacitor value = 1.2C = 144pF c) Minimum capacitor value = 0.8C = 31.2F Maximum capacitor value = 1.2C = 46.8F Problem 5.36 The capacitor in Figure P5.36a is 51 nF with a tolerance of 10%. Given the voltage waveform in Figure P5.36b graph the current i(t) for the minimum and maximum capacitor values. i(t) -60 v(t) C v(t)(V) 0 -60 0 1 2 3 4 time (ms) (b) 5 6 (a) Figure P5.36 Suggested Solution Maximum capacitor value = 1.1C = 56.1 nF Minimum capacitor value = 0.9C = 45.9 nF The capacitor voltage and current are related by the equation ( i (t ) = C dvdtt ) Problem 5.37 Find the possible inductance range of the following inductors a) 10 mH with a tolerance of 10% b) 2.0 nH with a tolerance of 5% c) 68 H with a tolerance of 10% Suggested Solution a) Minimum inductor value = 0.9L = 0.9 mH Maximum inductor value = 1.1L = 1.1 mH b) Minimum inductor value = 0.95 = 1.9 nH Maximum inductor value = 1.05L = 2.1 nH c) Minimum inductor value = 0.9L = 61.2 nH Maximum inductor value = 1.1L = 74.8 nH Problem 5.38 The inductor in Figure P5.38a is 330H with a tolerance of 5%. Given the current waveform in Figure P5.38b, graph the voltage v(t) for the minimum and maximum inductor values. i(t) v(t) L (a) Figure P5.38 Suggested Solution Maximum inductor value = 1.05L = 346.5 H Minimum inductor value = 0.95L = 313.5 H The inductor voltage and current are related by the equation ( v(t ) = L didtt ) Problem 5.39 The inductor in Figure P5.39a is 4.7 H with a tolerance of 20%. Given the current waveform in Figure P5.39b, graph the voltage v(t) for the minimum and maximum inductor values. i(t) v(t) L (a) Figure P5.39 Suggested Solution Maximum inductor value = 1.2L = 5.64 H Minimum inductor value = 0.8L = 3.76 H The inductor voltage and current are related by the equation ( v(t ) = L didtt ) Problem 5.40 What values of capacitance can be obtained by interconnecting a 4-F capacitor, a 6-F capacitor, and a 12-F capacitor? Suggested Solution Combo A C1 Combo C Combo B C1 C1 C2 C3 C2 C3 C2 C3 CA= 1 / ( 1/C1 + 1/C2 + 1/C3 ) = 2F CB = C1 + C2 + C3 = 22F CC = [ C1 (C2 + C3) ] / [C1 + (C2 + C3)] CC = 3.27F Combo D Combo E Combo F C2 C1 C3 C3 C2 C1 C1 C3 C2 CD = [ C2 (C1 + C3) ] / [C2 + (C1 + C3)] CD = 4.36F CC = [ C3 (C2 + C1) ] / [C3 + (C2 + C1)] CC = 5.45F CF = C3 + C1 C2 / (C1 + C2) = 14.4F Combo G Combo H Possibilities: C2 CA = 2F CB = 22F CC = 3.27F CD = 4.36 F CE = 4.45F CF = 14.4F CG = 9 F CH = 8 F C1 C2 C1 C3 C3 CG = C2 + C1 C3 / (C1 + C3) = 9F CH = C1 + C2 C3 / (C2+ C3) = 8F Problem 5.41 Given a 1, 3, and 4-F capacitor, can they be interconnected to obtain an equivalent 2-F capacitor? Suggested Solution C1 = 1F C2 = 3F C3 = 4F C1 C2 C3 Ceq = (C1 + C 2)C 3 (1 + 3 )4 = = 2 F C1 + C 2 + C 3 1 + 3 + 4 Problem 5.42 Given four 2-F capacitors, find the maximum value and minimum value that can be obtained by interconnecting the capacitors in series/parallel combinations. Suggested Solution Minimum combo C1 C2 Maximum combo C3 C4 C1 C2 C3 C4 C min = 1 C1 + 1 C2 1 = 0.5 F 1 + C 3 + C14 C max = C1 + C 2 + C 3 + C 4 = 8 F C min = 0.5 F C max = 8 F Problem 5.43 The two capacitors in Figure P5.42 were charged and then connected as shown. Determine the equivalent capacitance, the initial voltage at the terminals, and the total energy stored in the network. 4V + 6F 1V + 3F Figure 5.43 Suggested Solution + V1 V Ceq C1 = 6F V1 =- 4V + - C1 + V2 - C2 C2 = 3F V2 = 1V Ceq = C1C 2 /(C1 + C 2) = 2 F V = V 1 + V 2 = -3V W = 1 C1V 21 + 1 C1V 2 2 2 2 W = 49.5 J Ceq = 2 F V = -3V W = 49.5 J Problem 5.44 Two capacitors are connected in series as shown in Figure P5.44. Find Vo i(t) + - Vo C1 + 24V - C2 C1 = 12F C2 = 6F Suggested Solution i(t) + - Vo C1 + 24V - C2 C1 = 12F C2 = 6F vc = 1 C i(t )dt Since same current charged both caps, C1Vo = C2 (24) Vo = 24(C2 / C1 ) Vo = 12V Problem 5.45 Three capacitors are connected as shown in Figure P5.45. Find V1 and V2. V1 + + 8F + 12V V2 - 4F - Figure 5.45 Suggested Solution V1 + + C1=8F + 12V V2 - C2=4F - V 1 + V 2 = 12V Assuming C1 and C2 are charged by the same surrent, C1V 1 = C 2V 2 V 1 = 12 (same as Q1 = Q2) C2 = 4V C1 + C 2 C1 V 2 = 12 = 8V C1 + C 2 V 1 = 4V V 2 = 8V Problem 5.46 Select the value of C to produce the desired total capacitance of CT=2F in the circuit in Figure P5.46 C CT C1=2F C2=4F Suggested Solution C CT C1=2F C2=4F C (C1 + C 2) 6C 2= C + C1 + C 2 6+C 12 + 2C = 6C 4C = 12 CT = C = 3 F Problem 5.47 Select the value of C to produce the desired total capacitance of CT=1F in the circuit in Figure P5.47. C CT 1F 2F C 1F 1F Figure P5.57 Suggested Solution 3C (1 + C )(1) + 3 + C (1 + C ) + 1 3C 1+ C 1= + 3+C 2+C (3 + C )(2 + C ) = 3C (2 + C ) + (3 + C )(1 + C ) CT = 1 = 0 = 3C 2 + 5C - 3 C1, C 2 = -5 25 - 4(3)(-3) 2(3) C = 0.47 F Problem 5.48 Find the equivalent capacitance at terminals A-B in Figure P5.48 5F A 3F 2F 6F 2F B 6F Figure 5.48 12F 6F Suggested Solution C1 A C5 C2 C3 B C4 C1=5F C2=2F C3=2F C4=7F C6=6F C5=3F C7=6F C8=12F C8 C6 C7 Ceq1 = C 5C 6 = 2F C5 + C6 Ceq2=C2+Ceq1=4F C 7C 8 = 4F C 7 + C8 Ceq4=C3+Ceq3=6F 1 1 1 = + Ceq 5 = 2.4F Ceq 5 Ceq 2 Ceq 4 1 1 1 1 = + + CAB = 1.32F CAB C1 C 4 Ceq 5 Ceq 3 = Problem 5.49 Determine the total capacitance of the network in Figure P5.49 4F 1F 2F CT 4F 2F 12F Figure P5.49 4F Suggested Solution C1 C2 C3 C4 C5 C6 C7 C1=4F C2=4F C3=12F C4=1F C5=2F C6=2F C7=4F (C 4 + C 6)(C 5 + C 7) (C 4 + C 6) + (C 5 + C 7) CAB = 2 F 1 1 1 1 1 1 1 = + + = + + CT C 2 + CAB C 1 C 3 6 4 12 CT = 2 F CAB = Problem 5.50 Find CT in the network in Figure P5.50 if (a) the switch is open and (b) the switch is closed. 3F CT 6F 6F 12F Figure P5.50 Suggested Solution The networks can be reduced as follows: a) 3 6 2F 6 12 4F CT CT =(4||2)=6F b) 3 6 9F 18F 6 12 CT CT =9(18)/(9+18)=6F Problem 5.51 Find the total capacitance CT of the network in Figure P5.51. 4F CT 6F 1F 8F 3F Figure P5.51 Suggested Solution C1 = 4F C1 CT C4 C2 = 1F C3 = 8F C4 = 6F C3 C5 C5 = 3F C2 Ceq1 = C2 + C3 + C5 = 12F Ceq2 = C1Ceq1/(C1 + Ceq1) = 3F CT = C4 + Ceq2 = 9F CT = 9F Problem 5.52 Compute the equivalent capacitance of the network in Figure P5.52 if all the capacitors are 5F. Figure P5.52 Suggested Solution 4 8 8 Ceq = 4(8)/12 + 8 = 32/3 F Problem 5.53 If all the capacitors in Figure P5.53 are 6 F, find Ceq Ceq Figure P5.53 Suggested Solution 6 6 6 Ceq 6 6 6 6 6 Ceq 6 12 12 9 6 4 Ceq 12 + 54/15 = 15.6 Ceq = 4(15.6)/(19.6)=3.18F Problem 5.54 Given the capacitors in Figure 5.54 are C1=2.0uF with a tolerance of 2% and C2=2.0uF with a tolerance of 20%, find a) the nominal value of CEQ b) the minimum and maximum possible values of CEQ c) the percent errors of the minimum and maximum values C1 Ceq C2 Figure P5.54 Suggested Solution a ) The nominal value is CEQ = C1 + C 2 + C 3 = (0.1 + 0.33 + 1) x10-6 = 1.43 F b) The minimum value of CEQ is CEQ ,min = C1,min + C2,min + C3,min = (0.1*0.9 + 0.33*0.8 + 1*0.9) x10-6 = 1.254 F The maximum value of CEQ is CEQ ,max = C1,max + C2,max + C3,max = (0.1*1.1 + 0.33*1.2 + 1*1.1) x10-6 = 1.606 F c) The percent irror of the minimum value is 1.254 - 1.43 x100 = -12.3% 1.43 The percent irror of the maximum value is 1.606 - 1.43 x100 = 12.3% 1.43 Problem 5.55 Given the capacitors in Figure 5.55 are C1=0.1F with a tolerance of 2% and C2=0.33F with a tolerance of 20% and 1 F with a tolerance of 10% . Find the following. a) the nominal value of CEQ b) the minimum and maximum possible values of CEQ c) the percent errors of the minimum and maximum values Ceq C1 C2 C3 Figure P5.55 Suggested Solution a) The nominal value is C1C 2 (2.0)(2.0) = CEQ = x10-6 = 1.0 F C1 + C 2 2.0 + 2.0 b) The minimum value of CEQ is CEQ ,min = C1,min C2,min ( 2.0*0.98)(2.0*0.8) ( 2.0*1.02)(2.0*1.2 ) C1,min + C2,min (2.0*0.98) + (2.0*0.8) C1,max C2,max x10-6 = 0.881 F The maximum value of CEQ is CEQ ,max = C1,max + C2,max (2.0*1.02) + (2.0*1.2) x10-6 = 1.103 F c) The percent irror of the minimum value is 0.881 - 1.0 x100 = -11.9% 1.0 The percent irror of the maximum value is 1.103 - 1 x100 = 10.3% 1 Problem 5.56 Select the value of L that produces a total inductance of LT = 10mH in the circuit in Figure P5.56 12mH LT = 10mH L 8mH Figure P5.56 Suggested Solution LT = (L1 + L2) L / (L1 + L2 + L) L1 LT L 10 = 20 L / (20 + L) 200 + 10L = 20L L = 20mH L2 Problem 5.57 Find the value of L in the network in Figure P5.57 so that the total inductance LT will be 2 mH. 4mH 2mH L LT 6mH Figure P5.57 Suggested Solution 4mH 2mH All L's in mH L LT 6mH [ 66+LL + 2]4 =2 6L 6+ L + 2 + 4 24 L 6+ L 6L 6+ L +8 =2 +6 24 L + 8(6 + L) = (6 L + (6 + L))2 24 L + 48 + 8 L = 12 L + 72 + 12 L 8 L = 24 L = 3mH Problem 5.58 Find the value of L in the network in Figure 5.58 so that the value of LT will be 2 mH. 2mH 1mH 6mH LT L L 4mH Figure P5.58 Suggested Solution 2 L (4L) / (4+L) [ 44+LL + 2]L =2 4L +2+ L 4+ L 6 L2 + 8L = 2 L2 + 20 L + 16 ( L + 4)( L + 1) = 0 L = 4mH Problem 5.59 Determine the inductance at terminals A-B in the network in Figure P5.59. 1mH A 6mH 2mH 12mH 1mH B 4mH 2mH 2mH Figure 5.59 Suggested Solution L6 A L1 L4 L2 L5 L8 L7 L3 B L9 L1 = L3 = L6 = 1mH L2 = 12mH L4 = 6mH L5 = 4mH L7 = L8 = L9 = 2mh Leq1 = L6 + L7 = 3mH Leq 2 = L4 Leq1 /( L4 + Leq1 ) = 2mH Leq 3 = L8 + L9 + 4mH Leq 4 = Leq 3 L5 /( Leq 3 + L5 ) = 2mH Leq 5 = L2 ( Leq 2 + Leq 4 ) L2 + Leq 2 + Leq 4 = 3mH LAB = L1 + L3 + Leq 5 = 5mH LAB = 5mH Problem 5.60 Compute the equivalent inductance of the network in Figure P5.60 if all inductors are 5 mH Leq Figure 5.60 Suggested Solution Redrawing the network 4 4 6 4 2 2 2 Leq = 4(6) / 10 + 2 = 24 / 10 + 20 / 10 = 44 / 10 = 4.4mA Problem 5.61 Determine the inductance at terminals A-B in the network in Figure P5.61 1mH A 4mH 12mH 2mH 3mH 2mH B Figure 5.61 4mH Suggested Solution A L1 L2 L3 L6 L4 L5 B L1 = 1mH L2 = L7 = 4mH L3 = 12mH L4 = 3mH L5 = L6 = 2mH L7 Leq1 = L6 + L7 = 6mH Leq 2 = L2 L3 /( L2 + L3 ) = 3mH Leq 3 = Leq 2 + L4 = 6mH Leq 4 = Leq1 Leq 3 /( Leq1 + Leq 3 ) Leq 4 = 3mH LAB = L1 + Leq 4 + L5 LAB = 6mH Problem 5.62 Find the total inductance at the terminals of the network in Figure P5.62 6mH 3mH 6mH 4mH LT 10mH Figure 5.62 Suggested Solution L1 = 6mH L1 L3 L2 = 6mH L3 = 3mH L4 = 4mH L4 LT L5 L5 = 10mH L2 L1 is shorted out! Leq1 = L2 L3 /( L2 + L3 ) = 2mH Leq 2 = Leq1 + L5 = 10m + 2m = 12mH LT = L4 Leq 2 /( L4 + Leq 2 ) LT = 3mH Problem 5.63 Given the network shown in Figure P5.63 find (a) the equivalent inductance at terminals A-B with terminals C-D short circuited. And (b) the equivalent inductance at terminals C-D with terminals A-B open circuited. 12H A C 6H 2H B 2H Figure P5.63 D Suggested Solution A L1 C L3 D L1 = 12H L2 = 2H L3 = 6H L4 = 2H L2 L4 Leq1 = L1L3/(L1 + L3) = 4H Leq2=L2L4/(L2 + L4) = 1H LAB = Leq1 + Leq2 LAB = 5H B A L1 C L3 C Leq3 = L1 + L3 = 18H Leq4 = L2 + L4 = 4H 1/LCD = 1/Leq3 + 1/Leq4 L4 D LCD = 3.27 H L2 B Problem 5.64 For the network in Figure P5.64 choose C such that vo = -10 vs dt C Rs=10K + 70K + Vs RL vo - Figure P5.64 Suggested Solution C i2 i1 Rs 70K 0 + Vs RL vo - Rs = 10 K vo = -10 vs dt R e q = Rs + 70 K = 80 K Using ideal op-amp assumptions, i1 = i2 1 vs - 0 dv = -C o vo = - vs dt R eq R eq C dt So, R eq C = 1 C = 1.25 F 10 Problem 5.65 For the network in Figure P5.65, vs(t)=120cos377t V. Find Vo(t) 1K 1F 0 + Vs(t) vo(t) - Figure P5.65 Suggested Solution R C i1 0 i2 + Vs(t) vo(t) - C = 1 F R = 1K vs = 120 cos 377t V i1 = i2 (ideal op-amp assumptions) C dvs v dv = - o vo = - RC s dt R dt vo (t ) = + (1k )(1 )(120)(377) sin(377)t vo (t ) = 45.24sin(377t ) V Problem 5.66 For the network in Figure P5.66, vs(t)=115sin377t V. Find vo(t) 5F 5K 0 + Vs(t) vo(t) Figure P5.66 Suggested Solution 5F 5K R Vs(t) C + vo(t) - vo (t ) = - 1 vs dt RC 115 = cos(377t ) (5K )(5 )(377) vo (t ) = 12.20 cos(377t ) V Problem 5FE-1 Given three capacitors with values 2F, 4F and 6F, can the capacitors be interconnected so that the combination is an equivalent 3F? Suggested Solution Yes. 2F 3F 4F 6F Problem 5FE-2 The current pulse shown in Figure 5PFE-2 is applied to a 1F capacitor. Determine the charge on the capacitor and the energy stored. 6A t (sec) 0 1 Figure PFE-2 Suggested Solution The capacitor voltage is 1 1 v(t ) = i (t )dt = -6 C0 10 -6 T 10-6 6dt = 6V 0 Q = CV = 10 6 = 6 C 1 1 W = CV 2 = (10-6 )(6) 2 = 18 J 2 2 Problem 5FE-3 In the network shown in Figure 5PFE-3, determine the energy stored in the unknown capacitor Cx. + 8V 24V Cx 60F + Figure PFE-3 Suggested Solution + 8V 24V Cx 60F Q = CV = (60x10-6)(8) = 480C Vx = 24 - 8 = 16V Cx = Q/Vx = (480x10-6) / 16 = 30F W = 1/2 CV2 = 1/2 (30x10-6) (16)2 W = 3.84mJ + Problem 6.1 Use the differential equation approach to find VC(t) for t>0 in the circuit in Fig. P6.1 t=0 + 3Kr VCtt1 Suggested Solution Vc (0-) = 12( 3k ) = 4V 9k Vc (t ) - 12 cdvc (t ) + =0 6k dt OR dVc (t ) Vc (t ) + = 12 dt 0.6 Vc (t ) = k1 + k2 e -t / 0.6 Att = ok1 + k2 = 4 and at t = k1 = 12 k2 = -8 HenceVc (t ) = 12 - 8e- t / 0.6Vt > 0 Problem 6.2 Use the differential equation approach to find i(t) for t > 0 in the network in Fig P6.2 as shown. t=0 12V 6 i(t) 2H 6 Suggested Solution t=0 12V 6 i(t) 2H LL(0-) 6 iL (0-) = 2 A i (t ) + Ri (t ) = 0 dt i (t ) = K 2 e - R / 2t = K 2 e-3t Ld i (0) = 2 = K 2 i(t)=2e -3t A t>0 Problem 6.3 Use the differential equation approach to find VC(t) for t>0 in the circuit in Fig. P6.3 1K 3K 4K + 12V t=0 VC(t) 100F Suggested Solution 4K 8K ) = 6V VC (t ) V (t ) V (t ) V (t ) + Cd C = 0 d C + C = 0 4K dt dt 0.4 - t / 0.4 VC (t ) = K 2 e V and since VC (0) = 6 = K 2 VC (0-) = 12( VC (t ) = 6e - t / 0.4V t>0 Problem 6.4 Use the differential equation approach to find i0(t) for t>0 in the network in Fig P6.4 4K 2K 1K 2K 4A i0(t) 300F Suggested Solution 4K 2K 1K 2K 4A i0(t) 300F 6 6K 4 ( )= A K 9K K 4 VC (0-) = 2 K ( ) = 8V K dV (t ) V (t ) for t > 0 C c + c =0 dt 2K dVc (t ) Vc (t ) + =0 dt 0.6 Vc (t ) = K 2 e - t / 0.6V i0 (0-) Vc (t ) = 8 = K 2 Vc (t ) = 8- t / 0.6 V t>o Problem 6.5 In the network in Fig P6.5, find i0(t) for t>0 using the differential equation approach. 6 i0(t) 2H 4 t=0 2A 12 Suggested Solution 3 2 iL (0- ) = 2 = A 3+ 6 3 using current division L dio (t ) + 10io (t ) = 0 dt 10 = 6 + 4 10 10 = L 2 io (t ) = K 2 e -5t A So, io (0) = 5= 2 = K2 3 2 -5t e A, t > 0 3 Then, io (t ) = Problem 6.6 In the circuit in Fig P6.6, find i0(t) for t>0 using the differential equation approach. t=0 2 12V 3 2H 6 i0(t) Suggested Solution t=0 2 12V 3 2H 6 iL(t) iL (0-) = 12 6 ( ) = 2A 2 + 6 /13 3 + 6 di (t ) di (t ) 9 for t > 0 L L + RiL (t ) = 0 or L + iL (t ) = 0 dt c(t ) 2 i L (0) = 2 = K 2 -4.5t iL (t ) = K 2 e -4.5t A iL (t ) = 2e A but i 0 (t ) = -iL (t ) = -2e-4.5t A t>0 Problem 6.7 Use the differential equation approach to find V0(T) for t>0 in the circuit in Fig P6 and plot the response including the time interval just prior to switch action. t=0 5k 2k + 12V 200F 4k - Suggested Solution t=0 R1 R2 + 12V Vc(t) C R3 V0 (t) For t<0, VC (0-) = VC (0+ ) = 0 and V0 (0-) = 0V R1 = 5 K R 2 = 2 K R 3 = 4 K C=0.2mF For t 0 VC (0+ ) = 0V By KCL : V0 (0+ ) R3 = 0V V2 12-V2 dV2 = +C R2 + R3 R1 dt R3 ) R2 + R3 12=V2 [1 + R1 dV2 ] + RC R2 + R3 dt By KVL : V0 (t ) = V2 ( or V2 = 1.5 V0 (t) Assume, dV0 11 + V0 - 8 = 0 Eq.1 dt 6 -t V0 (t ) = K1 + K 2 e - 8 = 0 {K1 = 48 + K 2e 6 V Now, V0 (t ) = 11 -48 48 V0 (0+) = 0 = + K2 K2 = 11 11 48 V0 (t ) = (1 - e 11 -11t -11t 48 V 11 6 )V t>0 and V0 (t ) = 0 t<0 Problem 6.8 Use the differential equation approach to find i(t) for t > 0 in the circuit in Fig. P8 and plot the response including the time interval just prior to opening the switch. t=0 i(t) 4 6 2 36 V 8 2H Suggested Solution t=0 R3 4 i(t) R2 R1 iL(t) 36 V R4 8 L R1 = 2 R2 = 6 R3 = 4 R4 = 8 L = 2H For t<0 i L (0-)=i L (0+)=4A i(0-)=2A By KVL: i(t)[R 2 + R3 + R4 ] + L or, Assume Now, di(t) + 9i (t ) = 0 dt -t d i (t) =0 dt i(t)=K1 +K 2 e -K 2 -t -t e + 9 K1 + 9 K 2 e = 0 1 sec. Also, I(0+)=LK1 +K 2 =-4 K 2 =-4 9 So, K1 = 0 A and = i(t)=-4e-9t i(t)=2 t>0 t<0 Problem 6.9 Use the differential equation approach to find vc(t) for t > 0 in the circuit in Fig P6.9 and plot the response including the time interval just prior to opening the switch. 12k 12 k 12 V t=0 6 k 6 k + 200 F - vc(t) Suggested Solution 12 v R1 R2 12 k 12 V t=0 R3 R4 + C - v c(t) Problem 6.10 Use the differential equation approach to find IL (t) for t > 0 in the circuit in Fig. P6.10 and plot the response including the time interval just prior to opening the switch. 2H iL(t) 6 3 12 V t=0 12 V 6 Suggested Solution Problem 6.11 Use the differential equation approach to find i(t) for t>0 in the circuit in Fig. P6.4 and plot the response including the time interval just prior to switch movement. t=0 i(t) 10 mA 10 k 1 mH 1 k Suggested Solution t=0 i(t) 10 mA R1 L iL R1 = 10k R 2 = 1k L = 1 mH At t=0- and t=0+ i L (0- )=i L (0+ )=10 mA i(0- )=0A i(t) iL(t) L R2 iL(0+)=-iL(0+)=-10mA By KVL: R 2i(t)+L or, di(t) =0 dt di(t) R 2 + i(t)=0 dt L -t Assuming i(t)=K1 +K 2 e K2 -t -t K1 = 0 R2 K1 R2 K 2 + Now, - e + e =0 L = = 0.1 s L L R2 i (t ) = -10e -10 7t mA t>0 and i(t)=0 t<0 Problem 6.12 Use the differential equation approach to find VC(t) for t >0 in the circuit in Fig. P6.12 and plot the response including the time interval just prior to closing the switch. 4 k 12V 4 k VC(t) t=0 + - + - 4 k 100 F 4 k Suggested Solution R1 R2 R3 12V + t=0 + C VC(t) R4 All R=4k C = 0.1 mF R3 + R4 + VC (0- ) = VC (0 ) = 12 R1 + R2 + R3 + R4 VC (0) = 6V For t > 0 By KCL: VC (t ) dVC (t ) +C =0 8 dt k 3 dVC (t ) 15 + VC (t ) = 0 dt 4 -t Assume, VC (t ) = K1 + K 2 e -K 2 15 15 e + K1 + K 2 e = 0 4 4 4 = sec. K1 = 0 15 -t -t Now, -t VC (t ) = K 2 e VC (t ) = 6e VC (0+ ) = K 2 = 6 15t -( ) 4 so, v for t > 0 VC (t ) = 6 for t < 0 Problem 6.13 Use the differential equation approach to find Vc(t) for t >0 in the circuit in Fig. P.13 and plot the response including the time interval just prior to opening the switch. 100 F t=0 + 6 k V0(t) - 6 k 12V 6 k Suggested Solution 100 F V (0) +C t=0 + 6 k V0(t) - t=0 6 k 12V 6 k R2 12V R1 R3 + V0(0) - R1 = 6k R 2 =6k R 3 =6k C = 0.1mF For t < 0 Vc(0- )=VC (0+ ) R2 VC =12 R2 + R3 VC (0)=6V VC (0- )=6V At t>0 By KCL: or, Assuming -t C dVc VC VC + + =0 dt R 2 R 1 + R3 dVC (t) +2.5VC (t)=0 dt -t VC (t)=K1 +K 2 e , we find -t -K 2 e + 2.5 K1 + 2.5 K 2 e = 0 K1 =0 and = 0.45 -t VC (0 )=6V=K 2 e = K 2 + But , V0 (t)=-VC (t) [R 3 ] so VC (t)=6e-2.5tV R 1 +R 3 so, V0 (t)=-3e-2.5 tV t 0 V0 (t)=6V t < 0 +VC(t)- R2 R1 R3 -V0(t)t Problem 6.14 Use the differential equation approach to find i0(t) for t > 0 in the circuit in Fig. P6.14 and plot the response including the time interval just prior to closing the switch. 100 F 6 K 6 K 12V 6 K i0(t) t=0 6 K Suggested Solution VC(t) + R1 C R3 R4 12V R2 i0(t) t=0 All R = 6 K C=0.1mF For t < 0 i 0 (0- )= 12 = 1mA R1 +R 2 VC (0- )=VC (0+ ) =i 0 (0- )R 2 =6V For t > 0 KCL: or, -t 12-VC (t) VC (t) C dVC(t) = + R1 R2 dt dVC (t) 10 + VC (t)=20 dt 3 -t -t Assume So, VC(t)=K1 +K 2e . Now, -K 2 e + 10 K1 10 K 2 e + = 20 3 3 -t =0.3 s and K1 =6. Yields VC (t)=6+K 2 e VC (t)=6V. Note that i 0 (t)= VC (t) =1mA R2 But VC (0+ )=6=6+K 2e0 =6+K 2 K 2 =0! i 0 (t)=1mA for all t Problem 6.15 Use the differential equation approach to find i(t) for t > 0 in the circuit in Fig. P.16 and plot the response including the time interval just prior to switch movement. 6 K t=0 12V 0.2 mH i(t) 4 K 2 mA Suggested Solution t=0 R1 12V L i(t) R2 2 mA R1 = 6 K R2=4K L=0.2mH For t > 0 when t < 0, i(0- )=i(0+ )= 12 =2mA R1 KVL: L di1 (t) +R 2 (i1 -i 2 )=0 and i 2 =2mA dt di1 (t) R 2 R + i1 = 2 i 2 dt L L -t -t -t R R -K 2 R2 Assume i1 (t)=K1 +K 2 e , now, K1 + 2 K 2 e = 2 i2 e + L L L L = 50ns and K1 = 2mA. yields = R2 i1 (0) = -i (0) = -2mA = K1 + K 2 K 2 = -4mA. Also, i (t ) = -i1 (t ) i(t)= 4e-2x10 t - 2 mA for t 0 and i(t)=2mA t < 0 ( 6 ) R2 i(t) i2 2 mA Problem 6.16 Use the differential equation approach to find V0(t) for t > 0 in the circuit in Fig. P.16 and plot the response including the time interval just prior to switch action. 10 k 6 k t=0 6 k 10 k 12 V 8 k + 100 F V0(t) 8 k 6V Suggested Solution R1 R3 t=0 R6 R4 R1=R4=10k R2=R5=8k R3=R6=6k C=0.1mF R2 12 V C + V0(t) R5 6V Perform 2 Thevenin equivalents, one at each side of the switch, R R t=0 16V/3 + C (8/3)V _ V0(t) For t > 0 R= 8 - V0 3 R dV0 dt -t 94 k 9 By KCL: C Assume, V0 (t)=K1 + K 2 e Now, 8 9.4 8 9.4 K 2 - K1 - K 2 e = -( )e K1 = and = 3 9 3 9 16 8 = K1 + K 2 K 2 = 3 3 V0 (t)=8 1-e-0.96t V , t 0 V0 (t)= 16 V, t < 0 3 -t -t V0 (0+ ) = Problem 6.17 Use the differential equation approach to find V0(t) for t > 0 in the circuit in Fig. P.17 and plot the response including the time interval just prior to closing the switch. 2 k 100 F V0(t)10 mA 5 k t=0 4 k 6V 2 k Suggested Solution R2 V0(t) + C Is R1 t=0 R3 6V R4 R1=5k R2=2k R3=4k R4=2k C=0.1 mF Is=10 mA Perform 2 Thevenin equivalents-one at each side (left/right) of the switch. V0(t) + Ra 50V t=0 C 4V RB Ra=7k RB=4/3 k For t < 0 No current flow, V0(0-)=V0(0+)=54V For t > 0 _ 50V Ra V0(t) + C RB 4V By KCL: or, C dV0 -4-V0 (t) + =0 dt RB dV0 -7.5V0 (t)=30 dt -t If V0 (t)=K1 +K 2e , yields, = -K 2 e - 7.5 K1 - 7.5 K 2 e = 30 -t -t 1 and K1 =-4. Nor, V0 (0+ )=54=K1 +K 2 K 2 =58 7.5 V0 (t)=-4+58e-7.5tV t 0 = 54V t< 0 Problem 6.18 Use the differential equation approach to find V0(t) for t > 0 in the circuit in Fig. P6.18 and plot the response including the time interval just prior to opening the switch. t=0 24V 2 4 6 (1/4)H 4 8 + VC(t) - Suggested Solution R2 t=0 24V R4 + t=0 R2 + L L R1 R3 R5 - V0(t)=24V R R1 iL(t) - VX(t) R1 = 2 R 2 =R 4 +4 R 3 =6 1 R 5 =8 L= H 4 R=4 V0 = 2 x 3 At t < 0: i L (0- )=i L (0+ )= At t > 0 KVL: 24 =6A and Vx (0- )=0 R2 di LdiL + Rx iL = 0 L + 9.6iL = 0 dt dt -t 1 If i L (t)=K1 +K 2 e , then = S and K1 =0 9.6 i L (0+ )=6=K 2 i L (t)=6e-9.6t A Vx (t)=-R x i L (t)=-1-4e-9.6t V R x =R//(R1 +R 2 ) R x =2.4 V0 (t)=-9.6e-9.6t V =0 t>0 t<0 Problem 6.19 Use the differential equation approach to find V0(t) for t > 0 in the circuit in Fig. P.19 and plot the response including the time interval just prior to opening the switch. t=0 1 k 50 F 3 k 9V 3 k 3 k + V0(t) - Suggested Solution t=0 R1 C + VC R3 R2 R4 R1 9V V0(t) 2 k R4 R2 For t < 0 R3 + 9V 2K V0 (0- ) = 9 = 6V 2 K + R1 R2 VC (0- ) = V0 (0-) = 3V R2 + R3 VC (0- ) = VC (0- ) For t > 0 VC C R3 + R2 R4 - V0 KCL: C dVC VC VC dV + + = 0 C + 10 VC = 0 dt R3 R2 + R4 dt -t If VC (t)=K1 +K 2e , -K 2 e + 10 K1 + 10 K 2 e + 0 =0.15, K1 =0 VC (t)=3e-10tV t>0 t<0 -t -t VC (0+ )=3 =K 2 R4 V0 = -VC R4 + R2 V0 = - VC 2 V0 (t)=-1.5e-10t V = 6V Problem 6.20 Use the differential equation approach to find i0(t) for t > 0 in the circuit in Fig. P.20 and plot the response including the time interval just prior to opening the switch. 24V t=0 2H 4 6 i0(t) 12 2 Suggested Solution 24V iL(t) L R1 R2 i0(t) R3 R4 For t < 0 t=0 R1=4 R2=6 R3=12 R 1 R4=2 R5=2 L =2H iL 24V R2 i0 2 12V R3 R4 6A i0 (0 - ) = - 12 R2 12 = 3A R1 i0 (0 - ) = - 2 A iL (0 - ) = iL (0 + ) = For t > 0 KVL: L di X +(R+R1 )i X (t)=0 dt di X or, (t)+4i X (t)=0 dt -t If i X (t)=K1 +K 2 e , R = R2 // R3 = 4 i0 (t ) = iX (t ) R3 2iX = 3 R2 + R3 - K 2e -t + 4 K1 + 4 K 2 e = 0 -t yields, =0.25 sec. and K1 =0 i X (t)=K 2e-4t and i X (0+ )=K 2 =i L (0+ )=3A i 0 (t)=2e-4t A, t 0 i 0 (t)=-2 A, < 0 Problem 6.21 Use the differential equation approach to find i0(t) for t > 0 in the circuit in Fig. P.21 and plot the response including the time interval just prior opening the switch. t=0 50 F 3 k 4 k 12 k i0(t) 12 V 4 k 12 k 8 k Suggested Solution VC(t) + t=0 - - VC(t) + t=0 C R1 4 k i0(t) R2 R3 12 V R5 R6 R1 RA iX(t) 12 V RB R4 R1 = 3 k R 2 =R 5 =4 k R 3 =R 4 =12 k R 6 =8 k C = 0.05mF RA = 3k RB = 6k For t < 0: VC (t)= i X (0- )= For t > 0 : -12R A = -4V = VC (0- ) = VC (0+ ) (R A +R B ) i (0-) R3 VC (0-) -4 = mA i0 (0-) = X = -1 mA RA 3 ( R2 + R3 ) C dVC (t) VC (t) dVC 10 + =0 + =0 dt R 1 + RA dt 3 By KCL: For t > 0 C R1 RA iX(t) If VC (t ) = K1 + K 2 e , then = 0.3 S and K1 = 0 - t VC (0 ) = K 2 = -4V so, VC (t)=-4e + - 10 t 3 V 10 t - -VC (t) i (t)R 3 and i0 (t)= X But i X (t)= i 0 (t)=0.5e mA, t > 0 R 1 + RA R 2 + R3 =1 mA, t < 0 Problem 6.22 Use the differential equation approach to find i0(t) for t > 0 in the circuit in Fig. P.22 and plot the response including the time interval just prior to opening the switch. 8 k 4 k 4 k t=0 4 k 6 k 100 F 12V 4 k Suggested Solution i0(t) R4 R5 R4 i0(t) R1 + VC - t=0 R6 t=0 RA C + VC(t) _ 12V R8 R3 C R2 12V R1 = 8 k R 2 =R 4 =R 5 =R 6 =4 k R 3 =6 k C=0.1 mF R A =4 k R B =8 k For t < 0: VC (0- )=VC (0+ )= 12R A =6V i 0 (0- )=-1.5 mA (R A +R 4 ) R4 i0(t) t=0 RA C + VC(t) _ R8 KCL : C dVC VC VC dV 10 VC = 0 + + =0 C + dt RA R4 + RB dt 3 -t If VC (t)=K1 +K 2 e , =0.3s and K1 =0 -10t 3 VC (0+ )=6=K 2 So, i 0 (t)=0.5e mA, t> 0 =-1.5 mA, t < 0 Problem 6.23 Use the differential equation approach to find i0(t) for t > 0 in the circuit in Fig. P.23 and plot the response including the time interval just prior to opening the switch. 12V 2 2H 2 24V 4 t=0 i0(t) Suggested Solution 12V 2 2H For t =0- Use superposition iL= i0= 2 24 2 12 2 - =1.2A 2+(4//2) 2+4 2+(4//2) 2+ 4 24 4 12 + = 8.4A 2+(4//2) 2+4 2 +(4//2) 24V iL 4 t=0 i0(t) For t=0+ : i0 =-L=-L(0- )=1.2A 12=6 i 0 (t)+2 di0 (t) so, i0 (t)=K1 +K 2 e dt -t 12 1 -2K 2 Now, 12=6K1 +6K 2 e + = 2 and = 3 e K1 = 6 i 0 (0+ )=-1.2=K1 +K 2 K 2 =-3.2 t<0 -1.2A i0(t)= -3t 2-3.2e A t > 0 -t -t Problem 6.24 Use the differential equation approach to find V0(t) for t > 0 in the circuit in Fig. P.24 and plot the response including the time interval just prior to opening the switch. 12V + 12 k 24V t=0 12 k 12 k _ Suggested Solution + VC (t) C R2 - For t < 0 R2 VC (0- ) = VC (0+ ) = 12 = 6V R2 + R3 V0 (0- ) = 12 - VC (0- ) = 6V For t > 0 KCL: V0 V1+24 C d + 1(V1 + 24) = R3 dt dV1 V1 V or, + +20= 0 dt 1.2 1.2 V V Also, 1 + 0 =0 V1 =-V0 R1 R 3 + 12V 24V R3 R1 V0(t) - All R=12 k C=0.1 mF + VC (t) C R2 - V1+24 24V 24V R3 R1 + V0(t) - Now, dV0 V0 + = 20 dt 0.6 -t Assume -t V0 (t)=K1 +K 2 e -t substitute, K K e -K 2e + 1 + 2 =20 =0.65 and K1 =12 0.6 0.6 24-6 + V0 (0 )= =9V=K1 +K 2 K 2 =-3 2 -5t V0 (t)= 12-3e 3 V t > 0 V0 (t)=6 t < 0 Problem 6.25 Find VC(t) for t > 0 in the network in Fig. 6.25 using the step-by-step method. 2 k t=0 2 k 12V 2 k + V0(t) 100 F Suggested Solution 2 k t=0 2 k 12V 2 k + V0(t) 100 F Problem 6.26 Use the step-by-step method to find i0(t) for t > 0 in the circuit in Fig. P.26. 2 k 6 2H 12V i0(t) t=0 Suggested Solution iL (0- ) = 2 k 6 12V iL(0-) 12 = 6A 2 t=0 6 6A i0(0+)=-6A i0(0+) -t R 6 = = 3 sec. i 0 (t)=-6e 3 A, t >0 L L i- Problem 6.27 Find i0(t) for t > 0 in the network in Fig. P6.27 using the step-by-step method. t=0 4 6A 12 6 i0(t) t=0 1 Suggested Solution 4 6A 12 6 i0(t) t=0 iL(0-) 6A 4 iL(0-)=3A 4 4 6 4 i0(0+) iL(0-)=3A i0(0+)=-3A i0()=0 source is out i = R = 10 = 10 sec i 0 (t)=-3e-10t A, t>0 Problem 6.28 Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P6.28. t=0 200 F 6 k 6 k 4 mA 6 k i0(t) Suggested Solution - VC(0-) + t=0 6K 4K 6K VC (0- )= 12 4 =(6K//6K)=12V K i 0 (0+ )= 12 1 = A 12K K i0( )=0 Source is out of network -t 6K 6K i=RC=200 x 10-6 x 12 x 103 = 2.4 sec i0 (t ) = 1e 2.4 mA, t > 0 i0(0+) Problem 6.29 Use the step-by-step technique to find i0(t) for t > 0 in the network in Fig. P6.29. 2 k 4 k i0(t) 200 F 12 V t=0 6 k Suggested Solution R1 R2 i0(t) 12 V C + VC - R3 t=0 For t=0 R 2 +R 3 VC (0- )=VC (0+ )=12 = 10V R 1 +R 2 +R 3 R1 = 2 k R2 = 4 k R3 = 6k C = 0.2mF For t=0+ i0 (0+ )= For t i0 (t ) = 12 = 2mA = K1 -t VC (0+ ) =2.5 mA R2 K1 +K 2 =2.5mA K 2 =0.5mA i 0 (t)=K1 +K 2 e i0 (t ) = 2 + 0.5e -3.75t mA =C [ R1//R 2 ] = 0.267 sec. Problem 6.30 Use the step-by-step method to find V0 (t) for t > 0 in the network in Fig. P6.30. t=0 6 mA 2 k 100 F 6 k + V0(t) - Suggested Solution t=0 6 mA R1 + VC - C R2 + V0(t) - For t=0V0 (0- ) = VC (0+ ) = 6m( R1 // R2 ) = 9V For t=0+ R 1 =2 k R 2 =6 k C=0.1 mF -t VC (0+ )=V0 (0+ )=9V=K1 +K 2 For t V0 ( )=0=K1 K 2 =9 V0 (t ) = K1 + K 2 e V0 (t ) = 9e V 5t - 3 =C R 2 = 0.6 sec Problem 6.31 Use the step-by-step method to find i0(t) for t > 0 in the circuit in Fig. P6.31. 3 k 3 k 200 F + 12V t=0 6 k 2 k V0(t) i0(t) - Suggested Solution R1 R2 C + VC - 12V t=0 R3 + V0(t) - i0(t) R4 R1=R2=3K R3=6k R4=2K C=0.2 mF For t=0 R3 + VC (0- )=12 = 6V = TC (0 ) R 1 +R 2 +R 3 For t = 0+ 6V R2 R3 i0(t) 2K R1=R2=3K R3=6K R4=2K C=0.2 mF R4 For t 6 =1.5 mA 4K 3 i0=is = 0.5mA 3+6 is = i0 ( )=0=K1 =C Req Req=(R 2 //R 3 )+R 4 =0.8 s So, K1 +K 2 =0.5 mA=K 2 i0 (t)=0.5e -t mA 0.8 Problem 6.32 Find V0(t) for t > 0 in the network in Fig. P6.32 using the step-by-step technique. 3 k 3 k 200 F + 12V t=0 i0(t) 6 k Suggested Solution R1 R2 + VC - C 12V t=0 R3 R4 + V0(t) - For t=012 R 3 VC = =6V=VC (0+ ) R 1 +R 2 +R3 For t=0+ R1 = R2 = 3K R3 = 6 K R4 = 2 K C = 0.2mF -t VC (0+ )=6V R4 = -3V V0 (0+ )=-VC R 4 +(R 2 //R 3 ) -t V0 (t)=K1 +K 2 e = C [ R4 + ( R2 // R3 )] = 0.8 s V0 (t)=-3e 0.8V Problem 6.33 Find i0(t) for t > 0 in the network in Fig. P6.33 using the step-by-step method. t=0 5 k i0(t) 12V 5 k 10 mH Suggested Solution R1 i0(t) t=0 12V R2 L 10 mH R1=R2=5 k L=10 mH For t=0i0(0-)=i0(0+)=0A For t=0+ : For t=: i0 (0+ )=0=K1 +K 2 i0 ( )=12=2.4 mA=K1 K 2 =-2.4 mA L L = =4 s = Req R 1//R 2 i0 (t)=2.4 1-e-2.5x10 ( 5t ) mA Problem 6.34 Find V0(t) for t > 0 in the network in Fig. P6.34 using the step-by-step method. 4 2 t=0 1/2H + 3 12V - Suggested Solution R3 R1 is t=0 R2 12V R4 + V0(t) - iL L For t=0- : is = For t=0+ : V0 = iR 12 i L = s 2 =1A R1 +(R 2 //R 4 ) R 2 + R4 -i L (0+)R3 24V R4 = =K1 +K 2 R 3 +R 2 +R 4 13 24V =1.85V 13 For t : V0 =0=K1 K 2 = = R1 =2 R 2 =3 R 3 =4 R 4 =6 1 L= 2 L =0.18 sec {R 3 //(R 2 +R 4 )} V0 (t)=1.85e-5.54tV Problem 6.35 Use the step-by-step technique to find V0(t) for t > 0 in the network in Fig. P.35. 2 4 + 2H t=0 2 V0(t) 6V - Suggested Solution R2 R1=R3=2 R2=4 L=2H t=0 R3 + V0(t) - R1 L iL 6V For t = 0i L (0- )=i L (0+ )= -t 6 =3A R1 V0 (t)=K1 +K 2 e For t = 0+ i L (0+ )=3A, V0 (0+ )=-i L (0+ )R 3 =-6V=K1 +K 2 = L L 2 1 = = = sec Re q R1 + R2 + R3 8 4 V0 (t)=-6e-4t V Problem 6.36 Use the step-by-step technique to find i0(t) for t > 0 in the networking in Fig. P6.36. 12V 200 F 2 k t=0 2 k 4 k 2 k 2 k i0(t) Suggested Solution R1 6V R4 R1=R2=R3=R4=2 k C=0.2 mF For t=0t=0 R3 C + VC(t) - R2 i0(t) VC (0- ) = 6R3 = 3V R2 + R3 For t i 0 ()=K1 = so, K 2 =-0.25 mA 6 =2 mA 3k = C Re q = C 2k + ( i0 (t ) = 2 - e -3t mA 4 1k 1 ) = sec 2k 3 For t=0+ V (0+)=V (0-)=3V C C 6V 2k 2k 1k 3V i0(t) Superposition: i0 = 6 2 3 1 + 1) = 1.75mA 2k 2 + 2 8 k 1 + 2 3 + i0 (0 ) = K1 + K 2 = 1.75mA Problem 6.37 Find i0(t) for t > 0 in the networking in Fig P6.37 using the step-by-step method. 12V 200 F 2 k t=o 2 k 2 k 2 k 4 k i0(t) Suggested Solution 12V C R1 = R2 = R4 = R5 = 2 k R3 = 4 k , C = 0.2 mF -t For t=0VC (0- ) 12R 4 =4V R 1 +R 2 +R 4 + VC - R1 t=o 2 k 2 k i 0 (t)=K1 +K 2e R2 R3 i0(t) For t > 0 12V R=R3//(R1+R2=2 K 4V R R4 i0 RS 6M R i0 R9 2M RS i0 (0+ ) = K1 + K 2 = For t : 8M [ R // RS ] 8 = mA [ R // RS ] + R4 3 12 -1 =3 mA K 2 = mA R+R 4 3 i0 ( )=K1 = =CReq=C {R S + ( R // R4 )} = 0.6 sec i0 (t) = 3 - e - t 0.6 mA 3 Problem 6.38 Use the step-by-step technique to find i0(t) for t > 0 in the network in Fig. P6.38. t=0 4 k 3 k 150 F 2 k + 36V V0(t) i0(t) 3 k 3 k 12V Suggested Solution R1 iS + 36V R2 t=0 R3 C R5 + VC 12V R4 i0(t) R1 = 4k R 5 =2 k C=0.15 mF R 2 =R 3 =R 4 =3 k i 0 (t)=K1 +K 2 e -t For t = 0- iS = 36 = 6A R1 + R2 //( R3 + R4 ) i0 = iS R 2 =2 mA VC =i0 R 4 -12=-6V R 2 +R 3 +R 4 For t=0+ VC (0+ ) = VC (0- ) = -6V , so, i 0 (0+ )=K1 +K L =1 mA For t= -t 12 + VC (0+ ) [ R4 //( R2 + R3 )] R4i0 = = 3V R4 //( R2 + R3 ) + R5 i0 ( )=0=K1 K 2 =1 mA =CReq=C {R 5 + [ R4 //( R2 + R3 )]} = 0.6 sec i0 (t ) = 1 e 0.6 mA Problem 6.39 Find V0(t) for t > 0 in the circuit in Fig. P6.38 using the step-by-step method. Suggested Solution Problem 6.40 Find i0(t) for t > 0 in the network in Fig. P6.40 using the step-by-step method. t=0 5 k 2 k 10 mA 4 k 4 k 10 mH 2 k i0(t) Suggested Solution t=0 R1 R4 t=0 Is R2 R3 L R5 Ieq i0(t) R L i0 (t) R5 R1 = 5 k R2 = R3 = 4 k I5 = 10 mA R = R4 + ( R1//R2//R3) = 3.43 k L = 10mH R4 = R5 = 2 k Ieq = 5 May 2001 For t = 0- : i0 (0- ) = i0 (0+ ) = Ieq = 5 mA = k1 + k2 For t = 0+ : i0 (0+ ) = Ieq = 5 mA = k1 + k2 For t = : i0 () = 5 mA = k1 k2 i0 (t) = 5mA Problem 6.41 Find V0(t) for t > 0 in the network in Fig. P6.41 using the step-by-step method. 2 k t=0 50 F 24V 2 k + 4 k V0(t) 4 k 12V Suggested Solution t=0 v + 24 + vc + + 3/2 v0 v c R1 24V R4 R2 v0 (t) - R3 12V R2 = R 3 = 4 k , R1 = R 4 = 2 k , c = 50 F For t : v0 0 = k1 k2 = 4.37V = c R1 + R2 + ( R3 // R4 ) = 0.37 sec For t = 0- : v 0 = 0V , v c = 24V For t = 0+ : v = 6.55V , v v v + 24 - 12 + + =0 R1 + R2 R3 R4 2 v0 = v = 4.37V = k1 + k2 3 vo (t ) = 4.37e - t / 0.37V Problem 6.42 Find V0(t) for t > 0 in the network in Fig. P6.42 using the step-by-step method. 1 k 9V t=0 6 k 12 k + 50 F 2 k 2 k V0(t) - Suggested Solution R1 9V + v - + c vc t=0 R2 + v2 - R v0 R3 + - R1 =1 k , R 2 = R 3 = k , c = 50 F R= 6 k// 12 k= 4k For t : v 0 ( ) 0 = k1 k 2 = - 3.6V se c = c R1 + ( R3 // ( R + R2 ) ) = 8 R2 For t = 0 - : v c = 9 - v2 , v 2 = v R2 + R3 R // ( R2 + R3 ) v = 9 = 6V R // ( R2 + R3 ) + R1 So , v 2 = 3V , v c ( o - ) = 6V v = , 2 1 v 0 = - 3.6 e -8 tV Problem 6.43 Find i0(t) for t > 0 in the circuit in Fig. P6.43 using the step-by-step method. t=0 5 k i0(t) 12V 5 k 10 mH Suggested Solution R1 R1 = 4 k C 24V R2 = 2 k R3 = 10 k C = 200 F + Vc R2 t=0 R3 i0 4 So, k 2 = - mA 3 = c R1 // ( R2 + R3 ) = 0.6 se c 24 For t = 0- : i 0 = = 4mA, v c = i0 R2 = 8V R1 + R2 For t : i 0 0 = k1 , Fort = 0+ : i 0 = vc (0+ ) - 24 8 - 24 4 = = - mA 12k 3 R2 + R3 i 0 (t ) = -1.33e - t / 0.6 mA 4 io (0+ ) = - mA = k1 + k2 3 Problem 6.44 Find V0(t) for t > 0 in the network in Fig. P6.44 using the step-by-step method. 4 k t=0 2 k 2 k 4 k V0(t) 12V 100 F Suggested Solution for t=0 R4 vo = 12 R4 + ( R1 ( R2 + R3 )) R2 v2 = (12 - vo ) = 2V R2 + R3 vc = (12 - v2 ) = 10V for t=0+ (12 - vo ) vc (0+ ) - vo vo + = R1 R3 R4 vc (0+ ) = vc (0- ) = 10V 12 - vo - 2vo + 20 = vo vo (0+ ) = 8V = k1 + k2 for t : R4 vo = 12 = 6V = k1 k2 = 2V R4 + R1 vo (t ) = 6 + 2e -2.5t v Problem 6.45 Use the step-by-step method to find i0(t) for t > 0 in the network in Fig. P6. 45. 6 k 6 k i0(t) 6 k 12V t=0 50 F Suggested Solution for t=0 R2 vc = 12 = 6V R2 + R3 for t=0+ by superposition io = v (0+ ) -12 + c = -1 + 0.5 = -0.5mA R1 + R2 R1 + R2 i o (0+ ) = -0.5mA = k1 + k2 for t : i o () = 0 = k1 k2 = -0.5mA i o (t ) = -0.5e -5t mA Problem 6.46 Find V0(t) for t > 0 in the circuit in Fig. P6.46 using the step-by-step method. 4 k 4 k V0(t) + 8 k 12V 50 F Suggested Solution for t=0 R3 vo = 12 = 6V R3 + R1 vc = -vo = -6V = vc (0+ ) for t=0+ vo = -vc = -6V vo (0+ ) = 6 = k1 + k2 for t : vo () = 0 = k1 k2 = 6 vo (t ) = 6e 0.15 v -t Problem 6.47 Use the step-by-step technique to find V0(t) for t > 0 in the circuit in Fig. P6.47. 12V 6 12 t=0 + 2H 2 4 V0(t) - Suggested Solution for t=0is = iL = 12 = 3A R1 + ( R2 R) is R2 = 1.5 A = iL (0+ ) R2 + R L 1 = = s R2 + R 4 for t=0+ vo = -iL R2 = -6 = k1 + k2 for t : vo = 0 = k1 k2 = 6V vo (t ) = -6e -4t v Problem 6.48 Find V0(t) for t > 0 in the circuit in Fig. P6.48 using the step-by-step method. 2 t=0 3 + V0(t) 5H 24V 6 12V 3 1 Suggested Solution for t=0io = 12 = 3.2 A R3 + ( R4 R5 ) R5 iL = io = 2.4 A R4 + R5 use superposition, for t=0+ R3 R5 + vo = 12 + iL (0 ) R3 = 9.6 R3 + R5 R4 + R5 vo (0+ ) = 9.6 = k1 + k2 for t : R3 vo = 12 = 9.6 = k1 R3 + ( R4 R5 ) k2 = 0 vo = 9.6V Problem 6.49 Use the step-by-step method to find V0 (t) for t > 0 in the network in Fig. P6.49. 6 k t=0 6 F 2 k + 4 k 12V V0(t) 6 F 6 F - Suggested Solution vo (t ) = k1 + k2 e -t vo (0+ ) = 6V = RAC = 6k ( ) = 0.0245 vo (t ) = 4e 0.024 v t>0 -t 4 k Problem 6.50 Find V0(t) for t > 0 in the network in Fig. P6.50 using the step-by-step method. 12V 3 mH 6 k + 6 k V0(t) - 3 mH t=0 6 k 3 mH Suggested Solution vo (t ) = k1 + k2 e -t iL (0- ) = 12 12 4 + = A 6k 6k k -2 6k 6k = -4V k 6k + 12 L 4.5*10-6 = = Rth 4k vc (0+ ) = vo (t) = -4e -t Problem 6.51 Use the step-by-step method to find i0(t) for t > 0 in the network in Fig. P6.51 12V t=0 2 k 600 F 4 k 1 mA i0(t) 2 k 300 F Suggested Solution 12V t=0 2 k 600 F 4 k 1 mA i0(t) 2 k 300 F R1 = 4k , R 2 = R3 = 2k c = 200 + For t = 0+ , Use Superposition i012 v = -12 = -6mA R2 8 = 4mA R2 1 1 1 + 600 300 = 400 F i01mA = 1m(0) = 0 i0 v (0+ ) = c i0 (0+ ) = i0 = -2mA = K1 + K 2 t = 0- , Use Superposition vc = 12( R2 + R3 R1 ) + 1m( ) R1 + R2 + R3 R1 + R2 + R3 8 15 vc = 8v = vc (0+ ) = c[ R1 || R2 ] = for t i0 = 1m( i0 () = 2 R1 ) = mA 3 R1 + R2 2 mA = K1 3 8 K 2 = - mA 3 2 8 i0 (t ) = - e -15t / 8 mA 3 3 Problem 6.52 Use the step-by-step method to find i0(t) for t > 0 in the network in Fig. P6.52 4 t=0 4 6 i0(t) 12 12V 3H 3A 2H Suggested Solution 4 t=0 4 6 i0(t) 12 12V 3H 3A 2H R1 = 4 = R2 = 4 R3 = 6 R4 = 12 L= 1 1 1 + 2 3 = 1.2 H = L 2 = R2 || R3 + R1 || R3 9 4 2 -9t / 2 A - e 3 15 i0 = For t = 0- iL = For t = 0+ i0 = i2 (0+ )[ For t i0 = i0 = R2 || R4 24 [ ] R1 + ( R2 || R3 || R4 ) ( R2 || R4 ) + R3 4 -2 = K1 K 2 = 3 15 R2 ] = 1.2 A = K1 + K 2 R2 + R3 24 = 3 A = iL (0+ ) R1 + R2 Problem 6.53 Find i0(t) for t > 0 in the circuit in Fig. P6.53 using the step-by-step method. 12 6 k 4 k 100 F 3 k 300 F 3 k i0(t) t=0 12 V 100 F Suggested Solution 12 6 k 4 k 100 F 3 k 300 F 3 k i0(t) t=0 12 V 100 F R1 = 12k R 2 = R 4 = 3k R3 = 4k R5 = 6k c = 300 + 100 + 100 = 500 F for t = 0- i4 = 0 vc = -12v for t = 0+ vc = -12v i4 = 0 i0 = for t i4 = 0 vc = -12v i0 = -12 = -1.5mA R3 + ( R1 || R5 ) -12 = -1.5mA R3 + ( R1 || R5 ) i0 = -1.5mA Problem 6.54 Find i0(t) for t > 0 in the circuit in Fig. P6.54 using the step-by-step technique. 3 k 6 k 1 k 2 k 200 F 12V i0(t) t=0 6V 50 F 150 F 1 k Suggested Solution R1 = 3k , R2 = 6k , R3 = 2k = R c = 100 F for t i 0 = 12[ -6[ R3 || R 1 ]( ) R1 + ( R2 || R3 || R) R3 || R + R2 R1 || R3 1 ]( ) R1 + ( R1 || R || R2 ) R1 || R3 + R 2 1 i0 = mA = k1 9 1 k2 = mA 18 for t = 0- vc = 12[ R1 || R2 R || R ] - 6[ 1 2 ] R || R2 + R1 R1 || R2 + R vc = 1v = vc (0+ ) for t = 0+ v 1 i0 = c = mA R2 6 = c[ R1 || R2 || R3 || R] = i0 = 1 1 -15t + e mA 9 18 1 sec 15 3 k 6 k 1 k 2 k 200 F 12V i0(t) t=0 6V 50 F 150 F 1 k Problem 6.55 The differential equation that describes the current i0 (t) in a network is d 2i 0 (t) di (t ) + 6 0 +8i 0 (t ) = 0 2 dt dt Find (a) The characteristic equation of the network. (b) The network's natural frequencies. (c) The expression for i0 (t) Suggested Solution d 2i 0 (t) di (t ) + 6 0 +8i0 (t ) = 0 2 dt dt (a) The characteristic equation is S2+6S+8=0 (b) The natural frequencies are S=-2 and S=-4 (c) i0 (t)=K1e-2t+K2e-4t Problem 6.56 The terminal current in a network is described by the equation d 2i 0 (t) di (t ) + 10 0 +25i 0 (t ) = 0 2 dt dt Find (a) The characteristic equation of the network. (b) The network's natural frequency. (c) The equation for i0 (t). Suggested Solution i0(t): d 2i 0 (t) di (t ) + 10 0 +25i 0 (t ) = 0 2 dt dt (a) The characteristic equation is S2+10S+25=0 (b) The natural frequency is S=-5 (c) i0(t)=K1e-5t+K2+e-5t Problem 6.57 The voltage v1 (t) in a network is defined by the equation d 2i 0 (t) di (t ) + 10 0 +25i 0 (t ) = 0 2 dt dt Find (a) The characteristic equation of the network. (b) The circuit's natural frequencies. (c) The expression for v1 (t). Suggested Solution i0(t): d 2i 0 (t) di (t ) + 10 0 +25i 0 (t ) = 0 2 dt dt (a) The characteristic equation is S2+2S+5=0. (b) The natural frequencies are S=-1+2, S=-1-2 (c) v1 (t)=K1e-t cos 2t+K2 e-t sin 2t. Problem 6.58 The output voltage of a circuit is described by the differential equation d 2 v 0 (t) dv (t ) + 6 0 +10v 0 (t ) = 0 2 dt dt Find (a) The characteristic equation for the circuit. (b) The networks natural frequencies. (c) The equation for v0 (t). Suggested Solution d 2 v 0 (t) dv (t ) + 6 0 +10v 0 (t ) = 0 2 dt dt (a) The characteristic equation S2+6S+10=0. (b) Natural frequencies are S=-3 +. S=-3-8. (c) v0 (t)=K1e-3t cos t+K2e-3t sin t Problem 6.59 The parameters for a parallel RLC circuit are R = 1, L = 1/5 H, and C = F. Determine the type of damping exhibited by the circuit. Suggested Solution What type of damping occurs in this network? 1 .2H Characteristic Equation: 1S 1 + = 0 = S 2 + 4 S + 20 RC LC The Roots are S=-14 which are complex conjugates. The network is underdamped. S2 + Problem 6.60 A series RLC circuit contains a resistor R = 2 and a capacitor C = 1/x F. Select the value of the inductor so that the circuit is critically damped. Suggested Solution What inductance cause critical damping? 2 L 1/8F If S2 + R 1 S+ =0 L LC -2 4 32 2- 2 8 L L S2 + S+ =0 with roots S= L L L 2 4 32 for critical damping 2 - =0. L L 1 L= H 8 Problem 6.61 For the underdamped circuit shown in Fig. 6.61 determine the voltage v(t) if the initial conditions on the storage elements are iL(0)=1 A and v(0) = 10 V. iL (0) + - - Suggested Solution iL (0) + - - Find V(t) if i L (0)=1A and v c (0)=10v d 2 v(t ) 1 dv(t ) 1 + + v(t ) = 0 dt 2 RC dt LC The characteristic equation is s 2 + 8s + 20 = 0 s = -4 2 j so v(t ) = k1e -4t cos 2t + k2 e -4t sin 2t at t = 0, vc = 10 v(t ) = k1 = 10 then dv(t ) = -2k1e -4t sin 2t - 4k1e -4t cos 2t + 2k2 e-4t cos 2t - 4k2 e-4t sin 2t dt at t=0 dv(t ) = -4k1 + 2k2 = -40 + 2k2 dt also cdv(t ) v(t ) + + iL (t ) = 0 dt R at t=0 1 -10 dv(t ) 1 -v(t ) = ( - iL (t )) = ( - 1) dt c R c R if -40 + 2k2 = -120 finally v(t ) = 10e -4t cos 2t - 40e -4t sin 2t Problem 6.62 Given the circuit and the initial conditions of Problem 6.61, determine the current through the inductor. Suggested Solution Find i L (t) for the circuit The network is Underdamped i L (t ) = k3e -4t cos 2t + k4 e -4t sin 2t at t=0 iL = k3 = 1A diL (t ) = -2k3e -4t sin 2t - 4k3e-4t cos 2t + 2k4 e-4t cos 2t - 4k4 e-4t sin 2t dt at t=0 diL (t ) = - 4 k 3 + 2 k 4 = -4 + 2 k 4 dt sin ce LdiL (t ) diL (t ) v(0) = =5 , v(t ) = dt dt L so 1 9 (5 + 4 x3) = 2 2 9 iL (t ) = e -4t cos 2t + e-4t sin 2t 2 k4 = Problem 6.63 Find vC(t) for t > 0 in the circuit in Fig. P6.63 if vC(0) = 0. 1 k 100 mH t=0 12V 1 F + vc (t) - Suggested Solution 1 k 100 mH t=0 12V 1 F + vc (t) - R = 1k , c = 1 F , L = 100mH for t = 0- v0 = 0 iL = 0 for t = 0+ v0 = 0, iL = 0 for t>0 series RLC with constant forcing function char _ eq 1 R =0 s+ L LC s 2 + 104 s + 107 = 0 s2 + Roots _ are -r r s1 = -1127 , s2 = -8873 overdamped s s solution is i2 (t ) = k1e s1t + k2 e s2t + k3 iL (0+ ) = 0 = k1 + k2 + k3 = 0 iL () = 0 = k3 k2 = -k1 iL (t ) = k1 (e s1t - e s2t ) KVL 12 = RiL (t ) + L at t=0 12 = R(0) + L[ s1e0 - s2 e0 ]k1 + 0 k1 = 15.5mA now iL (t ) = 15.5(e s1t - e s2t )mA r0 (t ) = 1 15.5 1 e s1t e s2t [ - ]+ k iL (t ) = 100 10-6 s1 c s2 diL (t ) + v0 (t ) dt v0 () = 12 = k v0 (t ) = 12 - 13.75e -1127 t + 1.75e -8875t v t > 0 v0 (t ) = 0v t 0 Problem 6.64 Find v0(t) for t > 0 in the circuit in Fig. P6.64 and plot the response including the time interval just prior to moving the switch. 4 k 1 mH + 10V t=0 6.25 nF 1 k v0 (t) - Suggested Solution R1 1 mH + 10V t=0 Vc R2 v0 (t) - R1 = 4k , R2 = 1k , L = 1mH c = 6.25nF for t = 0- v0 = 0 iL = 0, vc = 10v for t = 0+ v0 = 0, iL = 0, vc = 10v for t>0 series RLC with constant forcing function char _ eq R2 1 = 0 s 2 + 106 s + 1.6 x1011 = 0 s+ L LC roots _ are s2 + s1 , s2 = -2 x105 and - 8 x105 r / s overdamped v0 (t ) = k1e s1t + k2 e s2t at t = 0+ v0 (0+ ) = 0 = k1 + k2 k2 = - k2 also at t = 0+ vc (0+ ) = L or dv0 / dt |t =0 + v0 (0+ ) = 10v R2 10-6 k1[ s1 - s2 ] = k1 (0.6) = 10 k1 = 16.67 v0 (t ) = 0v t<0 v0 (t ) = 16.67[e -2 x10 t - e-8 x10 t ]v t 0 5 5 Problem 6.65 Find vC(t) for t > 0 in the circuit in Fig. P6.65. + 1A 8 0.04F - t=0 vc (t) iL (t) 1H Suggested Solution + 1A 8 0.04F - t=0 vc (t) iL (t) 1H R = 8, c = 40mF , L = 1H for t = 0- iL = 0, iR = 1A, vc = 8v, ic = 0 for t = 0+ iL = 0, iR = 1A, vc = 8v, ic = 0 for t= iL = 1, vc = 0 Parallel RLC char eq is 1 1 =0 s+ RC LC roots s1 , s2 = 1.56 j 4.75 s2 + system is underdamped and vc (t ) = e t ( A cos t + B sin t ) + k where = -1.56 = 4.75 vc (0+ ) = 8 = A + K and vc () = 0 = k A=8 by KCL at t = 0+ vc (0+ ) dv (t ) dv (t ) + iL (0+ ) + c c |t =0+ = 1 + 0 + c c |t =0 1= R dt dt so - A dvc (t ) = 2.63 |t =0 = 0 = B + A B = dt vc (t ) = e -1.56t [8cos(4.75t ) + 2.63sin(4.75t )]v t>0 vc (t ) = 8v t 0 Problem 6.66 Find iL(t) for t > o in the circuit in Fig. 6.65. Suggested Solution From problem 6.65 for t>0 vc (t ) = e -1.56t [8cos t + 2.63sin t ] where = 4.75r / s KCL : v (t ) dv (t ) 1 = c + c c + iL (t ) R dt -1.56 t iL (t ) = 1 - e [cos t + 0.329sin t ] + [1.52sin t - 0.5cos t ]e-1.56t + 1.56e -1.56t [8cos t + 2.63sin t ] iL (t ) = 1 + e -1.56t [10.98cos(4.75t ) + 5.29sin(4.75t )] A Problem 6.67 Given the circuit in Fig. 6.67, find the equation for i(t), t > 0. i (t) t=0 3 12V 3/8F 1/3H Suggested Solution i (t) t=0 3 12V 3/8F 1/3H for t<0 iL = 12 / 3 = 4 A, vc = 0v for t = 0+ iL = 4 A, vc = 0v for t>0 Series RLC circuit R 1 s2 + s + = 0 s 2 + 9 s + s = 0 ( s + 8)( s + 9) = 0 L LC since roots are real and unequal, overdamped and i (t ) = k1e - t + k2 e-8t i (0) = 4 = k1 + k2 KCL : vc (0+ ) = Ri (0+ ) + L so k1 + k2 = 4 k1 + 8k2 = 36 -4 32 , k2 = 7 7 32 4 i (t ) = e -8t - e- t A for t>0 7 7 k1 = di (t ) 1 |t =0 0 = 12 + (-k1 - 3k2 ) dt 3 Problem 6.68 In the circuit shown in Fig. P6.68, find v(t), t > 0. t=0 12/5 H + 8V 2 1/12 F v (t) - Suggested Solution t=0 12/5 H + 8V 2 1/12 F v (t) - for t = 0- iL = 0, vc = v = 0v for t = 0+ iL = 0, v = 0 for t>0 Parallel RLC circuit, char eq is s 1 s2 + + = s 2 + 6s + 5 = 0 RC LC roots _ are s1 = -1, s2 = -5 overdamped v(t ) = k1e -t + k2 e -5t + k3 v(0+ ) = 0 = k1 + k2 + k3 v() = k3 = 8v KCL at t = 0+ iL (0+ ) = v(0+ ) dv(t ) dv(t ) +c |t =0+ | +=0 R dt dt t =0 or - k1 - 5k2 = 0 k1 + k2 = -8 k1 = -10, k2 = 2 v(t ) = 8 + 2e -5t - 10e- t v Problem 6.69 Find i0(t) for > 0 in the circuit in Fig. P6.69 and plot the response including the time interval just prior to opening the switch. iL (t) t=0 1A 1 1H 2/5 F i0 (t) 5 iL (t ) 2 Suggested Solution iL (t) t=0 1A 1 1H 2/5 F i0 (t) V R1 = 1, R2 = 5, L = 1H , c = for t = 0- 2 F 3 iL = 0, vc = 0, i0 = 0, iR1 = iR2 = 0, ic = 0 for t = 0+ iL = 0, vc = 0, i0 = 0, iR1 = 0, ic = 1A for t vc 0, i0 0 for t>0 KCL : 1= v v i + + ic + L R1 R2 2 dv 1 vdt + dt 2 L 1 = v(G1 + G2 ) + c i0 = so, v R2 di0 d 2i0 3di0 + 12.5 i0 dt 2 + 25 = 30i0 + 10 + 1.25 = 0 dt dt dt char _ q s 2 + 3s + 1.25 = 0 roots _ are s1 = -0.5, s2 = -2.5 overdamped i0 (t ) = k1e - t / 2 + k2 e-6t / 2 + k3 i0 () = 0 = k3 i0 (0+ ) = 0 = K1 + k2 k2 = - k1 ic (0+ ) = 1 = k1 ( so 1 A 4 1 i0 (t ) = (e - t / 2 - e -5t / 2 ) A 4 k1 = -1 5 + 2 2 Problem 6.70 Find v0(t) for t > 0 in the circuit in Fig. P6.70 and plot the response including the time interval just prior to closing the switch. 8 1/8 H t=0 1/25 F 1/8 H + 1.5A 4 6 1 v0 (t) - Suggested Solution R2 L1 C L2 + 1.5A R1 4 R3 6 R4 v0 (t) 1.5A R2 R1 R3 + Voc L + Vc + 4 2V t=0 i R4 Vo(t) - 1 1 R1 = 4, R2 = 8, R3 = 6, R4 = 1, L1 = L2 = H , c = F 8 25 R3 ] = 2.0v voc = 1.5[ R1 || ( R2 + R3 )][ R2 + R3 RTH = ( R1 + R2 ) || R3 = 4 For t = 0- iL = 0, vc = 0, v0 = 0 for t = 0+ iL = 0, vc = 0, v0 = 0, vL = 2 for t v0 = 0 L = L1 + L2 = for t>0 series RLC 1 R =0 s+ L LC s 2 + 20s + 100 = 0 roots : s2 + s1 = s2 = -10r / s critically _ damped v0 (t ) = k1e -10t + k2te-10t + k3 v0 () = 0 = k3 v0 (0+ ) = 0 = k1 so v0 = k2te -10t at , t = 0+ di (t ) |t =0 dt i (t ) = v0 (t ) / R4 = k2te-10t vL = 2 = L 2 = Lk2 k2 = 8 v0 = 8te -10t v 1 H 4 Problem 6.71 Find v0(t) for t > 0 in the circuit in Fig. P6.71 and plot the response including the time interval just prior to moving the switch. t=0 2.5 mH 8 k 1 k 2/15 nF 6 k 4 k 3 mA + 12V 6 k v0 (t) - Suggested Solution t=0 L R4 R1 2/15 nF R3 + R3 3 mA 12V R2 v0 (t) - R4 R5 3mA + Vx R3 + Voc - t=0 2.5 mH 1 k 2/15 nF 6 k RTH + 12V 6 k v0 (t) - Theve _ eq voc = vx R3 R3 + R4 vx = 3m[ Rs || ( R4 + R3 )] RTH = Rs || ( R4 + R3 ) = 4k , voc = 4v for t = 0- vc = -12v, v0 = 0, iL = 0 A for t = 0+ vc = -12v, iL = 0, v0 = 0, 4 + vL = vc + v0 vL = -16v for t v0 0 for t>0 series RLC circuit 1 R =0 s2 + s + L LC R = RTH + R2 s 2 + 4 x106 s + 3x1012 = 0 roots _ are s1 = -1x106 ands2 = -3x106 overdamped v0 (t ) = k1e s1t + k2 e s2t + k3 v0 () = 0 = k3 v0 (0+ ) = 0 = k1 + k2 k2 = - k1 now, v0 (t ) = k1 ( s s1t - e s2t ) vL (0+ ) = L diL |t =0 dt but iL (0+ ) = -v0 RL L k1 ( s1 - s2 ) = 16 k1 = 19.2 R2 v0 (t ) = 19.2[e -10 t - e -3 x10 t ]V , t > 0 v0 (t ) = 0v, t 0 6 6 Problem 6.72 Using the PSPICE Schematics editor, draw the circuit in Figure P6.72, and use the PROBE utility to plot vC(t) and determine the time constants for 0 < t < 1 ms and 1 ms < t < . Also, find the maximum voltage on the capacitor. R2 1 k C 0.33 F R3 t = 1ms 2 k + t=0 vc (t) R1 1 k I1 50 mA Suggested Solution R2 1 k C 0.33 F R3 t = 1ms 2 k + t=0 vc (t) R1 1 k I1 50 mA Problem 6.73 Using the PSPICE Schematics editor, draw the circuit in Figure P6.73, and use the PROBE utility to find the maximum values of vL(t), iC(t), and i(t). vL (t) i (t) 0.5 mH t = 1 s I1 6A R2 500 iC (t) R1 700 C 0.1 F Suggested Solution vL (t) i (t) 0.5 mH t = 1 s I1 6A R2 500 iC (t) R1 700 C 0.1 F Problem 6.74 Design a series RCL circuit with R>= 1Kohm that has characteristic equation. S^2 + 4*10^6S +4*10^14 = 0 Suggested Solution Series RLC where R 1K and characteristic eq. is, S2 + R 1 10-14 R S+ = 0 LC = ,L = L LC 4 4*107 choose R=40K. Now, L=1mH & C=2.5 F R=40K L=1mH & C=2.5 F Problem 6.75 Design a parallel RLC circuit with R>=1kohm that has the characteristic equation S^2 + 4*10^7S +3*10^14 = 0 Suggested Solution Parallel RLC where R 1K and characteristic eq. is, S2 + 1 10-14 1 S + = 0 LC = &C = 3 4*107 R RL LC choose R=2.5K. Now, L=333 H & C=10 F R=2.5K L=333 H & C=10 F Problem 6FE-1 In the circuit in Fig the switch, which has been closed for a long time, opens at t=0. Find the value of the capacitor voltage Vc(t) at t=2s. t=0 8K 12V 6K 6K 6K 100F Suggested Solution t=0 8K 12V 6K 12K 4V 12K FOR t<0 6K 12K=4K 4K VC (0) = 12 = 4V 4 K + 1K t > 0 SWITCH OPENS VC (t ) = 4e V -t = 100*10-6 *12*103 = 1.2SEC. VC (t ) = 4e -t 1.2 j AT t=2SEC VC (t ) = 4e1.2V = 0.76V -t Problem 6FE-2 In the network in fig, the switch closes at t=0. Find Vo(t) at t=1s. 12K 12V t=0 12K 4K + Vo(t) 100F - Suggested Solution 12K 6K 12V 6K 6V 6K 4K Vo(t) 6V VO (t ) = 6(1 - e ) -t = 10*103 *100*10-6 = 1SEC. VO (t ) = 6(1 - e )V AT t=1SEC, VO (t ) = 6(1 - e -1 ) = 3.79V -t Problem 6FE-3 Assume the switch, in the network in fig, has been closed for sometime. At t=0 the switch opens. Determine the time required for the capacitor voltage to decay to of its initially charged value. t=0 12V 12K + 6K Vc(t) 100F Suggested Solution t=0 12V 12K + Vc(t) 100F 6K 6K VC (0- ) = 12 = 4V 18K VC (t ) = 4e -t = 100*10-6 *6*103 = 0.6sec. VC (t ) = 4e V 1/2 OF INITIAL CHANGE VALUE = 2= 4e V -t 1 0.6 = e t = 0.42sec. 2 -t 0.6 -t 0.6 Problem 7.1 Given i(t)=5 cos(400t 120) A, determine the period of the current and the frequency in hertz. Suggested Solution 400 = = 63.6 Hz 2 2 1 T = = .016s f f = Problem 7.2 Determine the relative position of the two sine waves V1(t)=12 sin(377t-45) V2(t)=6 sin(377t+675) Suggested Solution V2(t) can be rewritten as V2(t)=6 sin(377t-45), so V1(t) and V2(t) are in phase. PHASE=0. Problem 7.3 Given the following currents: i1(t)=4 sin(377t 10) A i2(t)=-2 sin(377t 195) A i3(t)=-1 sin(377t 250) A Compute the phase angle between each pair of currents. Suggested Solution i1(t) leads i2(t) by 85. i2(t) leads i3(t) by 145. i1(t) by i3(t) by 60 Problem 7.4 Determine the phase angles by which v1(t) leads i1(t) and v1(t) leads i2(t), where: v1(t)=4 sin(377t + 25) i1(t)=0.05 cos(377t 10) i2(t)=-0.1 sin(377t + 75) Suggested Solution v1(t) leads i1(t) by 25-80=-55 v1(t) leads i2(t) by 25-255=-230 Problem 7.5 Calculate the current in the resistor if the voltage input is: i(t) (a) v1(t)=10 cos(377t+180) (b) v2(t)=12 sin(377t+45) + v(t) _ 2 Give the answers in both the time and frequency domains. Suggested Solution (a) i (t ) = v1 (t ) = 5cos(377t + 180) A = 5 180 A R v (t ) (b) i (t ) = 2 = 6sin(377t + 45) A = 6 cos(377t - 45) A = 6 -45 A R Problem 7.6 Calculate the current in the capacitor if the voltage input is: (a) v1(t)=16 cos(377t 22) (b) v2(t)=8 sin(377t + 64) i(t) + v(t) _ C=1326F Give the answers in both the time and frequency domains. Suggested Solution 1 1 1 = = -6 j C j (377)(1326 10 ) .5 j (a) v(t ) 16 -22 i (t ) = = = 8 68 A = 8cos(377t + 68) A 2 -90 Zc Zc = (b) i (t ) = v(t ) 8 64 = = 4 64 A = 4 cos(377t + 64) A Zc 2 -90 Problem 7.7 Calculate the current in the inductor if the voltage input is: (a) v1(t)=24 cos(377t + 12) (b) v2(t)=18 sin(377t 48) i(t) + v(t) _ L=10.61 mH Give the answers in both the time and frequency domains. Suggested Solution Z L = j L = j (377)(10.61 10-3 ) = 4 j (a) i (t ) = v(t ) 24 12 = = 6 -78 A ZL 4 90 v(t ) 18 42 = = 4.5 -48 A ZL 4 90 (b) i (t ) = Problem 7.8 Find the frequency domain impedance, Z, for the network shown. 1 z j2 -j1F Suggested Solution Z=1+j2-j1=1+j1 Problem 7.9 Find the frequency domain impedance, Z, for the network shown. z 3 j4 Suggested Solution Z = 3 || 4 j = 12 j 12 90 = = 2.4 37 = 1.92 + j1.44 3 + 4 j 2.4 53 Problem 7.10 Find the frequency domain impedance, Z, for the network shown. z 2 j2 -j4 Suggested Solution 1 1 1 1 1 = + - = - Z 2 j2 j4 2 4 4(2 + j ) = Z= 2 - j (2 - j )(2 + Z = 1.6 + j 0.8 1 1 1 j - = (2 - j) 2 4 4 8+ 4 j = = 1.6 + j 0.8 4 +1 j) Problem 7.11 Find the frequency domain impedance, Z, for the network shown. j5 z 3.75 -j5 Suggested Solution Z = 5 j + (-5 j || 3.75) = 5 j + 18.75 j 18.74 -90 =5j+ = 5 j + 3 -36.9 = 2.4 + j 3.2 = 4 53 3.75 - 5 j 6.25 -53.1 Problem 7.12 Find the frequency domain impedance, Z, for the network shown. 10 -j5 z -j10 5 Suggested Solution Z = 10 - 5 j + 5(-10 j ) 5(-10 j ) = 10 - 5 j + = 10 - 5 j + 4 - 2 j = 14 - 7 j = 15.65 -26.6 5 - 10 j 11.2 -23.4 Problem 7.13 Find Z(j) at a frequency of 60 Hz for the network shown. 2 4 z 10mH 500F Suggested Solution Z L = j L = j (377)(10 10-3 ) = j 3.77 1 106 = = - j 5.3 j C j 377(5) j 3.77(4 - j 5.3) j15.08 + 20 Z = 2+ = 2+ = 5.1 + j 4.96 j 3.77 + 4 - j 5.3 4 - j1.535 Zc = Problem 7.14 Calculate the equivalent impedance Z at the terminals A-B for the network shown. 2 A 3 z -j2 j4 6 B Suggested Solution Z AB = 2 + (3 - 2 j ) || (6 + 4 j ) = 2 + (3 - 2 j )(6 + 4 j ) 18 + 4 j + 26 44.18 5.19 = = = 4.8 -7.34 9+2j 9+2j 9.22 12.53 Problem 7.15 Find Z for the network shown. z 2 -j1 j2 -j4 4 j4 4 Suggested Solution Y = 1+ Z= 1 1 1 1 + + + = 1 + j + .25 j - .5 j - .25 j = 1 + .5 j = 1.12 26.57 - j -4 j 2 j 4 j 1 = .89 -26.57 Y Problem 7.16 Find Z for the network shown. 2 2 -j2 z -j1 1 j2 Suggested Solution Z = 2 + 1 + 2 || -2 j + 2 j || - j = 3 + 2 -4 j + = 4 - 3 j = 5 -37 2-2j j Problem 7.17 Find Z for the network shown. 2 1 -j1 j2 2 2 Suggested Solution Z = 2 + 2 j || (1 + 1|| j ) = 2 + 2 j || (1.5 + 0.5 j ) = 2 + 2j-4 = 2.8316.92 -1 + 4 j Problem 7.18 The impedance of the network shown is found to be purely real at f=60 Hz. What is the value of C? 3 Z 10mH Suggested Solution 1 -3 j 3.77 -j ( 3 + 3.77 j ) C + C 3 - j 3.77 - C 1 Z= || ( R + j L) = C = 2 -j j C 1 + 3 + 3.77 j 32 + 3.77 - C C THE IMAGINARY COMPONENT IS ZERO -9 j 3.77 j 1 1 - 3.77 - = -9 j - 3.77 j 3.77 - C C C C = 2 60 = 377 C = 431 F 0= 2 Problem 7.19 In the circuit shown determine the value of the inductance such that the current is in phase with the source voltage. 2 +1326F 24cos(377t+60)V L Suggested Solution = 377 f = 60 Hz L = 1 1 1 ,L = 2 = = 5.3mH 2 C C ( 377 )(1326 10-6 ) Problem 7.20 Draw the frequency domain circuit and calculate I(t) for the circuit shown if vs(t)=10 cos(377t+30) V. 2 i(t) +1 mF Vs(t) 10 mH Suggested Solution j L = j (377)(10 10-3 ) = j 3.77 1 -j = = - j 2.65 j C 377 10-3 2 I -+ Vs(t) 10 mH -j2.65 Vs 10 30 10 30 = = = 4.37 0.75A 2 + j 3.77 - j 265 2 + j1.12 2.29 29.25 i (t ) = 4.37 cos(377t + 0.75) A I= Problem 7.21 Draw the frequency domain circuit and calculate v(t) for the circuit shown if Is(t)=20 cos(377t+120) A. + is(t) V(t) _ 1 100 mH 10 mF Suggested Solution Z = j L = j 37.7 Zc = 1 1 = j C j 3.77 1 2 120 2 120 2 120 = = = = 0.52 44.91V V =I 1 + 1 + 1 1 - j + j 3.77 1 + j 3.74 3.84 75.03 R Z L ZC 37.7 v(t ) = 0.52 cos(377t + 44.97)V Problem 7.22 The voltages vR(t), vL(t), and vC(t) in the circuit shown can be drawn as phasors in a phasor diagram. Show that vR(t) + vL(t) + vC(t) = vS(t). Suggested Solution VR (t ) = i (t ) R = 2(4.37) cos(377t + 0.75) = 8.74 cos(377t + 0.75)V VR = 8.74 0.75V di = (10-2 )(4.37)(377)( - sin(377t + 0.75)) = 16.47 cos(377t + 90.75) dt VL = 16.47 90.75V VL (t ) = L 1 1 1 idt = 10-3 377 (4.37) sin(377t + 0.75) C VC (t ) = 11.59 cos(377t - 89.25)V VC (t ) = Imaginary j16 j14 j12 j8 j4 j0 -j4 -j8 -j12 -j14 -j16 VC 4 VS VR 8 VL VL+VC 12 16 20 24 28 32 Real Problem 7.23 The currents iR(t), iL(t), and iC(t) in the circuit shown can be drawn as phasors in a phasor diagram. Show that iR(t) + iL(t) + iC(t) = iS(t). Suggested Solution V (t ) = 0.52 cos(377t + 44.97) A R I R = 0.52 44.97A I R (t ) = dv = (10-2 )(0.52)(377)( - sin(377t + 44.97)) = 1.96 cos(377t + 134.97) dt I C = 1.96 134.97A I C (t ) = C 1 0.52 1 vdt = 0.1 377 sin(377t + 44.97) = 13.79 cos(377t - 45.03)mA L I L (t ) = 13.79 -45.03A I L (t ) = Imaginary IS j1.5 IC j1 j0.5 j0 -1.5 -1 -0.5 -j0.5 0.5 24 28 32 IR Real Problem 7.24 The voltages vR(t) and vL(t) in the circuit shown can be drawn as phasors in a phasor diagram. Use a phasor diagram to show that vR(t) + vL(t) = vS(t). 4 i(t) +- Vs(t) 10 mH Suggested Solution I = 0.36 -43.38A VR = 4 I = 1.44 -43.38A VL = j 3.77 I = 1.36 46.62V Imaginary j1.5 j1 j0.5 j0 -j0.5 -j1 VR VS VL VR+VL Real Problem 7.25 The currents iR(t) and iC(t) in the circuit shown can be drawn as phasors in a phasor diagram. Use a phasor diagram to show that iR(t) + iC(t) = iS(t). + iR(t) 1 ic (t) 100 F is(t) V(t) _ Suggested Solution V (t ) = 9.99 cos(377t + 27.84) A R I R = 9.99 27.84A I R (t ) = dv = (10-4 )(9.99)(377)( - sin(377t + 27.84)) = 0.38cos(377t + 117.84) dt I C = 0.38 117.84A I C (t ) = C Imaginary j5 j4 j3 j2 j1 I j0 C 2 5 8 10 VS Real IS IR IC Problem 7.26 The currents iL(t) and iC(t) of the inductor and capacitor in the circuit shown can be drawn as phasors in a phasor diagram. Use a phasor diagram to show that iL(t) + iC(t) = iS(t). 1 i(t) iL (t) Z 10 mH ic (t) 1 F Vs(t) -+ Suggested Solution V = VS - IR = 10 30 - 0.99 -54.33 = 9.95 35.67V IL = V 9.95 35.67 = = 1.00 54.33A j L 10 90 I C = V ( j C ) = [9.95 35.67]10-3 90 = 9.95 125.67mA Imaginary j1 j0.5 IC j0 -1.5 -1 -0.5 -j0.5 0.5 IL 1 Real Problem 7.27 In the currents shown determine the frequency at which i(t) is in phase with vs(t). 1 i(t) 2 +10mH Vs(t)=120 cos(377t+90) V 10 F Suggested Solution -2 K1 tan -1 C L L j2 - C 1 C C Z = (2 + j L) || = = 1 1 j C 2 + j L - L - C C K tan -1 1 2 For Z to be real -2 = C L - 2 1 C L -4 C = 3156 L -4 = ( L) 2 - f = = 502.3Hz 2 L = C Problem 7.28 Find the frequency domain voltage V0 as shown. 6 -j2 + +- 10-90V j10 Vo _ Suggested Solution V0 = 10 -90 10 j (10 -90)(10 90) = = 10 -53.1V 6 - 3 j + 10 j 10 53 Problem 7.29 Find the frequency domain voltage V0 as shown. 1 + 15 590A -j12 Vo _ Suggested Solution 15 = 3.75 66.9A 15 + 1 - 12 j V0 = - j12(3.75 66.9) = 45 -23.1V I 0 = 5 30 Problem 7.30 Find the frequency domain current I0 as shown. 2 j1 Io Z V = 20 60 Z + 2 + j1 - j2 2 -90 = = 0.89 -26.57 Z = 1 || - j 2 = 1 - j2 5 -64.43 0.89 -26.57 V0 = 20 60 = 20 60(0.31 -38.66) = 6.20 21.34V 0.89 -26.57 + 2 + j1 V I 0 = = 6.20 21.34A R + 2060V -j2 1 - Suggested Solution Problem 7.31 Find the frequency domain voltage V0 as shown. 1 + 1 j -j Z = || = 6 =-j j j 2 3 - 2 3 2 -30 -j V0 = 2 60 = = 2 15V 1- j 2 -45 + 260V j2 j3 Vo _ - Suggested Solution Problem 7.32 Draw the frequency domain network and calculate v0(t) in the circuit iS(t)=100cos(5000t+8.13)mA. Also, using a phasor diagram, show that iL(t)+ iR(t)= iS(t). shown if iL (t) iR(t) + 30 Vo (t) _ 8mH is(t) Suggested Solution iS (t ) = 100 8.13 j L = j 0.008(5000) = 40 j 30 - 40 j V0 (t ) = iS (t )(30 || 40 j ) = 100 8.13 = 2.4 45 30 + 40 j V0 (t ) = 2.4 cos(5000t + 45)V V0 (t ) = 60 -45mA 40 j V (t ) i R = 0 = 80 45mA 30 iL = IR IS IL Problem 7.33 Draw the frequency domain network and calculate v0(t) in the circuit iS(t)=100cos(5000t+8.13)mA. Also, using a phasor diagram, show that i1(t)+ i2(t)= iS(t). shown if i1 (t) 20 i2 (t) 3.33 F is(t) 6mH 10 Suggested Solution iS (t ) = 0.3 -135A j L = j 0.006(10000) = 60 j V0 (t ) = iS (t )((20 + 60 j ) || (10 - 30 j )) = 14.14 180 V0 (t ) = 14.14 cos(10000t - 45)V V0 (t ) = .224 251.6mA 20 + 60 j V (t ) i2 = 0 = .447 -108.4mA 10 - 30 i1 = I2 IS I1 Problem 7.34 Find I0 in the network shown. 120V -j1 2 -+ 2 Io j2 -j2 Suggested Solution Z = 2 j || (2 - 2 j ) + 2 || (1 - j ) = 2 + 2 j + 0.8 - 0.4 j = 2.8 + j1.6 = 3.22 30 IS = 12 0 = 3.73 -30 3.22 30 1- j I 0 = 3.73 -30 = 5.89 -48.4A .8 - .4 j Problem 7.35 Find I0 in the network shown. 120V -+ 2 Io -j2 2 2 j2 Suggested Solution I0 = 1 12 0 = -2.68 63.4A 2 1- 2 j 120V -+ 2 -j2 2 j2 2 Problem 7.36 In the circuit shown, if V0=445 V, find I1 2 j1 + 4 45 = 2 45A 2 V1 = I 2 (2 + j ) = 4.47 71.6 I2 = I1 = V1 4.47 71.6 = = 2.24 162A 2 -90 - j2 + -j2 I1 2 Vo _ - Suggested Solution Problem 7.37 Find VS in the network shown if V0=40 V. -j1 Vs j1 -+ + 1 2 V1 _ -j1 2 Suggested Solution 2(1 - j ) = 0.8 - j 0.4 3- j ( - j )(2 + j ) = 0.5 - j1 Z2 = 2 4 0 I= 0.8 - j 0.4 V2 = IZ 2 = 5 -36.86V Z1 = VS = -(4 0 + 5 -36.86V ) = -8 + j 3 = -8.54 -20.56V Problem 7.38 Find VS in the network shown if I0=20 A. i1(t) -j1 + 1 2 V1 _ Vs -+ 2 j1 2 Suggested Solution i1(t) -j1 + 1 2 I2 Vs -+ V1 _ I3 Z 2(2 + j ) 4+ j V1 = I1 (1 - j ) = 2 - j 2 Z= V1 = 1- j 2 I 3 = I1 + I 2 = 2 + 1 - j = 3 - j I2 = 4 + 2 j 14 + j 2 V2 = I 3Z = (3 - j ) = 4+ j 4+ j 14 + j 2 VS = V1 + V2 = 2 - 2 j + = 5.58 -14V 4+ j Problem 7.39 Find VS in the network shown if I0=20 A. Vs -+ 6 j1 2 I0 -j1 2 Suggested Solution I 0 = 2 0 I1 = 3 0 VS = (3 0)(2 + j ) = 6 + 3 j Problem 7.40 Find IS in the network shown if V0=20 V. Is 4 + 2 Vo _ j1 2 -j1 Suggested Solution Is 4 + 2 Vo _ j1 I1 I2 2 -j1 V1 = 8 0V 8 = 6A 4 3 10 37 V = 4.46 63.6 I2 = S = 2 - j 2.24 -26.6 I S = I1 + I 2 = 6 + 1.98 + 4 j = 8 + 4 j I1 = Problem 7.41 Find I0 in the network shown if V1=20 V. -j2 _ Vo 2 + j1 is 1 I0 Suggested Solution V1 = V0 - 4 = 4 0V V0 4 4 + + 1 2 -2 j I1 = 2 + 2 j = I S - V0 IS = V0 = j (2 + 2 j ) = 2 j - 2 V0 = V0 + V1 = 2 j - 2 + 4 = 2 + 2 j I 0 = V0 = 2 + 2 j = 2.828 45 -j2 _ Vo 2 V1 + j1 I1 is 1 I0 Problem 7.42 In the network shown V0=445 V: find Z. 2 -j1 + Z 1 Vo _ 120V -+ Suggested Solution V1 = 4 45(1 - j ) = 5.66 0V THEN , 12 - V1 V1 = + 4 45 Z 2 12 - 5.66 5.66 SO , = + 2.83 + j 2.83 Z 2 5.66 Z= = 2 83 2.85 -83 Problem 7.43 In the network shown V0=245 V: find Z. 1 j1 + Vo _ 60A Z 1 -j1 Suggested Solution 2 45 = 1.414 - j1.414 j I 2 = 6 + I1 = 7.414 - j1.414 I1 = V2 = 1I 2 = 7.414 - j1.414 V3 = V1 + V2 = 8.83 V3 = 8.83 1 I 4 = I1 + I 3 = 10.242 - j1.414 I3 = V4 = - jI 4 = -1.414 - j10.242 V4 + V3 + V0 = 0 V0 = -(V4 + V3 ) = -( -1.414 - j10.242 + 8.83) = -7.416 + j10.242 I 0 = I 2 + I 3 = 7.414 - j1.414 + 8.83 = 16.25 - j1.414 Z= V0 -(7.416 - j10.242) = = 0.78 130.87 I0 16.25 - j1.414 I3 1 j1 Vo _ 60A I1 I2 1 I0 + -j1 I4 Z Problem 7.44 Find I0 in the network shown if IS=120 A. i1(t) -2j 2 is 2 -2j 2j 2 i0(t) Suggested Solution 2 I1 = I S 2 + Z j2 I 0 = I1 = I1 (1 90) 2 - j2 + j2 Z = ( j 2 || (2 - j 2)) + (2 || - j 2) = j4 + 4 j4 j2 4 - j2 - = 2 + j2 - = 2 2 - j2 1- j 1- j 24(1 - j ) 12(1 - j ) 2 = = 4.69 -11.31A I1 = 12 0 = 6 - j4 3 - j2 2 + 4 - j2 1- j I 0 = (4.69 -11.31)(1 90) = 4.69 78.69A Problem 7.45 Use nodal analysis to find I0 in the circuit shown. 2 1 -+ 120V 20A 2 I0 -j1 40A Suggested Solution 2 V1 1 V2 -+ 120V 20A 2 I0 -j1 40A V1; 12 - V1 V -V V =2+ 1 2 + 1 2 1 2 8 = 4V1 - 2V2 V1 - V2 V +4= 2 1 - j1 4 + V1 V2 = 1 + j1 2(4 + V1 ) 8 = 4V1 - 1 + j1 4(2 + j ) V1 = 1 + j2 V 2(2 + j1) I0 = 1 = = 2 -36.87 2 1 + j2 V2 ; Problem 7.46 Use nodal analysis to find V0 in the circuit shown. 60V j2 ++ 2Vx 2 Vx _ -j1 1 + Vo _ Suggested Solution V x + 6 Vx V + + x =0 2 -2 1 + 2 j -2(1 + 2 j ) Vx = V 1+ 4 j 2Vx + 1 V0 = Vx = -.49 -76 1+ 2 j Problem 7.47 Find V0 in the network shown using nodal analysis. 2 -j1 V 1 j2 -+ 120V -+ + 2 Vo _ 20A 40V Suggested Solution 2 V V V0 = V = 2 45 = 1 + j1 2 + j2 3 + j2 12 - V 1 V V -4 =2+ + = V 1 + - 2 = V 2 + j2 - 2 2 - j1 2 + j2 1 2 + j2 (3 + 2 j )(2 - j1) 8 + j1 12 - V = V - 4 + j2 = V 2 + j2 - 4 + j2 2 + j2 8 + j1 2 + j 2 + 8 + j1 16 - j 2 = V 1 + =V 2 + j2 2 + j2 10 + j 3 16 - j 2 = V 2 + j2 4(8 - j1)(1 + j1) V= (10 + j 3) V 4(8 - j1) 32.25 -7.13 V0 = = = 1 + j1 10 + j 3 10.44 16.70 V0 = 3.09 -23.83V Problem 7.48 Use nodal analysis to determine I0 in the network shown. In addition, solve the problem using MATLAB. 120V -+ 2 -j1 -+ I0 60V 2 j2 Suggested Solution 12 V 120V -+ 2 -j1 -+ I0 Vo 2 j2 60V 18 V V0 ; V0 (1 + j1) = 12 - j18 V0 = 12 - j18 1 + j1 12 - V0 = 2 12 - 12 - V0 18 - V0 V0 + = = jV0 - j1 2 j2 12 - V0 - j18 + jV0 = j 2V0 I0 = 12 - j18 12 + j12 - 12 + j18 j15 1 + j1 = = = 10.64 45A 2 2 (1 + j1) 1 + j1 MATLAB SOLUTION 120V -+ 2 -j1 -+ V1 2 j2 I0 V3 60V V2 V1 = -12 0 V2 = 6 0 V3 V3 - V1 V3 - V2 + + =0 - j1 j2 2 0 V1 -12 1 0 0 1 0 V2 = 6 2 -1 -1 + j1 V3 0 >>g=[2 1 1+1j;1 0 0;0 1 0] >>i=[0;-12;6] >>v=inv(g)*I v= -12.0000 6.0000 -15.0000 15.0000i Since V3=-15-j15 Then I0=-V3/2=7.5+j7.5 A=10.61/_45 A Problem 7.49 Find V0 in the network shown. j1 120V -+ + 2 -j1 2 Vo -+ -4V 160V 20A _ Suggested Solution j1 120V -+ + 2 -j1 -16V 2 Vo 20A _ -+ 160V 2 V0 = -4 2 + j1 V0 = 3.58 153.43V Problem 7.50 Find the voltage across the indicator in the circuit shown using nodal analysis. V1 -j4 V2 1645V -+ 4 Ix j2 2Ix Suggested Solution KCL @ V2 ; Ix = V1 ; 4 V1 - V2 V2 = + Ix - j4 2 V1 = 16 45 j8V1 4 V1 - V2 = -2V2 - V1 (1 + j 2 ) = -V2 V2 = -V1 (1 + j 2 ) = (16 -135 )(1 + j 2 ) V2 = 35.78 -71.57V Problem 7.51 Use mesh analysis to find V0 in the circuit shown. -j2 645V +120V -+ + j4 I1 I2 2 Vo _ Suggested Solution V0 = 2 I 2 mesh equations 12 = I1 ( j 2 ) + I 2 ( - j 4 ) -6 45 = - j 4 I1 + I 2 ( 2 + j 4 ) matrix form - j 4 I1 12 j2 - j 4 2 + j 4 I = 6 -135 2 Solve for I 2 12 j2 - j 4 6 135 12 -45 + 48 90 = I2 = = 4.52 51.3A j2 - j4 j 4 - 8 + 16 - j 4 2 + j 4 V0 = 2 I 2 = 9.04 51.3V Problem 7.52 Use mesh analysis to find V0 in the circuit shown. 4 j2 -j4 120V I2 -+ I1 490A + 2 Vo _ Suggested Solution mesh equations 12 = I1 ( 4 - j 4 + 2 ) - I 2 ( 2 - j 4 ) I 2 = -4 90 = - j 4 A Substitute ( 2 ) into (1) 12 = I1 ( 6 - j 4 ) + j 4(2 - j 4) I1 ( 6 - j 4 ) = 12 - j8 - 16 = -4 - j8 I1 = 2( -1 - j 2) (3 - j2) (1) (2) I1 = 1.24 -82.87 A = ( 0.15 - j1.23) A V0 = ( I1 - I 2 ) 2 = ( 0.15 - j1.23 + j 4 ) 2 V0 = 5.55 -86.9V Problem 7.53 Using loop analysis, find I0 in the network shown. 120V -+ 20A 2 j1 -j2 2 I0 40A Suggested Solution 120V 2 -+ 20A I1 j1 -j2 2 I2 I0 I3 40A I 3 = -4 A I2 = 2 A Supermesh Equations 12 = I1 ( 2 + j1 + 2 - j 2 ) - j1I 3 - 2 I 3 + 2 I 2 - j 2 I 2 12 = I1 ( 4 - j1) - I 3 ( 2 + j1) + I 2 ( 2 - j 2 ) 12 = I1 (4 - j1) + 8 + j 4 + 4 - j 4 I1 = 0 I 0 = I1 + I 2 - I 3 = 0 + 2 - ( -4) = 6 A MATLAB SOLUTION 120V -+ 20A I1 2 j1 -j2 2 I2 I0 I3 40A 12 0 = 2 I 3 + j1 ( I 3 - I 2 ) + 2 ( I1 - I 2 ) - j 2 I1 1 0 I 1 -4 0 1 0 -1 I 2 = 2 2 - j 2 -2 - j1 2 + j1 I 3 12 >> z=[2-2j 2-1j 2+1j; 0 1 0; 1 0 1] >> v=[12;-4;2] >> i=inv(z)*v i = 2.0000 + 0.0000i -4.0000 -0.0000 + 0.0000i I0=I1-I2=2-(-4)=6A I 2 = -4 0I1 - I 3 = 2 0 Problem 7.54 Find V0 in the network shown. 40A 1 Ix + 2I x j1 1 Vo _ Suggested Solution 40A I1 1 Ix + 2I x I2 j1 I3 1 Vo _ I1 = 4, I 2 = 2 I x , j ( I 3 - I 2 ) + 1 ( I 3 - I1 ) + 1I 3 = 0 j ( I 3 - 2 I x ) + 1( I 3 - 4 ) + I 3 = 0 jI 3 - 2 jI x + I 3 - 4 + I 3 = 0 jI 3 - 2 ( I 3 + 4 ) + I 3 - 4 + I 3 = 0 I x = I3 - 4 I 3 ( j - 2 j + 1 + 1) = 4 - 8 j I3 = 4-8j = 4 -36.86A 2- j jI 3 - 2 jI 3 + 8 j + I 3 - 4 + I 3 = 0 Problem 7.55 Find V0 in the network shown. + 40A 1 Vo _ -j1 1 + j1 2Vx 1 V0 _ Suggested Solution + 40A I1 1 Vo _ -j1 I3 1 + j1 I2 2Vx 1 V0 _ Vx = - j ( I1 - I 3 ) For I 3 , I1 = 4 0 I 2 = 2Vx j ( I 2 - I 3 ) + 1 ( I 2 + I 3 - 4 ) - j ( I 3 - 2Vx ) + 2 I 3 = 0 4 + 12 j 5 V0 = .8 + 2.4 j Solving I 3 = Problem 7.56 Use superposition to find V0 in the network shown. 20V -+ 1 1 + 40A -j1 Vo _ Suggested Solution 20V 1 + 1 -j1 V1 _ -+ -j 2j 2 90 V1 = -2 = 2 - j = 2.236 -26.56 = 0.89 116.56V 2- j 1 + 1 40A -j1 V2 _ V2 = 4 (1) -4 j -4 90 = = 1.78 -63.4V (- j) = 2- j 2 - j 2.236 -26.56 V0 = V1 + V2 = -0.398 + j 0.396 + 0.797 - j1.59 = 0.399 - j 794 = 0.89 -63.6V Problem 7.57 Find V0 in the network shown using superposition. 1 20V 120V -+ + -j2 j4 2 Vo _ Suggested Solution 1 20V I2 120V -+ + -j2 I1 j4 2 Vo _ For the 2 A source, I1 = 2+2 2-2j+ 8j 2+4j = 4 (2 + 4 j ) 12 + 12 j 8j -32 j V0 = I1 = 12 + 12 j 2+4j For the 12V source, I2 = 12 ( 4 - 2 j ) 12 = 2 (2 - 2 j ) 12 + 12 j 4j+ 2+2-2j 2 ( 2 - 2 j ) 12 ( 4 - 4 j ) = 2+2-2j 12 + 12 j Summing -32 j + 48 - 48 j V0 = = 5.5 -104V 12 + 12 j V0 = I 2 Problem 7.58 Using superposition, find V0 in the circuit shown. 2 -j1 j2 -+ 120V + Vo _ 4 20A Suggested Solution 2 -j1 Vs j2 Is -+ 120V + Vo _ 4 IL 20A For the voltage source 4 48 V1 = (12 0 ) = 4 + 2 + j 2 - j1 6 + j1 For the current source -4 + j 2 2 - j1 IL = IS = 6 + j1 4 + 2 + j 2 - j1 -16 + j8 V2 = 4 I L = 6 + j1 Summing 32 + j8 V0 = V1 + V2 = = 5.41 4.57V 6 + j1 Problem 7.59 Use both superposition and MATLAB to determine V0 in the circuit shown. 1 -j1 j2 120V -+ + 2 40A Vo _ Suggested Solution 1 -j1 Vs 120V j2 -+ + 2 V1 _ 2 2 48 = 9.23 - j1.85V V1 = V0 = = 12 5 1 1 1 5 + j1 +j 2 + + j 2 2 2 2 1 -j1 I2 j2 + 2 40A V2 _ - j1 - j4 - j12 I2 = 4 = j1 + 2 = 2 + j 3 3 - j1 + j 2 + 2 3 - j8 2 = -1.85 - j1.23V V2 = ( I 2 ) = 3 2 + j3 V0 = V1 + V2 = 7.38 - j 3.08 = 8.00 -22.65V MATLAB SOLUTION 1 -j1 V1 V2 j2 120V -+ + 2 40A Vo _ V1 = 12 0V V1 - V2 V -V +4+ 0 2 =0 - j1 j2 V1 - V0 V2 - V0 V0 + - =0 j2 1 2 0 V1 12 1 0 -2 1 1 V2 = - j8 j 2 1 -1 - j 3 V3 0 >> g=[0+2j 1 1-3j;1 0 0;-2 1 1] >> i=[0;12;-8j] >> v=inv(g)*i v = 12.0000 0.0000i 16.6154 4.9231i 7.3846 3.0769i V0=7.3846-j3.0769=8.00/-22.62 V Problem 7.60 Use source exchange to determine V0 in the network shown. -j1 120V +-+ + 2 20A 2 Vo _ 60V Suggested Solution 120V ++ (2+j6)A Z 2 Vo _ 120V Z VEQ ++ 2 Vo _ -+ Z = 2 || - j = VEQ = (2 + j 6)( Z EQ ) = - j2 2 = 2 - j1 1 + j 2 2(2 + j 6) = 5.66 8.13V = (5.60 + j 0.80 )V 1 + j2 2 4 V0 = (VEQ - 12 ) = ( -6.4 + j 0.8 ) = 7.16 139.18V 3 + j2 2 + Z EQ Problem 7.61 Use source exchange to find the current I0 in the network shown. 2 1 120V -+ 20A 2 I0 -j1 20A Suggested Solution 1 40A 2||2=1 2I 0 -j1 40A 1 -j1 -+ 2I 0 40A 1 -j4 4+2-2j 1-1j 1 2 I0 = (6 - 2 j ) I 0 = 2 -37 1- j 2- j Problem 7.62 Use source transformation to determine I0 in the circuit shown. 2 1 -+ 120V 20A 2 I0 -j1 40A Suggested Solution 1 -j1 60A 2 -+ I0 20A 2 4-90V 40A 2 2 I0 1 -j1 22-45A Z EQ = 2 || (1 - j1) = 2 - j2 3 - j1 I EQ = 4 + 2 2 -45 = 6 - j 2 (6-j2)A ZEQ 2 I0 Z ( 6 - j 2 )( 2 - j 2 ) 2 - j2 I 0 = ( 6 - j 2 ) EQ = ( 6 - j 2 ) = 8 - j4 2 - j2 + 6 - j2 Z EQ + 2 ( 3 - j1)(1 - j1) = 2 -36.87A I0 = 2 - j1 Problem 7.63 Use Thevenin's theorem to find V0 in the circuit shown. 2 -j1 2 + j2 1 Vo _ 60A 120V -+ -j1 Suggested Solution 2 2 + j2 1 Vo _ 60A 120V -+ -j2 ZTH1 1 -+ + 120V ZTH1 V1 _ V1=12/0V ZTH1=-j1 120V -+ -+ 60V -j1 2 + j2 1 Vo _ ZTH2 -j1 2 + j2 ZTH1 V2 _ V2=24/0V ZTH2=2-j2 120V -+ 2 -j1 + 1 Vo _ V0=6.66/33.69V -+ Problem 7.64 Using Thevenin's theorem, find V0 in the network shown. -j2 645V +120V -+ + j4 2 Vo _ Suggested Solution Voc = 12 ( j 4 ) - 6 45 = 20.2 -12.1V j2 RTH = V0 = ( - j 2 )( j 4 ) = - j 4 j2 2 - j4 ( 20.2 -12.1 ) 2 = 9.03 51.3V Problem 7.65 Use Thevenin's theorem to find V0 in the circuit shown. 4 j2 -j4 120V I2 -+ I1 490A + 2 Vo _ Suggested Solution Voc = 12 + 4 ( 4 90 ) = 12 + j16V RTH = 4 - j 4 V0 = (12 + j16 ) 2 = 5.55 86.8V 2 + 4 - j4 Problem 7.66 Solve Problem 7.49 using Thevenin's theorem. j1 120V -+ + 2 -j1 2 Vo -+ 160V 20A _ Suggested Solution VTH = 12 0 - 16 0 = 4 180V Z TH = 0 2 V0 = 4 180 = 3.58 153.43V 2 + j1 Problem 7.67 Apply Thevenin's theorem twice to find V0 in the circuit shown. 1 2 1 + -j1 j1 1 Vo _ 120V -+ Suggested Solution PART I VOC = Z TH 12 - j -12 j = 1- j 1- j -j = 1- j PART II VOC -12 12 1- j = = (- j) = -j 3- 2 j +2+ j 1- j -j j2 + 1- j 3+ 2 j = = -j 3- 2 j j+2+ 1- j Z TH FINALLY 12 3- 2 j V0 = (1) = 1.312.5V 3+ 2 j 2+ 3- 2 j Problem 7.68 Find V0 in the network shown using Thevenin's theorem. 60V j2 ++ 2Vx 2 -j1 _ Vx 1 + Vo _ Suggested Solution VOC = VX 2VX + VOC + 6 VOC -6 + = 0 VOC = 2 5 + j2 - j1 -6 I SC = = -3 A 2 2 V Z TH = OC = I SC 5 + 2 j -6 ( 5 + j 2 ) -6 1 5 + j2 = = 2.61125.8V V0 = VOC = 3 + j12 1 + j 2 + Z TH 5 + j 2 (1 + j 2 )( 5 + j 2 ) + 2 Problem 7.69 Find the Thevenin equivalent for the network shown at the terminals A-B. + 40A 1 -j1 _ Vx A j1 2Vx B Suggested Solution A Vx _ 1 I1 + 40A -j1 j1 I2 2Vx B VX = 4 0 ( - j ) = -4 j 2VX = -8 j I1 = 4 + 8 j I2 = 4 + 8 j - 4 = 8 j For VOC 2j-1 Z -+ V1 = 8 j - j = -8 4j-4 VOC = -8 + ( 4 + 8 j ) - 4 j = -4 + 4 j For I SC j ( I SC + 2VX ) + 1 ( I SC + 2VX - 4 ) - j ( I SC - 4 ) = 0 VX = - j ( 4 - I SC ) YIELDS I SC = 2.4 + .8 j Z TH = -4 + 4 j = -1 + 2 j 2.4 + .8 j Problem 7.70 Find Vx in the circuit shown using Norton's theorem. 11.345V ++ 40A j4 10 -j3 Vo _ Suggested Solution 11.345V ++ 40A j4 10 -j3 Vo _ ZEQ Z EQ = j 4 4= I SC V + I SC where V = 11.3 45V j4 jV =4+ = 6 + j2 A 4 + 6+j2 A j4 10 -j3 Vo _ 1 I SC = 48.59 -21.37V VX = I SC [10 || j 4 || - j 3] = I SC = 1 + 1 + 1 0.10 + j 0.08 10 j 4 - j 3 Problem 7.71 Find I0 in the network shown using Norton's theorem. -j4 j2 245A 2 I0 40A Suggested Solution I SC = -2 45 Z TH 2j + 4 0 = 4 + 2 45 2j-4j = -4 j + 2 j = -2 j Z TH -2 j = ( 5.41 + 1.41 j ) = 4 -30.3A 2 + ZTH 2-2j I 0 = I SC Problem 7.72 Apply both Norton's theorem and MATLAB to find V0 in the network shown. -j1 2 + 2 j2 1 Vo 20A 1 _ 120V -+ Suggested Solution -j1 2 120V -+ I1 2 j2 I sc Isc 20A I2 1 mesh equations 12 = I1 ( 2 + j1) - 2 I SC - j 2 I 2 -2 = I 2 0 = I SC ( 3) - 2 I1 - I 2 substitute ( 2 ) into (1) & ( 3) I1 ( 2 + j1) - 2 I SC = 12 - j 4 3 I SC 2 substitute ( 5) into ( 4 ) I1 = 1 + (1) (2) ( 3) (4) ( 5) ( 2 + j1) 1 + 3 3 I SC - 2 I SC = 12 - j 4 I SC 1 + j = 10 - j5 2 2 10 ( 2 - j1) I SC = 2 + j3 2 + j3 j2 = Z EQ = 2 || ( j 2 - j1) + 1 = 1 + 2 + j1 2 + j1 I0 2 + Isc ZEQ 1 Vo _ Z 10 ( 2 - j1) 2 + j3 I 0 = I SC EQ = 3 + Z EQ 2 + j 3 3 ( 2 + j1) + 2 + j 3 10 ( 2 - j1) 5 ( 2 - j1) = V0 = 1I 0 = 8 + j6 4 + j3 V0 = 2.24 -63.43V MATLAB SOLUTION -j1 2 120V -+ I1 2 + 1 Vo _ j2 I sc 20A I2 1 12 = - j1I 2 + ( I 2 - I 3 ) + j 2 ( I 2 - I1 ) 0 = 2 I 3 + (1) I 3 + ( I 3 - I 2 )(1) + 2 ( I 3 - I 2 ) 0 0 I 1 -2 1 - j 2 2 + j1 -2 I = 12 2 -2 6 I3 0 -1 >> z=[1 0 0;-2j 2+1j 2;-1 2 6]; >> v=[-2;12;0]; >> i=inv(z)*v i = -2.0000 4.0000 6.0000i 1.0000 2.0000i I1 = 2 0A V0=I3(1)=1-j2=2.24/-63.43 V Problem 7.73 Find V0 using Norton's theorem for the circuit shown. -+ 40V -+ Vx _ 1 -j1 1 + 1 80V + Vo _ j1 + - 2Vx Suggested Solution VX = 2VX - VOC VOC = V1 - 4 VX = V1 - 4 V1 V1 - 2VX V1 + 4 - 2VX V1 - 4 - 2VX + + + =0 j1 1 1 - j1 V1 = 6V VOC = 2V VX = 0 8 4 4 + + + I SC = 0 - j1 1 j1 I SC = -4 - j 4 A Z TH = VOC 2 1 1 = =- + j I SC -4 - j 4 4 4 V0 = I 0 (1) = 2.53 -18.43V I 0 = 2.53 -18.43 A Problem 7.74 Use Norton's theorem to find V0 in the network shown. + 40A 1 -j1 Vx _ 1 + j1 2Vx 1 Vo _ Suggested Solution + 40A I1 + -j1 Vx _ Voc 1 j1 I2 2Vx _ VX = ( - j1) I1 = - j 4V I1 = 4 0 I 2 = 2VX I 2 = - j8 A I 2 (1 + j1) - I1 (1 - j1) + VOC = 0 VOC = j8 (1 + j1) + 4 (1 - j1) = -4 + j 4 V + 40A I1 -j1 Vx _ I SC 1 Isc j1 I1 2Vx I1 = 4 0 A I 2 - I SC = 2VX VX = ( - j1)( I1 - I SC ) I 2 (1 + j1) - I1 (1 - j1) + I SC ( - j1) = 0 I SC (1 + j 2 ) - j8 (1 + j1) - 4 (1 - j1) + I SC ( - j1) = 0 -4 + j 4 I SC = A -1 + j 2 Z I 0 = I SC TH Z TH + 2 Z TH = -1 + j 2 I0 = -4 + j 4 A 1 + j2 V0 = (1) I 0 = 2.53 71.57V Problem 7.75 Use MATLAB to find the node voltages in the network shown. 1 2 1 -j1 j1 20A 1 120V -+ 2 -j2 Suggested Solution V1 1 V2 2 1 -j1 j1 V4 2 V5 20A V3 120V 1 -+ -j2 V1 - V3 V1 - V4 V1 - V2 + + =0 - j1 2 1 V2 - V1 V2 + = 2 0 1 1 V3 = 12 0 V4 - V1 V4 - V3 V4 V4 - V5 + + + =0 - j1 j1 1 2 V5 - V4 V5 + = -2 0 - j2 j1 ALSO V1 (1.5 + j1) - V2 - .5V3 - j1V4 = 0 -V1 + 2V2 = 2 0 V3 = 12 0 - j1V1 - V3 + 1.5V4 + j1V5 = 0 j1V4 - j 0.5V5 = -2 0 Matrix Form 0 V1 0 1.5 + j1 -1 -.5 - j1 -1 2 0 0 0 V2 2 0 0 1 0 0 V3 = 12 0 0 j1 V4 0 0 -1 1.5 - j1 0 j1 - j 0.5 V5 -2 0 0 0 >>Y 0 0 >>I >>V V= = [ 1.5+j*1 1 -.5 j*1 0; -1 2 0 0 0; 0 0 1 0 0;-j*1 0 1 1.5 j*1; j*1 j*0.5] = [0;2;12;0;-2] = inv(Y)*I 6.5800 4.2900 12.0000 4.5200 9.0400 2.0600i 1.0300i 0.0000i 1.6400i 7.2800i Problem 7.76 Using the PSPICE Schematics editor, draw the circuit shown. At what frequency are the magnitudes of iC(t) and iL(t) equal? R1 100 R2 150 VIN(t) 5 cos (wt) V -+ iC(t ) L 100mH C 100F iL (t ) Suggested Solution This Schematics circuit was simulated over the frequency range 100 Hz to 300 Hz. Since current into pin markers were placed in the circuit, PROBE will automatically plot the required current magnitudes. PROBE results show that the voltage and current phases are roughly equal at 238.9 Hz Problem 7.77 Using the PSPICE Schematics editor, draw the circuit shown. At what frequency are the phases of il(t) and vx(t) equal? R1 2k Vl(t) 14 cos (wt+70)V L 1 mH R2 50k + Vx(t) _ il(t) 50 cos (wt) mA -+ C 10F Suggested Solution This Schematics circuit was simulated over the frequency range 1 kHz to 2 kHz. Since vphase and iphase markers were placed in the circuit, PROBE will automatically plot the required voltage and current phase angles. PROBE results show that the voltage and current phases are roughly equal at 1.63 kHz. Problem 7FE-1 Find V0 in the network shown. 120V j1 -+ + 2 20A -j1 1 Vo _ Suggested Solution 120V j1 -+ 2 20A V1 + -j1 1 Vo _ V1 - 12 V V -2+ 1 + 1 =0 2 - j1 1 + j1 1 1 1 V1 + + =8 2 - j 1+ j 1 V0 = V1 1+ V1 = 32 V 4 + j2 32 = 2 + j 6 = 5.06 -71.6 V j Problem 7FE-2 Find V0 in the circuit shown. 2Ix 1 2 + 120V Vo 20A _ -j1 Ix -+ Suggested Solution 2Ix V1 1 12 2 + 120V Vo 20A _ -j1 Ix V1 V1 - 12 + + 2I X = 0 -j 1 V0 - 12 - 2 - 2I X = 0 2 V IX = 1 -j V1 = 12 V 12 304 + j 48 0 -2 j = 30.78 8.97V = 8 V0 = 1+ 3 j 2 10 1+ 3 j -+ Problem 7FE-3 Find the average power dissipated in the 4-Ohm load resistor. 2 -j1 j2 120V -+ + 4 V0 _ + 2V0 Suggested Solution 2 -j1 j2 120V -+ + I1 I2 4 V0 _ + - 2V0 -12 + 2 I1 + j 2 ( I1 - I 2 ) + 2V0 = 0 -2V0 + j 2 ( I 2 - I1 ) - jI 2 + 4 I 2 = 0 V0 = 4 I 2 I1 (1 + j ) + I 2 ( 4 - j1) = 6 1 I1 ( - j ) + I 2 -2 + j = 0 2 1 -3 - j 5 = (1 + j ) -2 + j - ( - j )( 4 - j ) = 2 2 1 -4 + j 6 I1 2 -2 + j 60 - j 36 = 2.05 -31.03A 2 I = -3 + j 5 0 I 2 = 34 2 j 1+ j 1 2 P4 = ( 2.05) ( 4 ) = 8.41watts 2 Problem 7FE-4 Determine the mid-band (where the coupling capacitors can be ignored) gain of the single-stage amplifier shown. + 1k + 5k Vx _ 40*10-3Vx 6k 12k Vo Vs -+ _ Suggested Solution VX = VS ( 5k ) 5 -40 5 = VS and V0 = -40 10-3VX ( 6k ||12k ) = ( 4k )VS = -133.33VS 1k + 5k 6 1000 6 V0 = -133.33 VS Problem 8.1 Given the network in Figure P8.1 (a) Find the equations for va(t) and vb(t) (b) Find the equations for vc(t) and vd(t) M i1(t) va(t) + Figure P8.1 + vc(t) L1 L2 vb(t) i2(t) + vd(t) + Suggested Solution a) va (t ) = - L1 di1 - M dt vb (t ) = - L 2 di 2 - M dt b) vc (t ) = L1 di1 + M dt vd (t ) = L 2 di 2 + M dt di 2 dt di1 dt di 2 dt di1 dt Problem 8.2 Given the network in Figure P8.2 (a) write the equations for va(t) and vb(t) and (b) write the equations for vc(t) and vd(t). M i1(t) + va(t) + Figure P8.2 vc(t) L1 i2(t) L2 vd(t) + + vb(t) Suggested Solution a) va (t ) = L1 di1 + M dt b) vc (t ) = -va (t ) = - L1 di1 + M dt vd (t ) = -vb (t ) = L 2 di 2 + M dt di 2 dt di1 dt di 2 dt di 1 dt vb (t ) = - L 2 di 2 - M dt Problem 8.3 Given the network in Figure P8.3 (a) write the equations for va(t) and vb(t) and (b) write the equations for vc(t) and vd(t). M i1(t) + va(t) + vc(t) L1 L2 vd(t) + i2(t) + vb(t) - Figure P8.3 Suggested Solution a) va (t ) = - L1 di1 + M dt vb (t ) = - L 2 di 2 - M dt b) vc (t ) = -va (t ) = - L1 di1 + M dt vd (t ) = -vb (t ) = L 2 di 2 + M dt di 2 dt di1 dt di 2 dt di 1 dt Problem 8.4 Given the network in Figure P8.4 (a) write the equations for va(t) and vb(t) and (b) write the equations for vc(t) and vd(t). M i1(t) + va(t) + vc(t) L1 Figure P8.4 L2 vd(t) + i2(t) + vb(t) - Suggested Solution a) va (t ) = L1 di1 + M dt b) vc (t ) = -va (t ) = - L1 di1 - M dt vd (t ) = -vb (t ) = L 2 di 2 + M dt di 2 dt di1 dt di 2 dt di 1 dt vb (t ) = - L 2 di 2 - M dt Problem 8.5 Find the voltage gain Vo/Vs of the network shown in figure P8.5 jM = j2 4K - j2 + Vs j4 j3 Figure P8.5 2 Vo - I1 I2 Suggested Solution jM = j2 4K - j2 + Vs j4 j3 2 Vo - I1 I2 -Vs = 2(- I 1) + j 4(- I 1) + j 2 I 2 Vs = (2 + j 4) I 1 - j 2 I 2 0 = - j 2 I 1 + (6 + j1) I 2 Vo = 2 I 2 J 2 I1 = (6 + 1) I 2 then Vs = (2 + j 4)( 6j+21 ) I 1 - J 2 I 2 = (13 - j 6) I 2 I2 = Vs 3- j 6 or I1 = 6 +1 j2 ( I 2) or Vs 2 Vo = 2 I 2 = 132- j 6 = Vo = 13- j 6 = 0.14024.78 Vs Problem 8.6 Find the voltage gain Vo/Vs of the network shown in figure P8.6 j1 2 - j2 + j1 Figure P8.6 4 Vo - 240o j2 Suggested Solution j1 2 - j2 + j1 I2 4 Vo 24 = (2+j2)I1 + J1 I2 0 = JI1 + (6-j)I2 Vo = 4I2 240 o I1 j2 -j I1 = (6 - j)I2 or I1 = (1 + j6)I2 24 = (2 + j2)(1 + j6)I2 + jI2 I2 = 24 / (-10 + j5) Vo = 4x24/(-10 + j15) 5.33-123o V Problem 8.7 Find Vo in the network in Figure P8.7 1 100o j1 j1 2 j2 2 + Vo Figure P8.7 Suggested Solution 1 100o I1 j1 j1 2 j2 I2 2 + Vo I1 = I2 (4 + j2) / j1 = I2 (2 - j4) 10 = [(1 + j1)(2 - j4) - j1] I2 = (2 + 4 + j2 - j4 - j1)I2 = (6 - j3) I2 I2 = 10 / (6 - j3) Vo = 2 I2 = 20 / (6 - j3) Loop equations 10 = (1+j1)I1 - j1 I2 (1) 0 = -jI1 + (4 +2j)I2 (2) Solve (2) for I1 and substitute into (1) and get I2 Vo = 2.9826.57o V Problem 8.8 Given the network in Figure P8.8 find Vo 2 j1 1 100oV j2 Figure P8.8 j2 1 -j1 + Vo - Suggested Solution 2 j1 1 100oV I1 j2 j2 I2 1 -j1 + Vo 10 = (2 + j2)I1 + j1 I2 (1) (2) 0 = j1I1 + (2 +2j)I2 Solve (2) for I1 and substitute into (1) and get I2 I1 = - I2 (2 + j1) / j1 = I2 (-1 + j2) 10 = [j1 - (-1 + j2)(2 + j2)] I2 = (-2 - 4 - j2 + j4 + j1)I2 = (-6 + j3) I2 I2 = 10 / (- 6 + j3) = 10 / (- 6 + j3) Vo = 1.49-153.43o V Problem 8.9 Find Vo in the circuit in Figure P8.9 2 100oV j2 j2 Figure 8.9 1 j2 1 + Vo - Suggested Solution 2 100oV I1 j2 j2 I2 1 j2 1 + Vo 10 = (2 + j2)I1 + j2 I2 (1) (2) 0 = j2I1 + (2 +2j)I2 Solve (2) for I1 and substitute into (1) and get I2 I1 = - I2 (2 + j2) / j2 = - I2 (2 - j1) 10 = [j2 - (1 - j1)(2 + j2)] I2 = (j2 -2 - 2 + j2 - j2)I2 = (6 - j3) I2 I2 = 10 / (- 4 + j2) = 5 / (- 2 + j) Vo = 2.24-153.43o V Problem 8.10 Find Vo in the network in Figure P8.10 1 - j1 j1 1 j1 + 100oV j2 j2 1 Vo - Figure P8.10 Suggested Solution 1 - j1 j1 1 j1 + 100oV I1 j2 j2 I2 1 Vo - 10 = (1 + j1)I1 - j1 I2 (1) 0 = - j1I1 + (2 +j3)I2 (2) Solve (2) for I1 and substitute into (1) and get I2 I1 = I2 (2 + j3) / j1 = I2 (3 - j2) 10 = [- j1 - (1 + j1)(3 - j2)] I2 = (3 + 2 + j3 - j2 - j1)I2 = (5) I2 I2 = 2A Vo = (1) I2 = 20o V Problem 8.11 Find Vo in the network in Figure P8.11 - j1 6K j4 j6 240o V Figure 8.11 10 + 2 Vo j1 Suggested Solution - j1 6K I1 j4 j6 I2 240o V (6 + j3) I1 - jI2 = 0 -jI1 + (12 + j6)I2 = -24 Vo = 2I2 + 24 I1 = jI2/(6 + j3) => -j ( j / (6 + j3) I2 + (12 + j6) I2 = - 24 I2 = -24(6 + j3) / (55 + j72) = - 1.78-26.1o Vo = 2 I2 + 24 = 20.864.32o V 10 + 2 Vo j1 Problem 8.12 Find Vo in the network in Figure P8.12 - j1 3 j2 2 - j1 + 2 160 V o j4 Figure P8.12 j6 2 Vo - Suggested Solution - j1 3 j2 2 - j1 + 2 240 V o j4 j6 2 Vo - Forming a Thevenin equivalent at the input Voc = 16(2)/(2 - j) = 32 / (2 - j) 3 Zth 2 Voc and Zth = -2j / (2 - j) j2 2 - j1 + j4 j6 2 Vo - 32 / (2 - j) = I1 ((-2j) / (2 - j) + 3 + j4) - J2I2 0 = -j2I1 + (4 + 5j) I2 => I1 + (4 + 5j) I2 => I1 = [(4 + 5j) / J2] I2 32 = [ (30 + 3J)(4 + 5J) / J2 - (2 + j4)] I2 => I2 = 64j / (33 + j58) Vo = 2I2 = 1.9229.64oV Problem 8.13 Find Vo in the circuit in Figure P8.13 j2 2 240oV - j2 j4 6 8 j8 4 + Vo Figure P8.13 Suggested Solution j2 2 240oV - j2 j4 6 8 j8 4 + Vo Thevenin equivalent at the input Voc = 24(-j2) / (2 - j2) = - 24j (1 - j) and Zth = -4j / (2 - j2) = -2j / (1 - j) j2 Zth Voc I1 j4 6 8 j8 I2 4 + Vo - 24j / (1 - j) = I1 ( -2j / (1 - j) + 6 + j4) + j2 I2 0 = j2 I1 + (12 + j8) I2 => I1 = ( -4 + j6) I2 -24j = [(-4 + j6)(10-j4) + 2 + j2] I2 I2 = -24j / (-14 + j78) = 24j / (14 - j78) and Vo = 4I2 = 1.21169.8oV Problem 8.14 Find Io in the circuit in Figure 8.14 4 - j4 j2 240oV Io 2 - j2 j3 3 Figure P8.14 Suggested Solution 4 - j4 240oV I1 2 j2 - j2 j3 I2 Io 3 Figure P8.14 24 = (6 - j2) I1 - (2 + j1) I2 0 = - (2 + j1) I1 + (5 + j) I2 I1 = (5 + j) / (2 + j) I2 24 = [ (6 - j2)(5 + j) / ( 2 + j) ] I2 - (2 + j) I2 I2 = 1.7842oA Problem 8.15 Write the mesh equations for the network in Figure P8.15 R1 V1 I1 jL1 I2 j jL2 1/(jC1) Figure P8.15 Suggested Solution R1 V1 I1 jL1 I2 jL2 1/(jC1) V1 = R1 I1 + jL1(I1 - I2) - j2 0 = jL1(I2 - I1) + j2 + I2(1/(jC1)) + jC2I2 + j(2- I1) Problem 8.16 Write the mesh equations for the network shown in Figure P8.16 R1 I2 I3 jL2 1/(jC1) V1 I1 jL1 R2 j Suggested Solution R1 I2 I3 jL2 1/(jC1) V21 V1 I1 jL1 R2 j Induced voltages are V12 = + j3 and V21 = + j(I1 - I2) V1 = I1(R1 + jL1) - jL1I2 + I3 j 0 = -jL1I1 + I2 ( R2 + jL1 - j / (C1) ) - I3(R2 + j) 0 = j1 - (R2 + j)2 + 3 (R2 + jL2) Problem 8.17 Write the mesh equations for the network shown in Figure P8.17 1/(jC1) jL1 jL2 I1 I1 I1 jL3 j Suggested Solution 1/(jC1) jL1 jL2 jL3 + V21 - I1 I1 I1 j Induced voltages are V12 = - j3 and V21 = + jI1 - V1 = I1(R1 + jL1) - jI3 0 = I2 ( jL2 - j / (C1) ) - jL2I3 0 = + j1 - jL22 + 3 (R2 + jL2 + jL3) Problem 8.18 Write the mesh equations for the network shown in Figure P8.18 R1 jL1 1/(jC1) I2 1/(jC2) V1 I1 R2 I3 jL3 Figure P8.18 R4 j R3 R5 V2 Suggested Solution R1 jL1 1/(jC1) I2 1/(jC2) V1 I1 R2 I3 jL3 V1 - V2 = I1 ( R1 + R2 + R5 + 1/(jC1) ) - 1/(jC1)I2 - R2 I3 0 = - 1/(jC1) I1 + I2 ( R3 + jL1 + 1/(jC1) + 1/(jC2) ) + I3 ( j - 1/(jC2) ) V2 = - R2 I1 + I2 ( jM - 1/(jC2) + I3 (R2 + R4 + jL3 + 1/(jC2) ) R4 j R3 R5 V2 Problem 8.19 Write the mesh equations for the network shown in Figure P8.19 R1 jL1 j1 j2 jL3 1/(jC2) jL2 I3 R1 I2 1/(jC1) V I1 j3 Figure P8.19 Suggested Solution R1 jL1 j1 j2 jL3 1/(jC2) 1/(jC1) Induced voltages (positive @ dot) IN L1 L1 L2 L2 L3 L3 R1 VIA M1 M3 M1 M2 M2 M3 EXPRESION jM1(I3 - I2) jM1(I3 - I1) jM1(I1 - I2) jM1(I3 - I1) jM1(I3 - I2) jM1(I1 - I2) NAME V11 V13 V21 V22 V32 V33 I2 V I1 j3 jL2 I3 Mesh equations V = I1 (R1 + jL1 + jL3 +1/(jC2)) - I2jL1 - I3jL3 + V11 + V13 - V32 -V33 0 = - I1jL1 + I2(jL1 + jL2 + 1/(jC1)) - I3jL2 + -V11 -V13 - V21 -V22 0 = - I1jL1 + I2(jL2 + I3(R2 + jL2 + jL3 ) + V21 + V22 + V32 + V33 Let Z1 = R1 + jL2 + jL3 + 1/(jC2); Z2 = jL1 + jL2 + (1/jC1); Z3 = R2 +jL2 + jL3 Substituting from the table above: V = I1(Z1 - jZM3 + I2[jM(M2 + M3 - M1 - L1)] + I3[j(M1 + - M2 + M3 - L2)] 0 = I1[ j(M3 - M1 + M2 - L1)] + I2[Z2 + jM1 ) - I3[j(M1 + M2 + M3 + L2)] 0 = I1[ j(M1 - M2 + M3 - L2)] - I2[ j(M1 + M2 + M3 + L2)] + I3(Z3 + j2M2) Problem 8.20 Find Vo in the network in Figure P8.20 -j2 1K j4 j2 j4 -j1 10oA + Vo - Suggested Solution -j2 j2 1K I1 j4 + V12 + V21 j4 I2 I3 -j1 10oA + Vo Induced voltages: V12 = -j2I2 V21 = +j2I1 0 = I1 ( 4 + j2) - j2I2 and 0 = I2(j3) + j1I3 - j2I1 and I3 = - 10oA From 1st equation: I1 = I2[ 1 / (1 - j2) ] Now: = 0 I2[j3 - j2 (1 - j2) ] + j1I3 and I3 = - 1 or: I2 [ (6 + j3 - j2) / (1 - j2) ] = j1 or I2 = (2 + j1) / (6 + j1) = 0.3717.10oA Vo = (I2 - I3) ( -j1) = [ 0.3717.10o + 10o] (1-90o) Vo = 1.36-85.4oV Problem 8.21 Find Vo in the network in Figure P8.21 j1 -j1 100oA 1 2 j2 j1 1 1 -j1 + Vo - Figure P8.21 Suggested Solution j1 -j1 100oA I1 1 I2 2 j2 j1 I3 1 1 -j1 + Vo = (1)I3 I1 = 100oA (1) 0 = - I1 + I2(3 + j2) + J1I3 (2) 0 - +j1I2 + I3(2) (3) Solve (3) for I2 and substitute into (2) along with (1). I2 = (j2)I3 0 = -10 + [(3 + j2)(j2) + j1]I3 => I3 = 10 / (-4 + j7) = 1.24-119.74oA Vo = (1) I3 = 1.24-110.74oV Problem 8.22 Find Vo in the network in Figure P8.22 1 j2 -j1 100oA j1 1 j1 2 2 j2 + Vo - Figure P8.22 Suggested Solution 1 A j2 1 j1 j1 + -j1 100oV 2 2 Vo j2 A' Thevenin Eq. at A - A' Voc = (- j1)/(1 - j1) (10) = 7.07-45oV Zth = 1 || (-j1) = 1/2 - j1/2 1 100 V o + -j1 Voc - 1/2 -j1/2 j2 j1 j2 + 2 I2 2 Vo 7.07-45oV I1 j1 1 - 7.07-45o = I1 ( 1.5 + j2.5) - J1 I2 0 = - j1 I1 + I2 (4 + j2) (1) (2) Solve (2) for I1 and substitute into (1), solve for I2 7.07-45o = [ (2 - j4) (1.5 + j2.5) - j1)] I2 = [13 - j2] I2 I2 = (7.07-45o ) / (13 - j2) = 7.54-36.25oA Vo = 2I2 = 1.08-36.25oV Problem 8.23 Find Vo in the network in Figure P8.23 1 + - j1 10 A o 2 j2 - j2 j2 - Vo Figure P8.23 j1 Suggested Solution A 1 + - j1 10 A o 2 A' Thevenin Eq. at A-A' j1 j2 - j2 j2 - Vo + - j1 10oA Voc Zth Voc = (-j1) (1) = 1-90oV Zth = -j 1 New Circuit 1 1-90 V 2 j2 - j2 j2 Vo = 2 I o I + j1 1-90oV = I (3 - j1 + j2 - j2 + j2) - j1 I - j1 I (induced) I = 1-90o / (3 - j1) = 0.32-71.57oA Vo = 2I = 0.64-71.57oV Problem 8.24 Find Vo in the network in Figure P8.24 j1 2 j2 - j2 100oV Figure P8.24 - j2 2 j2 + Vo - Suggested Solution j1 2 A j2 - j2 100oV A' Thevenin Eq. at A - A' + - j2 New Circuit j1 1 - j1 j2 - j2 I1 7.07-45oV 7.07-45o = I1 ( 1 - j1 + j2 - j2 + j2 + 2) - j1 I1 - j1 I1 (induced voltages) I1 = (7.07-45o) / ( 3 - j1) = 2.24-26.57oA 2 j2 + Vo = 2 I1 Voc Zth - j2 2 j2 + Vo 2 100oV Voc = 10 ( - j2) / (2 - j2) = 7.07-45oV Zth = 2 || (- j2) = 1 - j1 Vo = 2I1 = 4.48-26.57o Problem 8.25 Find Vo in the network in Figure P8.25 j2 + 1000oV 2000oV j5 10 - j10 j2 2 Vo Figure P8.25 Suggested Solution A j2 1000 V o 2000 V j5 o j2 2 + Vo - 10 - j10 A' + 1000 V o I = (100 - 200) / (10 - j10) = 7.07-135oA Voc = 200 - j10 I = 150 + j50 V 2000 V Voc Zth New circuit j2 5 - j5 I1 j5 10 j2 I2 o I1 10 - j10 Zth = 10 || (-j10) = 5 - j5 + Vo = 2I2 - Voc Voc = I1 ( 5 - j5 + j5) - j2 I2 0 = - j2I1 + I2 (2 + j2) (1) (2) Solve (2) for I1 and substitute into (1) to get I2 I1 = I2 (1 - j1) Voc = [(1 - j1)5 - j2] I2 = (5 - j7) I2 I2 = (150 + j50) / (5 - j7) = 18.3872.90oA Vo = 2 I2 = 36.7672.90oV Problem 8.26 Find Vo in the network in Figure P8.26 j2 j4 2 - j2 j2 480oV 120oV Figure P8.26 j4 2 4 6 Vo + Suggested Solution j2 j4 2 - j2 j2 480oV 120oV j4 2 4 6 Vo + Thevenin equivalent at the input Voc = [ (48 - 12) / (2 - j2 + j2) ] (j2 + 12) = 12 + j36 Zth = (2 - j2) (j2) /2 = 2 + j2 j2 2 12 + j36 j2 2 j4 4 j4 6 + Vo - 12 + j36 = (4 + j6)I1 - j2 I2 0 = -j2I1 + (10 + j4)I2 I1 = [ (10 + j4) / j2 ] I2 = (2 - j5) I2 12 + j6 = [ (4 + j6)(2 - j15) - j2 } I2 I2 = (12 + j36) / (38 - j10) Vo = 6I2 = 6 (12 + j36) / (38 -j10) V = 5.7986.31oV Problem 8.27 Find Vo in the network in Figure P8.27 4 - j4 100oV 2030oV j2 4 j4 j2 1 Vo j1 2 + Figure P8.27 Suggested Solution Thevenin Eq. - j4 + 100oV I 4 j4 2030oV Voc I = (100o - 2030o) / 4 = 12.39/4 -126.21o Voc = 2030o j4I = 27.455.60o Zth = j4 || (4 - j4) = 4 + j4 New circuit 10 j4 1 j1 2 I2 1 + Vo - Voc I1 j2 j2 Voc = I1 (5 + j6) - j1 I2 and 0 = (-j1)I1 + I2 (3 + j2) Solve (2) for I1, substitute into (1) to get I2 and Vo I1 = I2 (2 - j3) => Voc = [ (2 - j3)(5 + j6) - j1] I2 I2 = Voc / (28 - j4) = 0.9713.73o Vo = (1) and (2) 0.9713.73oV Problem 8.28 Find Vo in the network in Figure P8.28 j2 1 100oV 1 1330oA - j1 j2 1 1 j1 1 Vo - j1 + Figure P8.28 Suggested Solution 100o j2 1 Source 1330oA Transform - j1 1 1330 A - j1 o 1 100o j1 + Zeq Voc Zth j2 1 Zeq Ieq = 1030o + 100o = 19.3215o Ieq = (1 - j1) || 1 = (1 - j1) / (2 - j1) Voc = Ieq Zeq = 12.22-3.43oV Zth = Zeq - j1 + Voc I1 j2 j1 I2 1 Vo 1 Voc = I1 (Zth + 1 + j2) + j1 I2 and 0 = jI1 + 2I2 I1 = j2 I2 and Voc = [j2 (Zth + 1 + j2) + j1] I2 I2 = [Voc / (-3 + j12) ](2 - j1) and I2 = 2.2134.1oA and Vo = 2.21-134.1oV Vo = 2.21-134.1oV Problem 8.29 Find Vo in the network in Figure P8.29 j2 2 - j2 100oV Figure P8.29 j1 1030oV 1 + - j1 Vo j1 j2 Suggested Solution j2 2 - j2 j1 1030 V j1 I1 j2 I2 I3 100oV o 1 + - j1 Vo 0 = I1 ( 2 + j3) - j1 I2 -1030o = - j1 I1 + I2(-j1) (1) From (3) and (4) (2) we see that Vo is independent of I1 and I2 ! 1030o - 100o = I3 (1 - j1) (3) Vo = (-j1)I3 + 100o (4) From (3), I3 = (1030o - 100o) / ((2)-45o) = 3.66150o A From (4), Vo = 3.6660o - 100o Vo = 8.76158.80oV Problem 8.30 Find Io in the circuit in Figure P8.30 j1 1 j2 j2 100oA 1 2 Io Figure P8.30 Suggested Solution j1 1 j2 I1 2 j2 I2 1 10 = I1(3 - j2) - J1I2 - 2I2 0 = I2(3 + j2) - j 1I1 - 2I1 Solve (2) for I1 Io I1 = I2(3 + j2) / (2 + j1) (3) (1) (2) Substitute (3) into (1) to find I2 10 = [ (3 + j2)(3 + j2) / (2 + j1) - 2 - j1] I2 = I2 [ (9 - 4 + j12 - (2 + j1)(2 + j1) ) / (2 + j1)] I2 = Io = 10 (2 + j1) / (2 + j8) 2 = =2.71-49.40o A Problem 8.31 Find Vo in the circuit in Figure P8.31 j1 1 j2 1230 A 100oV 2 o 1 j2 + Vo - Figure P8.31 Suggested Solution Try to simplify the left side of the circuit 100oV 1 1030 A o Source 100oA transform 1 1030oA Add current sources Ieq 1 Ieq = 100 o + 1030o Ieq = 19.3215oA Source 1 19.3215oV transform New Circuit j1 1 j2 j2 19.3215 V o 1 + I2 Vo - I1 2 Vs = I1 (2 + j2) + j1 I2 - I2 (1) 0 = j1 I1 - I1 + I2 (3 + 1) (2) Solve (2) for I1, substitute into (1) to get I2 and Vo I1 = I2 (3 + j1) / (1 - j1) => Vs = [ (2 + j2) (3 + j1) / (1 - j1) + j1 -1] I2 Vs = I2 [ j2(3 + j1) + j1 - 1] = I2 [ -3 + j7] 2 = 2.54-98.20o A Problem 8.32 Find Io in the circuit in Figure P8.32 - j1 100oA 1 - j2 Figure P8.32 j2 j2 3 j2 - j3 Io Suggested Solution 1 - j1 100 V I1 - j2 o j2 - j3 j2 j2 Io 3 Io 10 = I1 ( 1 - j1) + Io (j2 + j1) 0 = I1 ( j2 + j1) + Io (3 - j3) (1) (2) Solve (2) form Io substitute inti (1) to get Io I1 = Io (1 + j1) => 10 = Io [ (1 -j1)(1 + j1) + j3] = Io (2 + j3) Io = 10 / ( 2 + j3) Io = 2.78-56.31oA Problem 8.33 Find Vo in the network in Figure P8.33 j2 4 j2 100oV 50oA j2 + - j4 2 50oV 1 Vo - j1 Figure P8.33 Suggested Solution Thevenin Eq. of left-side of circuit 4 By superposition: + 50oV 102135oV - j1 Voc Zth Voc = 102135o [ -j4 / (4 - j4)] + 5 [4 || -j4] Voc = 100oV Zth = 4 || -j4 = 2 - j2 New circuit 4 j2 100oV I1 j2 + - j4 2 50oV I2 1 Vo - j2 j2 - j1 5 = I1 (4) - I2 (j2 + 2) (1) 5 = - I1 (2 + j2) + I2 (3 + j1) (2) Solve (1) form I1, substitute into (2) to get I2 and Vo I1 = [5 + I2 (2 + j2)] / 4 => 5 = I2 [ (3 + j1) - (2 + j2)(2 + j2) / 4 ] - (2 + j2)(5/4) I2 (3 -j1) = 7.5 + j2.5 => I2 = (7.5 + j2.5) / (3 - j1) = 2.536.87oA Vo = (1) I2 = 2.536.87o V Problem 8.34 Find Vo in the network in Figure P8.34 j1 j1 1 100oA 2 Figure P8.34 j1 1 100oA Vo - j1 + Suggested Solution Thevenin eq. for left and right side of circuit + 100oA Voc Zth New circuit j1 1 j1 100oV I1 2 j1 I2 - j1 + Vo 100oV 10 = I1(3 + j1) - I2 (2 + j1) - 10 = - I1(2 + j1) + I2(3) solve (2) for I1 I1 = (10 + 3I2) / (j1 + 2) 10 = (10 + 3I2)[ (3 + j1) / (2 + j1) ] - I2 (2 + j1) = 10 [ (3 + j1) / (2 + j1) ] + I2 (6 - j1) / (2 + j1) I2 = [ 10(2 + j1) - 10(3 + j1)] / (6 - j1) = 1.64-170.54oA (1) (2) 1 Voc = 100oV Zth = 1 Vo = (1)I2 + 10 = 8.38-1.85o V Problem 8.35 Determine the impedance seen by the source in the network shown in Figure P8.35 1 - j1 1200 V o j2 2 j1 j4 - j2 j2 j4 Figure P8.35 Suggested Solution 1 - j1 1200 V o j2 2 j1 j4 I3 - j2 I1 j2 I2 j4 120 = I1 (1 + j2) - j2 I2 ; 0 = -j2I1 + j5I2 + j2 I3 ; 0 = j2I2 + I3(2 + j3) solve (3) for I2 and usbstitute into (1) and (2) (1) (2) and (3) I2 = I3(-3/2 + j1) => 120 = I1(1 + j2) + I3(2 + j3); 0 = -j2I1 + I3(-5 -11/2 j) (4) and (5) Solve (5) for I3 and substitute into (4) I3 = +j2I1 / (-5 -j11/2) = -j4I1 / (10 + j11) 120 = I1 [ 1 + j2 + (2 + j3)(-j4) / (10 + j11)] = I1[(10 + j11)(1 + j2) + 12 - j8] / (10 + j11) 120 = I1 [ 1 + j2 + (2 + j3)(-J4) / (10 + J11)] = I1 [(10 + J11)(1 + J2) + 12 - J8] / (10 + J11) Zsource = 120 / I1 = ( 10 - 22 + j11 + j20 + 12 - j8) / (10 + j11) = (j 23) / (10 + j11) Zsource = 1.5642.27o Problem 8.36 Determine the impedance seen by the source in the network shown in Figure P8.36 1 - j1 1200oV j1 j2 Figure P8.36 j2 j1 j2 1 Suggested Solution 1 - j1 1200oV I1 j1 I2 j2 j2 I3 j1 j1 1 120 = I1 (1 + j1) - j1 I2 ; 0 = -j1I1 + j2I2 - j1 I3 ; 0 = j1I2 + I3(1 + j3) solve (3) for I2 and usbstitute into (1) and (2) (1) (2) and (3) I2 = I3(-3 - j1) => 120 = I1(1 + j1) + I3(-1 - j3); 0 = -j1I1 + I3(2 5 j) (4) and (5) Solve (5) for I3 and substitute into (4) I3 = +j1I1 / (2 + j5) 120 = I1 [ 1 + j1 - (1 + j3)(j1) / (2 + j5)] = I1[j6 / (2 + j5) Zsource = 120 / I1 = (j 6) / (2 + j5) Zsource = 1.1121.8o Problem 8.37 Determine the input impedance Zin of the circuit in Figure P8.37 1 j1 Zin j2 - j2 2 j1 1 Figure P8.37 Suggested Solution Thevenin's eq. at right side of circuit 1 Zth = 1 + (2 || -j2) = 1 - (j4) / (2 - j2) = 1 + 1 - j1 = 2 - j1 Zth - j2 2 New circuit 1 j1 j1 Vx Ix j2 I - j1 Vx = Ix (1 + j3) - j2I + j1Ix + j1(Ix - I) 0 = I (2 + j1) -j2Ix - j1Ix 2 Solve (2) for I and substitute into (1) I = j3Ix / (2 + j1) => Vx = Ix [ (1 + j3 + 2) (-j3)(j3) / (2 + j1) ] Vx = Ix [ 1 + j5 + (9) / (2 + j1) ] = [ (2 + j10 + j1 -5 + 9) / (2 + j1) ] Ix = (6 + j11) / (2 + j1) Ix Zin = Vx / Ix = (6 + j11) / (2 + j1) = 5.6034.820o Problem 8.38 Determine the input impedance Zin in the network in Figure P8.38 1 Zin - j2 - j1 1 j2 j1 j2 Figure P8.38 Suggested Solution Zin 1 Vx Ix - j2 j2 I2 - j1 j1 j2 I3 1 Zin = Vx / Ix Vx = Ix (1 - j2) + j2 I2 ; 0 = j2Ix - j1I2 + j2 I3 ; 0 = j2I2 + I3(1 + j1) solve (3) for I2 and usbstitute into (1) and (2) (1) (2) and (3) I2 = I3(-1 + j1) / 2 => Vx = Ix (1 - j2) + I3(-1 - j1); 0 = j2Ix + I3(1/2 +5/2 j) (4) and (5) Solve (5) for I3 and substitute into (4) I3 = +j4Ix / (1 + j5) => Vx = Ix [ 1 - j2 + (1 + j1)(j4)/(1 + j5) ] Zin = Vx / Ix = [(1 - j2)(1 + j5) - 4 + j4 ] / (1 + j5) = (7 + j 7) / (1 + j5) Zin = 1.94-33.69o Problem 8.39 Given the network shown in Figure P8.39 determine the value of the capacitor C that will cause the impedance seen by the 240 V voltage source to be purely resistive. f = 60Hz. j = j6 12 4 1/(jC) 240oV j1 j50 j6 10 Figure P8.39 Suggested Solution Zin 12 j = j6 4 1/(jC) 100oV j1 I1 j50 I2 j6 10 Zin = 24 / I1 24 = I1 (12 + j1) - j6 I2 and 0 = -j6I1 + I2 (14 + j(56 -Xc)) (1) and (2) Solve (2) for I2 and substitute into (1) I2 = j6I1 / [14 + j(56-Xc)] => 24 = I1 [ 12 + j1 + 36 / (14 + j(56 -Xc) ] Zin = 24 / I1 = 12 + j1 + 36 / [14 + j(56 - Xc)] = Rin + j0 Thus, 36 / [14 + j(56 - Xc)] = R - J1 where Rin = 12 + R (3) and (4) 36 = 14R + 56 -Xc and 0 = R(56 - Xc) - 14 Solve (3) for Xc, substitute into (4) to find R Xc = 14R + 20 => 0= R(56 - 14R - 20) - 14 => 14R2 - 36R + 14 = 0 R = {2.094 , 0.478 } Xc = {49.316 , 26.692 } C = {53.79F, 99.38F} Rin = {14.09 , 14.48 } Problem 8.40 Analyze the network in Figure P8.40 and determine if a value of Xc can be found such that the output voltage is equal to twice the input voltage. 1 j1 V I1 j2 I2 - jXc Vo Figure P8.40 j2 + Suggested Solution 1 j1 V I1 j2 I2 - jXc Vo V = I1 (1 + j2) - J2 I2 + j1 I2 0 = - j2I1 + I2(j4 - jXc) + j1 (I1 - I2) (j1)I1 = I2 j(3 -Xc) => V = I2[(3 - Xc)(1 + j2) -j1] But I2 - Vo / (-jXc) so V = [ Vo / (-jXc) ] (3 + j6 -Xc - j2Xc - j1) And Vo/V = (-jXc) / (3 - Xc + j(5 - 2Xc) ) = Xc / [-5 + 2Xc + j(3 - Xc) ] For no phase shift between Vo and V1 Then 3 - Xc = 0 => Xc = 3 Vo / V = Xc / (-5 + 2Xc) = 3 / (-5 + 6) = 3 No value of Xc exists such that Vo = 2V If we allow phase shift, then, | Vo/V | = 2 = Xc / (-5 + 2Xc)2 + (3 -Xc)2 => 4 (34 - 26Xc + 5X2c) = X2c 19 X2c - 104Xc + 136 = 0 and Xc = 3.31 and 2.14 There are 2 values of Xc for |Vo/V| = 2 Vo/V = (3.31) / (1.62 -0.31j) = 210.8o and Vo/V = (2.161) / (-0.678 + j0.839) = 2128.9o j2 + Vo = I2 (-j Xc) Problem 8.41 Two coils in a network are positioned such that there is 100% coupling between them. If the inductance of one coil is 10 mH and the mutual inductance is 6mH, compute the inductance of the other coil Suggested Solution k= L2 = M L1 L2 1 L1 k = 1, M = 6mH , L1 = 10mH ( M ) 2 = 3.6mH k Problem 8.42 The currents in the magnetically coupled inductors shown in Figure P8.42 are known to be i1(t)=8cos(377t-20)mA and i2(t)=4cos(377t-50)mA. The inductor values are L1 = 2H, L2 = 1H, and k=0.6. Determine v1 (t) and v2(t) i1(t) + v1(t) Figure P8.42 i2(t) L1 L2 + v2(t) - Suggested Solution For the coupled inductors, i1(t) = 8cos(377t - 20) mA, and i2(t) = 4cos(377t - 50) mA, L1 = 2 L2 = 1,K = 0.6, and v1(t), v2(t). A + v1(t) i2(t) L1 i1(t) L2 + v2(t) B Mutual coupling Then v1(t ) = L1di1( t ) dt M = 0.6 2.1 = 0.85 -M di 2 ( t ) dt d d v1(t ) = 2 dt (8cos(377t - 20)) - 0.85 dt (4 cos(377t - 50))mA v1(t ) = -6sin(377t - 20) + 1.3sin(377t - 50) v 2(t ) = - L 2 di 2 ( t ) dt +M di 1( t ) dt v 2(t ) = 1.6sin(377t - 50) - 2.6sin(377t - 20) Problem 8.43 Determine the energy stored in the coupled inductors in Problem 8.42 at t=1ms Suggested Solution i1(t) + A v1(t) i2(t) Find energy stored in inductors of P8.42 at t=1ms At t = 1ms, t = (377)(0.001) = 0.337rad = 21.6 then, i1(t=0.001) = 8cos(21.6 - 20) = 8.0 mA i2(t=0.001) = 4cos(21.6 - 50) = 3.52mA and (t=0.001) = 1/2 (2) (2mA)2 + 1/2(1)(3.52mA)2 - 0.85(8)(3.53mA)2 L1 L2 + v2(t) B (t=0.001) = 46.3 J Problem 8.44 If L1 = L2 = 4H and k = 0.8, find i1(t) and I2(t) in the circuit in Figure P8.44 i1(t) 200 1cos(100t)V L1 Figure P8.44 L2 320 M i2(t) Suggested Solution Convert to frequency domain! 1cos(100t) -> 10 ; i1(t) -> I1 ; i2(t) -> I2 M = k (L1L2)1/2 = 3.2H j320 i1(t) 200 10 I1 j400 i2(t) j400 I2 320 10 = -I1 (200 + j400) + j320 I2 and 0 = I2(320 + j400) -j320I1 10 = [(-1.25 + j1)(200 + j400) + j320] I2 I1 = I2 (1.25 -j1) I2 = 1 / (-650 + j20 ) = 1.54-178.24 mA I1 = I2 (1.25 - j1) = I2 (1.60-38.66) = 2.46143.10 mA i2(t) = 1.54 cos(100t -178.24) mA i1(t) = 2.46 cos(100t +143.10) mA Problem 8.45 Determine the energy stored in the coupled inductors in the network in P8.44 Suggested Solution M i1(t) L1 L2 i2(t) L1 = L2 = 4H k = 0.8 M = 3.2H From problem 11.32, i1(t) and i2(t) as defined here are i1(t) = - 2.46 cos(100t + 143.10) mA i2(t) = 1.54 cos (100t - 178.24) mA (t) = 1/2 L1 i21(t) + 1/2 L2 i22(t) + Mi1(t)i2(t) Evaluate at t = 2ms (0.002) = 3.71 J Problem 8.46 Find all currents and voltages in the network in figure P8.46 I1 2 2430 V o 1:2 + V1 + V2 - I2 12 j4 Figure P8.46 Suggested Solution 1:2 Find I1, I2, V1, V2 For the secondary Z = (12 + 4j) / (22) = 3 + j 2430oV I1 In the primary, I1 = (2430o) / (2 + 3 + j) = 4.718.7oA Then V1 = I1Z1 = (4.718.7o)(3.218.4o) = 1537.1oV And V2 = 2V1 = 3032.1oV, I2 = I1/2 = 2.3518.7oA V1 2 + + V2 j4 12 I2 Problem 8.47 Determine I1, I2, V1, and V2 in the network in Figure P8.47 I1 2 1030oV + V1 Ideal Figure P8.47 2:1 + V2 2 I2 Suggested Solution 1 I1 + V1 1030oV Ideal I1 = (1030o) / (1 + Z1) = (10/9)30oA = 1.1130oA I2 = - I1 / n = - 2.2230oA = 2.22-150oA 2:1 + V2 2 ZL = 2 , n = L/2, Z1 = Zl/n2 = 8 V1 = 1030o[Z1 / (Z1 + 1)] = 8.8930oV I2 V2 = - n V1 = 4.44-150oV Problem 8.48 Determine Vo in the circuit in Figure P8.48 1:2 1 1030 V j2 Figure P8.48 Ideal o I2 1 + -j1 Suggested Solution 1:2 1 1030 V j2 Ideal o I2 1 + n=2 -j1 4 1030oV -j4 1 j2 + Vo - Vo = -2030o [j2 / (5 - j2) ] Vo = 7.43-38.20oV Problem 8.49 Determine I1, I2, V1 in the network in Figure P8.49 I1 2 1030oV -j1 Ideal Figure P8.49 + V1 1:2 I2 + V2 j4 4 Suggested Solution A I 1 2 + V1 1030oV -j1 1:2 I2 + V2 j4 4 Zl = 4 + j4 n = 2 Z1 = Zl / n2 = 1 + 1 A' Thevenin Eq. at A-A' Voc = 1030o[ -j1/(2 - j1) ] = 4.47-33.43oV Zth = 2 || (-j1 ) = (-j2)/(2 - j1) = 0.89-63.43o Zth 2 -j1 1030oV + Voc - New Circuit I1 Zth Voc Z1 V1 = I1 Z1 = 4.473.44oV V2 = n V1 = 8.943.44oV I2 = - I1 / n = 1.58138.44oA + I1 = Voc / (Zth + Z1) = Voc / (0.4 -j0.8 + 1 + j1) = Voc / (1.4 + j0.2) V1 I1 = 3.16-41.56oA Problem 8.50 Determine I1, I2, and V1 in the network in Figure P8.50 1 I1 + -j1 100oV V1 Ideal Figure P8.50 1 j1 1:2 2 + V2 -j1 I2 Suggested Solution 1 I1 A 1 + -j1 100oV A' V1 + VA Ideal 1:2 + VB 2 j1 + V2 -j1 I2 ZL = 2 + (j1 || (-j1) = 2 n=2 ZA = ZL/n2 = 1/2 Thevenin equivalent at A - A' 1 + 100oV -j1 Zth Voc Voc = 10[ (-j1) / (1 - j1) = 5 -j5 = 7.07-45oV Zth = 1 || (-j1) = (-j1) / (1 - j1) = 1/2 -j 1/2 New circuit 2 Zth + ZA Voc V1 VA IA + IA = Voc / (Zth + 1 + ZA) = 7.07-45o / (2 - 1/2j) IA = 3.43-30.96oA V1 = IA(1 + ZA) = 3/2 IA = 5.15-30.96oV I1 = V1 / (-j1) + IA = 5.1559.04o + 3.43-30.96o = 6.1925.380oA VB = -nVA = -2 (IA ZA) = 3.43-149.04oV IB = -IA / n = 1.72149.040oA In the secondary since j1 || (-j1) = , V2 = VB = 3.43149.04oA I2 = V2 / (-j1) = 3.43-120.96oA Problem 8.51 Determine I1, I2, V1, and V2 in the network in Figure P8.51 I1 3 1000oV V1 + 1:4 I2 + V2 -j80 32 2200oV Ideal Figure P8.51 Suggested Solution n=4 2200oV -j80 I2 V2 = 220 + 32I2 = 3080oV 48 4400oV + V2 32 I2 = (440 - 220) / (80) = 2.750oA V1 = V2/n = 770oV I1 = -nI2 = 11180oA Problem 8.52 Determine the input impedance seen by the source in the circuit in Figure P8.52 1 1:2 + Vs V1 + V2 j1 2 Ideal Figure P8.52 Suggested Solution 1 1:2 Vs j1 2 ZL = 2 + j1 n = 2 Z1 = ZL / n2 = (2 + j1) / 4 = 0.5 + j0.25 Zin = 1 + Z1 = 1.5 + j0.25 Zin Problem 8.53 Determine the input impedance seen by the source in the circuit in Figure P8.53 I1 1 Vs -j1 + V1 Ideal Figure P8.53 4:1 + V2 -j2 2 4 Suggested Solution I1 1 Vs -j1 + V1 Ideal Zin Zin = 1 + (Z1 || (-j1) ) = 1 + (-16 -j32) / (32 - j17) = (16 - j49) / (32 - j17) 4:1 + V2 -j2 2 4 ZL = 1 + 2 || (-j2) = 2 - j1 n = 1/4 Z1 = ZL/n2 = 32 - j16 Zin = 1.42-43.94o Problem 8.54 Determine the input impedance seen by the source in the network shown in Figure P8.54 20 2:1 Vs j1 Ideal Figure P8.54 Ideal 2 -j2 1:4 2 Suggested Solution Zs 20 2:1 Vs j1 Ideal Z1 ZL2 Z2 Ideal ZL1 2 -j2 1:4 2 ZL1 = 2 + j1 n2 = 1/2 Z2 = ZL1/n22 = (2 + j1)/(1/4) = 8 + j4 ZL2 = 2 - j2 + Z2 = 10 + j2 n1 = 1/4 Z1 = ZL2 / n12 = 160 + j32 Zs = 20 + Z1 = 180 + j32 Problem 8.55 Determine the input impedance seen by the source in the network shown in Figure P8.55 -j1 2:1 Vs 2 Ideal Figure P8.55 Ideal -j16 32 j1 1:4 Suggested Solution -j1 2:1 Vs Ideal ZS Z1 ZL1 Z2 Ideal ZL2 -j16 32 n2 = 4 n1 = 1/2 j1 1:4 ZL2 = 32 - j16 Z2 = ZL2 / n22 = 2 - j1 ZL1 = 2 + j1 + Z2 = 4 Z1 = ZL1/n12 = 16 Zs = 16 - j1 Problem 8.56 The output stage on an amplifier in an old radio is to be matched to the impedance of a speaker as shown in Figure P8.56. If the impedance of the speaker is 8 ohms and the amplifier requires a load impedance of 3.2 K ohms determine the turns ratio of the ideal transformer. n:1 Amplifier Ideal Figure P5.56 Suggested Solution n:1 Speaker = 8 R1 = (RSpeaker)n2 = 3200 n2 = 400 n= 20 Problem 8.57 Given that Vo = 4830 V in the circuit shown in Figure P8.57, determine Vs I1 6 -j6 1:2 + V1 Ideal I2 + V1 24 + V - Figure P8.57 Suggested Solution 6 -j6 Vs I1 If Vo = 48.3030oV, find Vs If Vo = 48.3030o, I2 = 230o V1 = -V2 / 2 = - 2430o, I1 = -2I2 = - 430oV Then Vs = V1 + I1 (6 - 6j) 53.66-175.7oV 1:2 24 I2 + Vo - Problem 8.58 If the voltage source Vs in the circuit of Problem 8.57 is 500V, determine Vo. Suggested Solution 6 -j6 Vs I1 If Vs = 500oV, find Vo I1 = (500o) / (6 - 6j + 24/4) = 3.7326.6o Then I2 = - I1/2 = - 1.8626.6o Vo = 24 I2 44.72-153oV 1:2 24 I2 + Vo - Problem 8.59 Determine Vs in the circuit in Figure P8.59 1:3 1 Vs j2 1 j1 + Vo=145oV Ideal - Figure P8.59 Suggested Solution 1:3 1 Vs I1 j2 + V1 Ideal + V2 1 j1 + Vo=145oV n=3 - I2 = Vo / (j1) = 1-45oA V2 = I2(1 + j1) = 1.41V V1 = - V2/n = 0.47180oV I1 = - nI2 = 3135oA Vs = I1 (1 + j2) + V1 = 6.71-161.57o + 0.47180oV Vs = 7.16-162.76oV Problem 8.60 Determine Is in the circuit in Figure P8.60 1:2 -j1 Is 1 Ideal Figure P8.60 1 j1 I2=1-45oA Suggested Solution 1:2 + Is Vs 1 -j1 + V1 I1 V2 = (1 + j1)I2 = 1.410oV I1 = nI2 = 2-45oA Vs = V1 - j1 I1 = 0.7070o + 2-135o = 1.58-116.57oV Is = Vs / 1 + I1 o = 2.91-75.95oA V1 = V2/n = 0.7070oV + V2 1 j1 I2=1-45oA Problem 8.61 Find the current I in the network in Figure P8.61 1:2 8 4 4 4-45oA j2 -j4 I 1260oV Figure P6.61 Suggested Solution 1:2 8 4 4 4-45oA j2 -j4 I 1260oV Refer primary to secondary Isec = Iprim / n Zsec = Zprim n2 8 4 8 2-45oA j8 -j4 I Source transformations 8 j8 8 16(2) 1/20oV -j4 I 360oA Another source transformation [16(2) 1/2]/(16 +j8) 360 A o 16 4 j8 I -j4 Ieq Zeq Zeq = 4 / (16 +j8) = 3.324.76o Ieq = [16(2) 1/2]/(16 +j8) - 360o = 3.19-96.64oA I = Ieq [ Zeq / (Zeq - j4 ] = 2.13-43.50o I = 2.13-43.50oA Problem 8.62 Find the voltage Vo in the network in Figure P8.62 -j4 2:1 240oV 2 Ideal Figure 8.62 Ideal -j8 16 j1 1:4 + Vo - Suggested Solution -j4 2:1 240oV j1 1:4 16 + Vo 2 T1 T2 -j8 Refer primary circuit of T1 to secondary: n1 = 1/2 Vsec = Vprim n1 -j1 Zsec = Zprim n12 j1 120oV 2 T2 Refer primary circuit of T2 to secondary: n2 = 4 (Dots reversed!) 32 Vo = - 48 [ 16 / (48 - j8) ] = -96 / (6 - j1) + 16 -j8 Vo Vo = 15.78189.46oV 16 -j8 1:4 + Vo - Problem 8.63 Find Vo in the network in Figure P8.63 -j6 2:1 4 1200oV Ideal -j8 2:1 4 + Vo 4 j8 Figure P8.63 Suggested Solution -j6 2:1 4 1200oV T1 -j8 Refer primary of T1 to secondary and secondary of T2 to primary. T2 2:1 + 4 Vo n1 = 1/2 n2 = 1/2 4 j8 1 -j2 -j6 + Vo = (-60) (4) / (6 - j2 + j2 - j6) = (-40) / (1 - j1) Vo = 28.28-135o V 600oV 1 j2 4 - Vo Problem 8.64 Find Vo in the circuit in Figure P8.64 -j1.2 1:2 + 1.6 -j8 Ideal 1200oV -j4 Figure P8.64 6 2 Vo -j12 4 Suggested Solution Thevenin Eq. + Voc 4 - Zth Voc = 120 [ 4 / (8 - j4) ] = 120 / (2 - j1) Zth = 4 || (4 - j4) = (4 - j4) / (2 - j1) = 2.4 - j0.8 4 -j4 New circuit -j1.2 1:2 1.6 Zth 6 Voc Refer secondary circuit to primary ZTH 1.6 Voc 3/2 -j1.2 -j3 1/2 + Vo/2 2 - j12 Vo/2 = -Voc [ 0.5 / (3.6 - j4.2 + Zth) ] Vo = 6.81-113.630o V Problem 8.65 Find Vo in the circuit in Figure P8.65 1:2 - j2 1 Ideal 2 240oV 2 2 2 j2 - j1 1.6 + Vo - Figure P8.65 Suggested Solution Two thevenin eq. circuits + Vo Zth2 Zth1 Voc = 120oV Zth1 = 2 || 2 = 1 2 240oV 2 2 2 Zth2 = 2 || (2 + j2) Zth2 = (2 + j2) / (2 + j1) 2 New circuit 1:2 - j1 - j2 1 1 + 2 Zth2 Vo - 120oV 4 -j8 -j1 + 240oV Zth2 1 Vo - Vo = - 24 [ 1 / (9 - j9 + Zth2 ) ] = (-24)(2 + j1) / [ (9 - j9)(2 + j1) + 2 + 2 ] = ( -24 ( 2 + j1) ) / (18 + 9 + 2 + j9 - j18 + j2) Vo = (-24) (2 + j1) / (29 - j7) Vo = 1.80-139.86oV Problem 8.66 In the circuit if Figure P8.66, if Ix = 645A, find Vo. + Vo 4 2 Vs 2 1:2 - j2 4 1:2 4 - j4 Figure P8.66 Suggested Solution 4 2 Vs 2 T1 T2 1:2 - j2 + Vo 4 1:2 4 N2 = 2 N1 = 2 - j4 Ix = 645o Refer primary of T1 to secondary and secondary of T2 to primary 8 2Vs 4 - j2 4 -j1 2 Ix 1 Vx = Ix (2 - j2) = 645o(2 - j2) = 16.9711.31o V Vo = Vx[ 4 / (4 + 1 - j1) ] = 4 Vx / (5 - j1) Vo = 13.3111.31o V Problem 8FE-1 In the network in Figure 8PFE-1 find the impedance seen by the source. 4 1H 24 cos(2t + 0)V k = 0.5 Figure 8PFE-1 M 4H 5 100mF Suggested Solution Since M = k(L1L2)1/2 =(0.5)(2) = 1H 4 j2 24 cos(2t + 0)V j2 j8 5 100mF V1 = (4 + j2)I1 - J2I2 0 = -j2I1 + (5 + j3)I2 I2 = (j2I1)/(5 + j3) V1 = (4 + j2)I1 - j2(j2) / (5 + j3) I1= [4 + j2 + 4/(5 + j3)] Zs = V1/I1 = (18 + j22)/(5 + j3) Zs = 4.8819 Problem 8FE-2 In the circuit in Figure 8PFE-2, select the value of the transformer's turns ratio n = N2/N1 to achieve impedance matching or maximum power transfer. Using this value of n calculate the power absorbed by the 3-Ohm resistor. N1:N2 48 1200oV ` Ideal Figure P8PFE-2 j32 2 j2 Suggested Solution The reflected impedance is Z = ZL/n2 The value of n that will make the reflected impedance the complex conjugate of the impedance in the primary is: n = i.e. (1/4)2 (48 + j32) = (3 j2)* and P3 = (1/2)(120/96)2(48) = 37.5W Problem 8FE-3 In the circuit in Figure 8FE-3, select the turns ratio of the ideal transformer that will match the output of the transistor amplifier to the speaker represented by the 16-Ohm load. 1K + Vs 5K Vx 10K Ideal 0.04Vx a:1 16 (speaker) Figure P8PFE-3 Suggested Solution a2(16) = 10K => a = 25 Problem 8FE-4 Find the power absorbed in the 1-Ohm load in the network in Figure 8PFE-4 2:1 2 120oVrms ` Ideal Figure PFE-4 j4 1 Suggested Solution 2 120oVrms ` j4 4 I = 120 / (6 + j4) = 1.664-33.69oAmps P = (1.664)2(4) = 11.1W Problem 8FE-5 Determine the average power absorbed by the 1-Ohm resistor in the network in Figure 8PFE-5. -j2 2:1 12 120oV ` Ideal Figure PFE-5 j16 1 Suggested Solution 12 120oV ` I j16 -j8 4 I = 12 / (16 + j8) = 0.07-26.57oAmps P = 0.5(0.67)2(4) = 0.9W Problem 9.1 Determine the equations for the current and the instantaneous power in the network in Figure P 9.1 I 4 12 75V j 3 Figure P 9.1 Suggested Solution I= 12 75 12 75 = = 2.4 38.1 A 4 + j 3 5 36.90 L ( t ) = 2.4 cos( t + 38.1 ) A 12 2.4 P (t ) = (cos 36.9 + cos(2 t + 75 + 38.1 )) 2 = 11.51 + 14.4 cos(2 t + 113.1 )W Problem 9.2 Determine the equations for the voltage and instantaneous Power in the network in Figure P 9.2 + 4 225A V j+ _ Figure P 9.2 Suggested Solution V = IZ = (2 25 )(4 + j 4) = 11.3 70 V L ( t ) = 2 cos( t + 25 ) A & V (t ) = 11.3cos( t + 7025 )(4 + j 4) = 11.3 70 )V 2 11.3 P (t ) = [cos 45 + cos(2 t + 95 )] 2 = 8 + 11.3cos(2 t + 95 )W Problem 9.3 Find the average power absorbed by the network shown in Figure P 9.3 12 4 230 A j4 Figure P 9.3 Suggested Solution Using Current division (2 30 )(- j 4 + j 2) = .89 -33.43 A 4 - j4 + j2 Then 1 1 P = IM 2 R = (.892 )(4) = 1.58W 2 2 IR = Problem 9.4 Given the network in Fig. P 9.4, find the power supplied and the average power absorbed by each element. + 2 2 60A VP j2 _ Figure 9.4 -j1 Suggested Solution (2 + 2 j )(2 - j ) 6 + j 2 = 2 + j2 + 2 - j 4+ j 6.32 18.43 ) = 9.2 4.39 V 4.12 14.04 ZT = VT = ITZT = (6 0 )( PS = 1 (9.2)(6) cos(4.39 - 0 ) = 27.52W 2 (6 0 )(2 - j ) I1 = = 3.26 -40.61 A 4 + j1 (6 0 )(2 + j 2) = 4.12 30.96 A 4 + j1 1 1 2 1 PA = I12 R1 + I 2 R 2 = [(3.26) 2 (2) + (4.12) 2 (2)] 2 2 2 PA = 27.61W I2 = Problem 9.5 Given the circuit in Fig. 9.5, find the average power supplied and the average power absorbed by each element. _ 4 40 A -j2 j1 2 + V Figure P 9.5 Suggested Solution 4 -2j j1 Z1 Z1 = [(4 - j2)(j1)]/ (4-j2 +j1) = (2+ j4) / (4-j) ZT = 2( Z 1) 4 + j8 = 2 + Z 1 10 + j 2 V = (4 0 )( ZT ) = 3.51 52.12 V 1 PS = (4)(V ) cos(52.12 - 0 ) = 4.31W 2 4 0 ( Z 1) I 2 = = 1.75 52.12 A Z1 + 2 4 0 (2) j1 I 4 = ( ) = .78 78.69 A Z1 + 2 4 - j 1 2 1 P 2 = I 2 (2) = (1.75) 2 (2) = 3.06W 2 2 1 2 1 P 4 = I 2 (4) = (.78) 2 (4) = 1.23W 2 2 Problem 9.6 Compute the average power absorbed by the elements to the right of the dashed line in the network shown in Fig. P 9.6 2 j4 1230V 40 V -j1 Figure P 9.6 4 Suggested Solution 2 j4 1230V PL = 0 W 40 V PR1 = (I2R1)/2 I PC = 0 W PR1 = (I2R2)/2 -j1 4 II = 12 30 - 4 0 12 30 - 4 0 8.77 43.19 = = A 2 + 4 + j 4 - j1 6 + j3 6.71 26.57 II = 1.3116.62 A => I = 1.31A 1 PR1 = ( )(1.31) 2 (2) = 1.72W 2 1 PR1 = ( )(1.31) 2 (4) = 3.43W 2 PL = 0W PC = 0W PR1 = 1.72W PR 2 = 3.43W Problem 9.7 Given the network in Fig. 9.7, determine which elements are supplying power, which ones are absorbing power, and how much power is being supplied and absorbed. 1 120 V 2 -j2 j1 60 V Figure P 9.7 Suggested Solution The mesh currents are obtained from the equations: ( I - j 2) I 1 + j 2 I 2 = 12 0 j 2 I 1 + (2 - j ) I 2 = -6 0 1 2 - j - j 2 12 0 or [ I1 ] = [ ][ ] I 2 4 - j 5 - j 2 1 - j 2 -6 0 3.75 51.39 =[ ] 2.1 -65.16 1 P12 0 = (12 )( 3.75 ) cos(0 - 51.39 ) = 14.04W Sup 2 1 P1 = (3.75) 2 (1) = 7.03W Abs 2 1 P 2 = (2.10) 2 (2) = 4.41W Abs 2 1 P 6 0 = (6)(2.10) cos(0 + 65.16 ) = 2.65W Abs 2 Problem 9.8 Given the circuit in Fig P 9.8 find the average power absorbed by the network. 2 120 V -j1 j2 60 A 4 Figure P 9.8 Suggested Solution V1 The nodal equation is: (V1 -12)/2+ V1/ (-j) + V1/ (4+ j2) = 60 Solving yields: V 1 = 6.46 - j8.3V = 10.52 -52.12 V 12 - 6.46 + j8.3 I 2 = = 4.99 56.28 A 2 10.50 -52.12 = 2.35 -78.69 A I 4 = 4 + j2 1 1 PS = (12)(4.99) cos(0 - 56.28 ) + (10.52)(6) cos(-52.12 - 0 ) 2 2 = 16.62 + 19.38 = 36W 1 1 Pabs = (4.99) 2 (2) + (2.35) 2 (4) 2 2 = 24.9 + 11.05 = 35.95W Problem 9.9 Given the network in Fig 9.9, find the average power supplied to the circuit. 2 1 120 A -j2 j1 40 V Figure P 9.9 Suggested Solution V1 V1/ (1-j2) + (V1 - 4)/ (2 + j1) = 12 V1 = 21.54 -21.81 V I 1 = 21.54 -21.81 = 9.63 41.62 A 1- j2 21.54 -21.81 - 4 = 8 -53.1 A 2 + j1 I 2 = PS = 1 (12)(21.54) cos(-21.81 - 0 ) = 119.99W 2 1 1 1 Pabs = (9.36) 2 (1) + (8) 2 (2) + (4)(8) cos 53.13 2 2 2 = 46.37 + 64 + 9.6 = 119.97W Problem 9.10 Given the circuit in Fig. P 9.10, determine the amount of average power supplied to the network. 120 V 40 A 2 j1 4 Figure P 9.10 Suggested Solution V1+12 V1 (V1 + 12)/ 2 + V1 / j + V1/ 4 = 4 V1 = (-8)/ (3 - 4j) = -1.653.13 V V 1 + 12 = 5.56 -6.61 A 2 V1 I 4 = = -.4 53.13 A 2 I 12 0 = I 2 - 4 0 = 1.65 -22.83 A 1 1 PS = (12)(1.65) cos 22.83 + (11.11)(4) cos(-6.61 ) = 31.19W 2 2 1 1 Pabs = (5.56) 2 (2) + (.4) 2 (4) = 30.91 + .32 = 31.23W 2 2 I 2 = Problem 9.11 Determine the average power absorbed by the 4- resistor in the network shown in Fig. P 9.11 2 -j4 4 40 A j2 1290 Figure P 9.11 Suggested Solution Let V1 be the voltage across the 4 resistor. V 1 V 1 V 1 - 12 90 + + =4 4 - j4 2 V 1 = 9.13 37.88 V V1 = 2.28 37.88 A I 4 = 4 1 P 4 = (2.28) 2 (4) = 10.4W 2 Problem 9.12 Given the network in Fig. P 9.12, show that the power supplied by the source is equal to the power absorbed by the passive elements. j2 -j3 40 A 2 1230 V Figure P 9.12 Suggested Solution I1 + 40 A V _ 2 I2 Vs 1230 V -j2 j3 Mesh Equations I1 = 4 0 A -12 30 = I 2(2 + j1) - 2 I 1 Find I2 2 I 1 - 12 30 8 0 - 12 30 = = 2.89 -138.30 A 2 + j1 2 + j1 Find V I2 = V = ( I 1 - I 2)(2) = (4 0 - 2.89 -138.30 )(2) = 12.90 17.30 V Power Delivered by current source 1 1 PI = VI 1 cos = (12.90)(4) cos(17.35 - 0) = 2 2 PI = 24.63W Power Delivered by voltage source 1 1 PV = - VSI 2 cos = - (12)(2.89) cos(30 - (-138.3) = 2 2 PV = +16.98W Power consumed by R PL = 0W 1 1 12.90 PV = V 2 / R = 2 2 2 Note PR = PI + PV PR = 41.60W 2 PC = 0W Problem 9.13 Calculate the average power absorbed by the 1- resistor in the network shown in Fig. P 9.13 I1 j2 -j3 120 V 2 1 Figure P9.13 Suggested Solution Therenin's Equation j2 -j3 + 2 120 V VOC _ ZTH VOC = 12 0 [ VOC = ZTH 2 ] 2 + j2 VOC = 8.49 -45 V 24 0 2 2 45 = - j 3 + (2 // j 2) ZTH = - j 3 + 4 90 j4 = - j3 + = - j 3 + 2 45 = - j 3 + 1 - j1 2 + j2 2 2 45 1 -j2 I 1 PR=I2R/2 ZTH = (1 - j 2) 8.4845 V I= 8.49 -45 8.49 -45 = =30 A 2 - j2 2 2 -45 1 2 1 I R = ( )(3) 2 (1) = 4.5W 2 2 PR = Problem 9.14 Determine the average power absorbed by the 4- resistor in the network shown in Fig 9.14 2 -j4 4 40 A j2 1290 V Figure P 9.14 Suggested Solution 2 -j4 4 40 A j2 1290 V Norton's Equivalent: 2 ZTH= 2 I5C V5 1290 V j2 I5 40 A Find I5C by superposition I5C1 due to I5: I 5C1 = I 5 = 4 0 A V5 = 6 90 A I5C2 due to V5: I 5C 2 = 2 I 5C = I 5C1 + I 5C 2 = 4 + j 6 A = 7.21 56.31 Norton Equation of Circuit ZTH 2 -j4 R2 4 + V _ I5C=7.2156.31 A V = I 5C[2 // 4 // - j 4] = 9.12 37.90 V P 4 = 1 | V |2 = 10.40W 2 R2 P 4 = 10.40W Problem 9.15 Find the average power absorbed by each element in the network shown in Fig. P 9.15 2 120 V j1 -j1 j1 1 60 Figure P 9.15 Suggested Solution Find I 2 using source transformation I1 2 120 V C L1 j1 -j1 60 IS L2 j1 R2 I2 1 PC = 0 W PL1 = PL2 = 0 W Let Z1 = (2 // - j1) j1 Z1 j1 I2 60 1 6Z1 Let Z2 =Z1+j1 j1 I2 6Z1/ Z2 Z2 60 A 1 Let Ix = 6Z1/ Z2 + 60 j1 Z2 I2 IxZ2 1 I2 = I x Z2 Z 2 + j1 + 1 I 2 = 3.26 -77.4 A Finding I1 using source transformation: 2 120 -j1 60 1 Let Z3 = 1+ j1 I1 2 120 -j1 6 Z3 j1 j1 1 Let Z4 = 1+ j2 I1 2 120 -j1 Z4 6 Z3/Z4 Let Z5 = Z4//(-j1) I1 Z5 2 (6 Z3Z5)/Z4 120 12 - I1 = 6Z3 Z5 Z4 2 + Z5 I1 = 4.60 57.53 A Power Absorbed by Each Element 1 1 PR1 = I12 R1 = (4.6) 2 (2) = PR1 = 21.16W 2 2 1 2 1 PR 2 = I 2 R2 = (3.26) 2 (1) = PR 2 = 5.31W 2 2 1 1 P = VS I1 cos(0 - 57.53 ) = (12)(4.6)(.54) VS 2 2 = 14.82W supplied PIS = - P + PR1 + PR 2 = 11.65W supplied VS I1 2 120 V C L1 j1 -j1 IS L2 j1 60 R2 I2 1 PC = 0 W PL1 = PL2 = 0 W 1 2 1 I1 R1 = (4.6) 2 (2) = PR1 = 21.16W 2 2 1 1 PR 2 = I 22 R2 = (3.26) 2 (1) = PR 2 = 5.31W 2 2 1 1 P = VS I1 cos(0 - 57.53 ) = (12)(4.6)(.54) = 14.82W supplied VS 2 2 PR1 = PIS = - P + PR1 + PR 2 = 11.65W supplied VS Problem 9.16 Determine the average power supplied to the network shown in Fig. P 9.16 10 V j1 20 A -j1 1 Figure P 9.16 Suggested Solution 10 V I 20A j2 -j1 1 Find I using source transformation j1 10 V j2 V -j1 1 I Let Veq = -1 + j2 V I 1 -j1 I j1 Veq Another source transformation I Veq/j1 j1 -j1 1 I Ieq (2+j1)A 1 PR =(1|Ieq|2R)/2 |Ieq| = 5 PR = 2.5 W Problem 9.17 Determine the average power supplied to the network in Fig 9.17 j1 1 -j1 2 10 A 1 j1 Figure P 9.17 Suggested Solution j1 V2 V1 -j1 I3 2 10 V3 1 1 j1 Nodal Equations @ V2 : 1 = V2 - V1 V3 - V2 + 2 - j1 or 2 = V2 (2 + j 2) - j 2V1 - V3 V1 V3 + 1 j1 or 1=V1 - j1V3 (1) @ and : 1 = (2) V1 = 1 + j1V3 @ V1 : V3 - V1 V2 - V1 V1 = + 1 + j1 1 - j1 Solve (2) for V1 and substitute into (1) and (3) 2 = V2 (2 + j 2) - j 2 + 2V3 - V3 => 2 + j 2 = V2 (2 + j 2) + V3 (4) V3 = 1 + j 2 + V3 (-2 + j1) + V2 (1 - j1) => 1 + j 2 = -V2 (1 - j1) + V3 (3 - j1) (5) Solve (4) for V3 and substitute into (5) (1 + j 2) = -V2 (1 - j1) + (3 - j1)[2 + j 2 - 2V2 - j 2V2 ] or V2 = PIS = 7 + j2 = 0.64 -21.93 V 9 + j7 1 | I S | | V2 | cos(-21.93 - 0) = 0.297W 2 Power delivered = 0.297 W Problem 9.18 Find the average power absorbed by the 2- resistor in the circuit shown in Fig 9.18 Ix j2 + 2Ix 2 -j3 120 A _ Figure P 9.18 Suggested Solution V1 V2 V1 V1 - V2 + - 2I x = 0 2 j2 -(V1 - V2 ) ; V2 = 12 0 V Ix = j2 Solving yields: V1 = IR = 18 -90 = 11.39 -18.43 V 1.58 -71.57 V1 = 5.7 -18.43 A 2 1 PR = (5.7) 2 (2) = 32.49W 2 Problem 9.19 Given the network in Fig P 9.19 determine which elements are supplying power, which ones are absorbing power, and how much power is being supplied and absorbed. 1 -j1 240 V 4V0 j1 1 1 + V0 _ Figure P 9.19 Suggested Solution I1 I3 + I2 V _ I4 I5 V1 - 4V0 - 24 0 V - 4V0 V V + + + =0 1 -3 j 2 V Where V0 = 2 Solving for V yields: V = 11.65 104.04 V V - 4V0 = -11.65 104.04 V SO 24 + V = 24 28.09 A 1 -V I2 = = 11.3 + j 2.83 A 1 -90 I1 = I 3 = I1 - I 2 = 13.01 40.63 A I4 = I5 = V = 11.3 + j 2.83 A 1 90 V = -1.41 + j 5.66 A 2 1 Then Pabs = (24)2 (1) = 288W 2 1 P2 = (5.83) 2 (2) = 34W 2 Psup = P24 0 + P4V0 1 (24)(24) cos(-28.09) 2 1 + (2)(11.56)(13.01) cos(104.04 - 40.63) 2 = 67.85 + 254 = 321.85W = Problem 9.20 Determine the impedance Z L for maximum average power transfer and the value of the maximum power transferred to the Z L for the circuit shown in Fig P 9.20 130 A 1 ZL Figure P 9.20 Suggested Solution 130 A 1 ZTH=1+j1 * Z L = ZTH = 1 - j1 Z L = 1 - j1 VOC = (1 30 )(1 0 ) = 1 30 V 1 V PMAX = ( OC ) RL = 0.125W 2 2 RTH Problem 9.21 Determine the impedance Z L for maximum average power transfer and the value of the maximum average power transferred to Z L for the circuit shown in Fig P 9.21 1 -j1 4 j1 1230V 40 A ZL Figure P 9.21 Suggested Solution 1 4 j1 1230V 40 A ZTH= 1230 + 40(1-j) = 14.53 1 4 j1 ZTH= 5 ZL= 5 1 14.53 2 PL = ( ) (5) = 5.28W 2 10 Problem 9.22 Determine the impedance Z L for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P 9.22 j2 20 V 4 245 A -j4 ZL Figure P 9.22 Suggested Solution j2 20 V 4 245 A -j4 + VOC= [(845- 20)/ (4- j2)](-j4) =6.03 6.32 V _ 4 j2 -j4 + _ ZTH= (4+ j2)(-j4) / (4- j2) =3.2 - j2.4 Z L = 3.2 + j 2.4 and PL = 1 6.03 2 ( ) (3.2) = 1.42W 2 3.2 + 3.2 Problem 9.23 Determine the impedance Z L for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P 9.23 j2 20 V 4 245 A -j4 Z4 Figure P 9.23 Suggested Solution 10 V j1 -j1 j1 ==> 1 ZTH ==> 1 1 1 ZTH ==> -j1 ZTH = [ j1//(- j1)] + (1//1) = ZTH = PMax = 0 1 1 + = j1 - j1 2 Problem 9.24 Determine the impedance Z L for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P 9.24 1 -j1 4 j1 1230 V 40 A ZL Figure P 9.24 Suggested Solution -j1 -j1 1 2 j1 20 ZTH ==> j1 1 2 ZTH ZTH = [(1// j1) - j1] // 2 j1 1 2 /(1 + j1) 2 =[ - j1] // 2 = ( ) // 2 = = 1 + j1 1 + j1 2 + 1/(1 + j1) 1 + 2 + j 2 2 ZTH = = 0.55 -33.69 3 + j2 ZTH = 0.55 -33.69 Find VOC Z S = j1 + 1//(2 - j1) = 1.14 52.1 IS = 20 = 1.75 -52.1 A 1.14 52.1 VOC = VS - I 2 (2 0 ) = 2(1 - 0.56 -33.7 ) = 1.24 29.7 V PMax = (1.24 ) 2 (8)(0.46) = 0.42W Problem 9.25 Determine the impedance Z L for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P 9.25 1 1230 V -j1 Z4 j1 20 A 1 Figure P 9.25 Suggested Solution + -j1 VOC 1 1230 V j1 20 A 1 12 30 - (1 - j ) I1 - j ( I1 - I 2 ) = 0 _ I2 = 2 0 A I1 = 13.11 37.6 A VOC = - jI1 + I 2 = 14.42 -46.09 V ZTH is determined from the network below 1 -j1 ZTH = {[(1+j)(-j)] / (1+j-j)} - 1= 2-j j1 1 I2 + VOC _ j1 2 -j1 2 I2= (14.42-46.09)/4 = 3.6-46.09 PL = [1(3.6)2(2)]/2=13W Problem 9.26 Repeat problem 9.24 for the network in Fig P 9.27 120 V j1 2 -j1 ZL Figure 9.26 Suggested Solution 120 V j1 + -j1 VOC _ 1 2 VOC = 12 0 12 0 (2) - (- j ) = 3.79 18.43 V 2+ j 1- j 2 ZTH j1 -j1 2 ZTH = = -j 2j + 2 + j 1- j 3 = .9 + j.3 3- j Z L = .9 - j.3 1 3.79 PL = (.9) = 2W 2 1.8 2 Problem 9.27 Repeat problem 9.24 for the network in Fig P 9.27 1 1 j1 -j1 120 V Z4 Figure 9.27 Suggested Solution 1 1 IS -j1 120 V j1 I1 + VOC _ Figure 9.27 12 0 12(2 + j ) = A (1 + j )(1) 2- j -j 1+ j +1 12 I (1) I1 = 3 = A 2+ j 2- j IS = VOC = 12 0 - I1 ( j ) = 15.15 -18.43 V 1 1 j1 -j1 ZTH -j ZTH = ( + 1) //( j ) 1- j 2+ j = 2- j 1 15.15 2 ) (.6) = 47.82W PL = ( 2 1.2 Problem 9.28 Repeat problem 9.24 for the network in Fig P 9.28 1 j1 10 V ZL 1 -j1 290 A j1 Figure P 9.28 Suggested Solution 1 j1 -j1 10 V ZTH 1 j1 1 j1 290 A ZTH -j1 open j1 1 j1 1 -j1 ZTH = -j1+j1+1=1 ZL = ZTH* = 1 + 10V VOC j1 290 A 1 VOC = (1 + j )(2 90 ) + 1 0 = -1 + j 2 = 2.23116.57 V 1 2.23 PL = (1) = 0.622W 2 2 2 g Problem 9.29 Repeat problem 9.24 for the network in Fig P 9.29 Z4 2 20 A -j2 1 Figure P 9.29 Suggested Solution ZTH 2 20 A -j2 1 60 A 2 ZT H -j2 1 ZTH = 2 + (1// - j 2) = 2 - * j2 2 - j6 = = 2.83 -8.13 1- j2 1- j2 Z L = ZTH = 2.83 -8.13 I1 = -6 - j 4 = 3.22 -82.9 A 1- j2 VOC = (2 0 )(2) + (3.22 -82.9 )(1) = 5.44 -36 V PMax ( 5.44 ) = 2 (8)(2.8) = 1.32W Problem 9.30 Determine the impedance Z for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P. 9.30 120 V 1 2 Ix 2Ix j2 -j1 ZL Figure P 9.30 Suggested Solution VOC VOC - 2 I x - 2 VOC - 2 I x VOC + + =0 1 2 -j 4.8 V = 9.6 -53.13 V and I x = OC VOC = 4 3 + j4 2Ix' 2Ix' 1 2 Ix ' IT + 4Ix' = 1 Ix' = 1/4 -j1 10 _ IT = ZTH 1 3 + j4 ' + 3I x = 4 - j1 1 3 1 = = = .8 -53.13 = .48 - j.64 PL = (10) 2 (.48) = 24W IT 3 + j 4 2 + .48 9.6-53.13 -j.64 j2 ZL ZTH Total = .48+ j1.36 ZL = .48 - j1.36 IL = 9.6 / (.48 + .48) = 10 A _ 1 PL = (10) 2 (.48) = 24W 2 Problem 9.31 Determine the impedance ZL for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P. 9.31 1 120 V 2 2V -j1 + V _ ZL Figure P 9.31 Suggested Solution 120 V 2 2VOC -j1 + VOC _ VOC VOC + 12 + + 2VOC = 0 -j 2 -12 VOC = = -2.23 -21.8 V 5+ 2 j IT + V _ + 10 _ 2 2V -j1 IT = ZTH 1 1 5 + +2= j+ -j 2 2 1 1 = = = .37 -21.8 = .34 - j.14 IT 5 + j 2 + .34 -2.23-21.8 -j.14 1 ZL ZL = 1.34 + j.14 IL = 2.23 / [2(1.34)] = .83 A PL = 1(.83)2(1.34) / 2 = .46 W IL _ Problem 9.32 Find the impedance Z L for the maximum average power transfer and the value of the maximum average power transferred to Z L for the circuit shown in Fig P 9.32 Ix 2 j1 -j2 120 V 2Ix ZL Figure P 9.32 Suggested Solution 2 V j1 Ix 120 V -j2 ZL 2Ix Solving for VOC : 12 0 - V V - 2 I x = -2 j 2 12 - V Ix = 2 Solving, V = 7.6 -18.4 = VOC Then, we find ISC : 12 - V 2 I x - V V + = 2 -2 j j Solving, I SC = 17.0 45 Then, ZTH = 7.6 -18.4 = .45 -63.4 = .2 - .4 j 17.0 45 So, Z L = .2 + .4 j P= 1 (17) 2 (.2) = 28.9W 2 Problem 9.33 Determine the impedance Z L for the maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig P 9.33 120V 2Vx 2 -j1 1 + _ Vx ZL Figure P 9.33 Suggested Solution 1 120V + + ZVx' j1 Vx' 1 _ VOC _ VOC - 12 - 2VOC VOC - 12 VOC - + =0 1 1 j1 VOC = 12 + j12V Vx"= 0 1 120V + ZVx" j1 Vx" 1 _ Ix 1 j1 12 ISC I SC = ZTH 12 12(1 + j ) 12 + j12 = = j j j 1+ j V = OC = j1 Then I SC j1 12+j12 1 ZL ZL = 1- j IL = 12 + j12 = 8.4853 45 1+ j +1- j 1 And Pload = (8.4853) 2 (1) = 36W 2 Problem 9.34 Compute the rms value of the voltage given by the waveform shown in Fig P 9.34 v(t) (V) 4 0 2 6 8 Figure P 9.34 12 t(s) Suggested Solution VRMS = VRMS = 1 v 2 (t )dt 0 T T 1[ 42 dt ] 0 2 4 where v(t ) = { 0 0t 2 2t 6 } &T =6 6 T 2 16(t |0 ) 32 = = 6 6 VRMS = 2.31VRMS VRMS = VRMS = 1 v 2 (t )dt 0 4 where v(t ) = { 0 0t 2 2t 6 T 1[ 42 dt ] 0 2 } &T =6 6 2 16(t |0 ) 32 = = 6 6 VRMS = 2.31VRMS Problem 9.35 Find the rms values of the waveform shown in Fig P 9.35 v(t) V 3 0 1 Figure P 9.35 4 5 8 t(s) Suggested Solution 1 4 1 1 2 2 Vrms = [ ( (3t ) dt + (4 - t ) dt ] 2 1 4 0 t3 4 1 1 = [ (3t 3 |1 + (16t - 4t 2 + )|1 )] 2 0 4 3 = 3V Problem 9.36 Calculate the rms value of the waveform shown in Fig P 9.36 v(t) (V) 2 0 1 2 3 4 5 6 7 t(s) Figure P 9.36 Suggested Solution 1[ (2t ) 2 dt + 22 dt + (6 - 2t ) 2 dt 0 1 2 1 2 3 I rms = 4 4 4 1[ t 3 |1 + 4 + (36t - 12t 2 + t 3 )|3 0 2 3 3 = 4 = 1.29 A Problem 9.37 Calculate the rms value of the waveform shown in Fig P 9.37 v(t) (V) 4 0 1 2 ' 3 4 5 6 t(s) Figure P 9.37 Suggested Solution I rms = = 2 1 1 [ (4t ) 2 dt + (8 - 4t ) 2 dt ] 1 4 0 1 16 64t 2 16t 3 2 + 64t - + |1 = 1.63V 4 3 2 3 Problem 9.38 The current waveform in Fig P 9.38 is flowing through a 5- resistor. Find the average power absorbed by the resistor. v(t) (V) 4 2 0 -2 -4 Figure P 9.38 2 4 6 8 t(s) Suggested Solution A 5 resistor passes current shown in Fig P 9.38 I rms = = 4 1 2 2 2 0 (2t ) dt + 2 -4 dt 4 14 2 4 + 3|0 + 16t|2 = 3.27 A 43 P = (3.27) 2 (5) = 53.3W Then, Problem 9.39 Calculate the rms value of the waveform shown in Fig P 9.39 v(t) (V) 0 2 3 5 6 t(s) -4 Figure P 9.39 Suggested Solution Find rms value. Vrms = = 3 1 2 2 2 0 (-2t ) dt + 2 -4 dt 3 1 4 3 2 3 t |0 + 16t |2 = 2.98V 3 3 Problem 9.40 Compute the rms value of the waveform in Fig 9.40 v(t) (V) 4 2 0 -2 -4 Figure P 9.40 2 4 6 8 10 t(s) Suggested Solution Find rms value Vrms = = = 4 1 2 2 -4 dt + (8 - 2t ) 2 dt 0 2 6 1 4 3 4 2 2 16t |0 + 64t - 16t + 3 t |2 6 1 128 = 2.67V 6 3 Problem 9.41 Calculate the rms value of the waveform shown in Fig 9.41 v(t) (V) +1 0 1 -2 Figure P 9.41 2 3 4 5 t(s) Suggested Solution Find rms value: I rms = = = 2 1 1 2 2 0 (2t - 2) dt + 1 1 dt 3 1 4 3 8t 2 2 + 4t |1 +t |1 t - 0 3 3 2 1 7 = 0.88 A 3 3 Problem 9.42 Calculate the rms value of the waveform shown in Fig 9.42 v(t) (V) 2 1 0 1 2 3 4 5 t(s) Figure P 9.42 Suggested Solution Find rms value: I rms = = 2 1 1 2 2 0 (2t ) dt + 1 1 dt 3 1 4 2 3 t 3 |1 +t |1 0 3 = 0.88 A Problem 9.43 An industrial load consumes 100 kW at 0.8 pf lagging. If an ammeter in the transmission line indicates that the load current is 284 A rms, find the load voltage. Suggested Solution A load consumes 100 kW at pf=.8 lag I load = 284 A rms. Find load voltage. Vrms = 100, 000W = 440V (.8)(284 A) Problem 9.44 An industrial load that consumes 80 kW is supplied by the power company through a transmission line with 0.1 resistance, with 84kW. If the voltage at the load is 440 Vrms, find the power factor at the load. Suggested Solution A load consumes 80 kW. 1 B source supplies 84 kW. Rline = 0.1,Vload = 440Vrms . And load p.f. PS = Pload + I 2 Rline = 84kW = I 2 (.1) + 80kW Yeilds I = 200 A rms . Then, PF = 80,000 = .91Lag (440)(200) Problem 9.45 The power company supplies 80 kW to an industrial load. The load draws 220 A rms from the transmission line. If the load voltage is 440 V rms and the load power factor is 0.8 lagging, find the losses in the transmission line. Suggested Solution PS = 80kW , I L = 220 Arms , Vload = 440Vrms with p.f.= 0.8 lag. Find line loss. PS = Vrms I rms pf = (440)(220)(.8) = 72.44kW Pline = PS - PL = 80k - 72.44k = 2.56kW Problem 9.46 An industrial plant with an inductive load consumes 10 kW of power from a 220 V rms line. If the load power factor is 0.8, what is the angle by which the load voltage leads the load current? Suggested Solution = cos -1 0.8 = 36.87 Problem 9.47 The power company must generate 100 kW in order to supply an industrial load with 94 kW through a transmission line with 0.09 resistance. If the load power factor is 0.83 lagging, find the load voltage. Suggested Solution PS = I 2 Rline + PL 100kW = I 2 (.09) + 94kW 600 I2 = .09 I rms = 258.2 A Vrms = PL 94000 = = 440V I rms Pf (258.2)(.83) Problem 9.48 An industrial load operates at 30 kW, 0.8 pf lagging. The load voltage is 2200 V rms. The real and reactive power losses in the transmission-line feeder are 1.8 kW and 2.4 kvar, respectively. Find the inpedance of the transmission line and the input voltage to the line. Suggested Solution Pload = 30kW as pf =.8 lag. VL = 2200Vrms Pline = 1.8kW + 2.4k var . Find R line , VS IL = PL 30000 = = 170.45 cos -1 .8 Vrms . pf 220 .8 = 170.45 - 36.9 Then, Rline = P I2 1800 = .062 Rline = 170.452 P 2400 = .083 j Lline = 2 = I 170.452 Z line = .062 + .083 j = 2371.19 VS = Vload + I L Z L = 2200 + (170.45 - 36.9) Z L Problem 9.49 A transmission line with impedance 0.08 j 0.25 is used to deliver power to a load. The load is inductive and the3 load voltage is 220 0 V rms at 60 Hz. If the load requires Suggested Solution Z line = .08 + j.25 VLoad = 2200 at 60 Hz Pload = 12kW 2 PLine = 560kW Find pf at load. Pline = 560 = I (.08) yeilds I = 83.67 A Pload = IV cos = 12kW = (220)(83.67) cos = cos -1 12000 = .65 = pf lagging (220)(83.67) Problem 9.50 Find the source voltage in the network shown in Fig P 9.50 0.08 VS j0.2 40kW 0.8 pf lagging 2200 V rms 30kW .707 pf lagging Figure P 9.50 Suggested Solution I L1 = 40000 (220)(.8) = 227.27 A I L2 = 30000 (220)(.707) = 177.1A 1 = cos -1 (.8) = 36.87 2 = cos -1 (.707) = 45 I L = I L1 + I L2 = 227.27 - 36.87 + 177.1 - 45 = 403 - 40.43 A Vline = I L (.8 + j 2) = (403.37 - 40.43)(.2268.2) = 88.7427.77V VS = Vline + 2200 = 301.377.89 Problem 9.51 Given the network in Fig P 9.51, compute the input source voltage and the input power factor. 0.08 VS j0.2 0.01 40kW 0.86 pf lagging j0.05 18 kW 0.8 pf lagging + 2200 V rms _ Figure P 9.51 Suggested Solution L = cos -1 .8 = 36.87 80000 = 102.27 A (220)(.8) I L2 = 102.27 - 36.87 A I L2 = VL1 = (.01 + j.05)(102.27 - 36.87) + 2200 = 223.91.89V 1 = cos -1 .86 = 30.68lagging I L1 = 40000 = 207.72 A ( 223.91)(.86 ) I L1 = 207.72 ( -30.68 + .89 ) = 207.72 - 29.79 A I L = I L1 + I L2 = 207.72 - 29.79 + 102.27 - 36.87 A = 309.47 - 32.12 A VS = ( 309.47 - 32.12 )(.08 + j.2 ) + 223.91.89 = 281.028.75V S = VS - iS = 8.75 - ( -32.12 ) = 40.87 PFsource = cos ( 40.87 ) = .756lagging L = cos -1 .8 = 36.87 80000 = 102.27 A (220)(.8) I L2 = 102.27 - 36.87 A I L2 = VL1 = (.01 + j.05)(102.27 - 36.87) + 2200 = 223.91.89V 1 = cos -1 .86 = 30.68lagging I L1 = 40000 = 207.72 A ( 223.91)(.86 ) I L1 = 207.72 ( -30.68 + .89 ) = 207.72 - 29.79 A I L = I L1 + I L2 = 207.72 - 29.79 + 102.27 - 36.87 A = 309.47 - 32.12 A VS = ( 309.47 - 32.12 )(.08 + j.2 ) + 223.91.89 = 281.028.75V S = VS - iS = 8.75 - ( -32.12 ) = 40.87 PFsource = cos ( 40.87 ) = .756lagging Problem 9.52 Given the network in Fig P 9.52, determine the input voltage Vs. 0.1 VS j0.3 24kW 0.82 pf lagging 36 kW 0.88 pf lagging + 2400 V rms _ Figure P 9.52 Suggested Solution IS 0.1 VS j0.3 I1 24kW 0.82 pf lagging 36 kW 0.88 pf lagging I2 + 2400 V rms = VL _ VS = VL + I S (.01 + j.03) I S = I1 + I 2 Find I1 : I1 = Find I 2 : I 2 = P 1 - cos -1 (.82) = 121.95 - 34.92 Arms | VL | ( pf1 ) P2 - cos -1 (.88) = 170.45 - 28.36 Arms | VL | ( pf 2 ) Find I S : I S = I1 + I 2 = 250 - j150.76 Find VS : VS = VL + I S ( 0.1 + j 0.3) = VL + I S [ 0.31671.57] VS = 240 + 92.2540.48Vrms VS = 315.9010.93Vrms I S = 291.94 - 31.09 Arms Problem 9.53 Find the input source voltage and the power factor of the source for the network shown in Fig P 9.53 j0.3 j0.1 24 kW 0.9 pf lagging 0.1 VS 0.03 12 kW 0.87 pf lagging + 2400 V rms _ Figure P 9.53 Suggested Solution IS 0.1 VS j0.3 I1 ZL1 0.03 j0.1 ZL2 I2 + VL=2400 V rms _ + V1 _ Load 1: 12kW 0.87 pf lagging Load 1: 24kW 0.90 pf lagging VS = VL + I 2 [ 0.03 + j 0.1] + I S [ 0.1 + j 0.3] I S = I1 + I 2 I2 = P2 - cos -1 ( pf 2 ) = 111.11 - 25.84 Arms VL ( pf 2 ) V1 = VL + (0.03 + j 0.1) I 2 = 240 + ( 0.10473.30 )(111.11 - 25.84 ) = 240 + 11.6047.46 = 247.991.98Vrms VS = V1 + (0.1 + j 0.3) I S | I1 |= 1 I S = I1 + I 2 P 12000 1 = = 55.62 Arms V1 ( pf1 ) ( 247.99 )( 0.87 ) 1 1 V - I = cos -1 ( 0.87 ) = 29.54 I = -27.56 I1 = 55.62 - 27.56 Arms I S = 166.71 - 26.41 Arms VS = 247.991.98 + ( 0.1 + j 0.3)166.71 - 26.41 VS = 288.168.48Vrms S = Vs - is = 8.48 - (-26.41) = 34.89 pf source = cos ( 34.89 ) = 0.82lagging Problem 9.54 Use Kirchhoff's laws to compute the source voltage of the network shown in Fig P 9.54. 0.09 VS j0.25 50kW 2400 V rms 0.85 pf lagging 20kW .78 pf lagging Figure P 9.54 Suggested Solution IS 0.09 VS j0.25 I1 50kW 0.85 pf lagging 20kW .78 pf lagging I2 + 2400 V rms = VL _ Find I1 : | I1 |= 1 P 50, 000 1 = = 245.10 | VL | ( pf1 ) 240 ( 0.85 ) I = - cos -1 (0.85) = -31.79 I1 = 245.10 - 31.79 Arms Find I1 : | I 2 |= 2 P2 20, 000 = = 106.84 | VL | ( pf 2 ) 240 ( 0.78 ) I = - cos -1 (0.78) = -38.74 I 2 = 106.84 - 38.74 Arms Find IS : I S = I1 + I 2 = 351.40 - 33.90 Arms Find VS : VS = I S (0.09 + j 0.25) + 240 = 93.3736.30 + 240 VS = 320.069.95Vrms = ( 351.40 - 33.90 )( 0.26670.20 ) + 240 Problem 9.55 Given the network in Fig P9.55, determine the input voltage V5. 0.1 VS j0.1 30kW 0.9 pf lagging 40 kW 0.95 pf lagging + 2400 V rms _ Figure P 9.55 Suggested Solution IS 0.1 VS j0.1 I1 ZL1 + V1 _ ZL2 I2 + VL=2400 V rms _ Load 1: 30 kVA Load 2: 40 kVA 0.9 pf lagging 0.95 pf lagging I S = I1 + I 2 VS = VL + I S ( 0.1 + j 0.1) Find I1 : 1 |I1 |= | S | 30, 000 P 1 = 1 = = 125 Arms | VL | ( pf1 ) | VL | 240 I = - cos -1 ( 0.9 ) = -25.84 I1 = 125 - 25.84 Arms Find I 2 : |I 2 |= P2 - cos -1 ( pf 2 ) = 175.44 - 18.19 Arms | VL | ( pf 2 ) Find I S : I S = I1 + I 2 = 299.79 - 21.38 Arms VS = 240 + (0.1 + j 0.1) I S = 240 + 0.1 245 [ 299.79 - 21.38] = 279.243.48Vrms VS = 279.243.48Vrms Problem 9.56 What value of capacitance must be placed in parallel with the 18-kW load in Problem 9.51 in order to raise the power factor of this load to 0.9 lagging? Suggested Solution Pold = 18000W old = cos -1 .8 = 36.87 Qold = Pold tan = 18000 tan 36.87 = 13500VAR Sold = 18000 + j13500VA PFnew = .9 new = 25.84 S new = 18000 + j18000 tan 25.84 = 18000 + j8718 Scap = S new - Sold = - j 4782 C= 4782 = 262 F ( 377 )( 220 ) 2 Problem 9.57 An industrial load is supplied through a transmission line that has a line impedance of 0.1 + j0.2 . The 60-Hz line voltage at the load is 4800 V rms. The load consumes 124 kW at 0.75pf lagging. What value of capacitance when placed in parallel with the load will change the power factor to 0.9 lagging? Suggested Solution Pold = 124kW old = cos -1 .75 = 41.41 Qold = Pold tan = 124000 tan 41.41 = 109358VAR Sold = 124000 + j109358VA PFnew = .9 new = 25.84 S new = 124000 + j124000 tan 25.84 = 124000 + j 60056 Scap = S new - Sold = - j 49302 C= 49302 = 567.6 F ( 377 ) ( 480 ) 2 Problem 9.58 A plant consumes 60 kW at a power factor of 0.5 lagging from a 220-V rms 60-Hz line. Determine the value of the capacitor which when placed in parallel with the load will change the load power factor to 0.9 lagging. Suggested Solution IL = 60, 000 = 545.45 cos -1 0.5 A ( 220 )( 0.5) Qold = 60, 000 tan ( -60 ) = 103,923VAR Sold = 60, 000 + j103,923VA PFnew = .9 new = 25.84 S new = 60, 000 + j 60, 000 tan 25.84 = 60000 + j 29056 Scap = S new - Sold = 29056 - 103,923 = -74866 C= 74866 = 4103 F ( 377 )( 220 ) 2 Problem 9.59 A small plant has a bank of induction motors that consume 64 kW at a pf of 0.68 lagging. The 60-Hz line voltage across the motors is 2200V rms. The local power company has tolad the plant to raise the pf to 0.92 lagging. What value of capacitance is required? Suggested Solution Pold = 64kW 47.2 Qold = Pold tan 47.2 = 69, 000VAR Sold = 64000 + 69, 000 j Qnew = tan -1 .42 = 23.1 S new = 64000 + j 64000 tan 21.3 = 64000 + 27, 264 j Then, Scap = S new - Sold = -41.744 j C= Scap 41.744 = 2, 288 F V 2 rms ( 377 )( 220 ) 2 Problem 9.60 The 60-Hz line voltage for a 60-kW, 0.76-pf lagging industrial load is 4400 V rms. Find the value of the capacitance which when placed in parallel with the load will raise the power factor to 0.9 lagging. Suggested Solution Pold = 60, 00040.5 Qold = 60, 000 tan 40.5 = 51,300VAR Sold = 60, 000 + j 51,300VA With the capacitor in place, pf=.9 =25.8 S new = 60, 000 + j 60, 000 tan 25.8 = 60, 000 + j 29, 000VA Scap = S new - Sold = - j 22,300 C= 22,300 ( 377 )( 440 ) 2 = 305 F Problem 9.61 A particular load has a pf of 0.8 lagging. The power delivered to the load is 40 kW from a 220-V rms 60-Hz line. What value of capacitance placed in parallel with the load will raise the pf to 0.9 lagging? Suggested Solution IL = 40, 000 = 227.3 - 36.9 A ( 220 )( 0.8) Qold = 40, 000 tan ( -36.9 ) = 30, 000VAR Sold = 40, 000 + j 30, 000 With the capacitor, PFnew = .9, new = 25.8 S new = 40, 000 + j 40, 000 tan 25.8 = 40, 000 + j 20, 000 So, Scap = S new - Sold = -10, 000 j C= Scap 10, 000 = 586 F (Vrms ) 2 = ( 377 )( 220 ) 2 Problem 9.62 An industrial load consumes 44 kW at 0.82 pf lagging from a 2200 V rms, 60-Hz line. A bank of capacitors totaling 900 F is available. If these capacitors are placed in parallel with the load, what is the new power factor of the total load? Suggested Solution Scap = - j ( 220 ) ( 377 )( 900 F ) = - j16, 422 2 Then, Sold = 44, 000 + j 44, 000 tan ( cos -1 .82 ) 44, 000 + j 30, 712VA S new = Sold + Scap = 44, 000 + j14,300VA ( ) new = tan -1 pf new 14,300 = 18 44, 000 = .95lag Problem 9.63 A bank of induction motors consumes 36 kW at 0.78 pf lagging from a 60-Hz, 2200 V-rms line. If 500F of capacitors is placed in parallel with the load, find the new power factor of the combined load. Suggested Solution IL = IC = 36kW = 209.8 - 38.7 ( 2200 )(.78) 2200 = 41.5 j 1 jC Then, IT = I L + I C = 209.8 - 38.7 + 41.5 j = 186.7 - 28.7 Finally, pf = cos ( 28.7 ) = .88 ( lag ) Problem 9.64 A single-phase three-wire 60-Hz circuit serves three loads, as shown in Fig P9.64. I aA, I nN , I C , and the energy use over a 24-hour period in kilowatt-hours. IaA A IC Determine a 1200V rms InN 100W n N 1 kVA pf=0.9 lagging 1200V rms 100W b Figure P9.64 Suggested Solution I AN = I NB = PAN 100 = = 0.8330 Arms VAN 1200 PNB 100 = = 0.8330 Arms VNB 1200 I nN = I NB - I AN = 0 Arms S 1000 I C = C - cos -1 ( 0.9 ) = - 25.84 = 4.17 - 25.84 Arms 240 240 I aA = I C + I AN = 4.17 - 25.84 + 0.8330 I aA = 4.86 - 21.60 Arms I nN = 0 Arms I C = 4.17 - 25.84 Arms Power usage = 100 + 100 + 1000 ( 0.9 ) = 1.1kW E = Energy = p ( t ) dt = p ( t )t t = 24hours E = (1.1)(24) = 26.4kW hr E = 26.4kW hr Problem 9.65 A man and his son are flying a kite. The kite becomes entangled in a 7200-V power line close to a power pole. The man crawls up the pole to remove the kite. While trying to remove the kite, the man accidentally touches the 7200-V line. Assuming the power pole is well grounded, what is the potential current through the man's body? Suggested Solution 7200 Vrms Rarm 100 Rtrunk 200 Rarm 100 7200V I Assuming shoes provided a high resistance! I= 7200 = 18 A 400 Problem 9.66 A number of 120-V household fixtures are to be used to provide lighting for a large room. The total lighting load is 8 kW. The National Electric Code requires that no circuit breaker be large than 20 A with a 25% safety margin. Determine the number of identical branch circuits needed for this requirement. Suggested Solution PL = 8kW # of branches = VL = 129Vrms IL = PL = 66.7 Arms VL IL = 3.33 20 # of breakers = (1.25 )( # of branches ) = 4.17 # of breakers = 5 Problem 9.67 A 5.1-kW household range is designed to operate on a 240-V rms sinusoidal voltage, as shown in Fig. P9.67. However, the electrician has mistakenly connected the range to 120 V rms, as shown in Fig P9.67. What is the effect of this error? a A a 1200 V rms Range 5100 W 1200 V rms A Range 1200 V rms n N b (a) B (b) Figure P 9.67 Suggested Solution V 2 2402 = = 11.29 P 5100 Now using 120V and the same RHE . RHE = 1202 = 1275W 11.29 The heating element does not get as hot! P= Problem 9.68 In order to test a light socket, a woman, while standing on cushions that insulate her from the ground, sticks her finger into the socket, as shown in Fig P9.68. The tip of her finger makes contact with one side of the line, and the side of her finger makes contact with the other side of the line. Assuming that any portion of a limb has a resistance of 95 , is there any current in the body? Is there any current in the vicinity of the heart? Figure P 9.68 Suggested Solution Iouch Iouch= 120 / 95 = 1.26 Arms No current near the heart! 1200 V rms Rfinger 95 OA Rrest of body Problem 9.69 An inexperienced mechanic is installing a 12-V battery in a car. The negative terminal has been connected. He is currently tightening the bolts on the positive terminal. With a tight grip on the wrench, he turns it so that the gold ring on his finger makes contact with the frame of the car. This situation is modeled in Fig. P9.69, where we assume that the resistance of the wrench is negligible and the resistance of the contact is as follows: R1 = Rbolt to wrench = 0.012 R3 = Rring = 0.012 R4 = Rring to frame = 0.012 R2 = Rwrench to ring = 0.012 What power is quickly dissipated in the gold ring, and what is the impact of this power dissipation? Suggested Solution 12m 12V 12m Ring 12m I 12m I= 12 = 250 A 48 10-3 Pring = I 2 Rring = 750W Ring will get very hot! Finger will be burned! Ring may spot weld to frame! Problem 9FE-1 An industrial load consumes 120 kW at 0.707 pf lagging and is connected to a 4800 V rms line. Determine the value of the capacitor which, when connected in parallel with the load, will raise the power factor to 0.95 lagging. Suggested Solution old = cos -1 .707 = 45 Qold = Pold tan old = 120, 000(1) = 120, 000 So, Sold = 120, 000 + j120, 000VA new = cos -1 ( 0.95 ) = 18.19 Qnew = Pnew tan new = 120, 000 tan18.19 = 39.431VAR 2 Qcap = Qnew - Qold = - CVrms = 39, 431 - 120, 000 = - CV 2 80569 ( 377 )( 480 ) 2 = C = 927.6 F Problem 9FE-2 Determine the average and rms values of the following waveform. v(t) V 2 1 0 1 2 3 4 5 t(s) Figure 9FE-2 Suggested Solution Vave = Vrms 1+ 2 +1 = 1V 4 1 = 4 ( 1dt + 1 0 1 2 1 2 4dt + 1dt 2 3 ) 1 1 2 = (1 + 4 + 1) = 1.22Vrms 4 Problem 9FE-3 Find the impedance Z L in the network in Fig 9PFE-3 for maximum average power transfer. 120 V 2 20 A ZL -j1 j2 Figure 9PFE-3 Suggested Solution 2 -j1 ZTH j2 ZTH = = ( 2 )( - j1) + j 2 2 - j1 - j 2 ( 2 + j1) + j2 5 - j4 + 2 = + j2 5 = 0.4 + j1.2 Z L = 0.4 - j1.2 Problem 9FE-4 A rms-reading volmeter is connected to the output of the op-amp shown in Fig. 9PFE-4. Determine the meter reading. 36 k 12 k 1.414cost V RMS-reading voltmeter Figure 9PFE-4 Suggested Solution Gain = -36k = -3 12k 1.414 ( -3) = 3 volts rms 2 Volmeter reading = Problem 9FE-5 Determine the average power delivered to the resistor in Fig. 9PFE-5a if the current waveform is shown in Fig. 9PFE-5b. v(t) V i(t) 1 0 (a) 1 2 3 4 t(s) -2 Figure 9PFE-5 Suggested Solution I rms 1 = 2 ( (t ) dt + 1 2 0 1 2 1 2 ( -2 ) dt 2 ) 1 1 t3 1 2 2 = + 4t 1 2 3 0 = 1.47 A Then 2 Pave = I rms (4) = 8.67W Problem 10.1 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Van = 10045 V rms . Label all magnitudes and assume an abc-phase sequence. Suggested Solution Phase voltages: Van = 10045 V rms Vbn = 100 - 75 V rms Vcn = 100 - 195 V rms Line voltages: Magnitude = 3 phase voltage magnitude Angle = phase voltage angle +30 Vab = 173.275 V rms Vbc = 173.2 - 45 V rms Vca = 173.2 - 165 V rms Imaginary Vab = 173.275 V rms Van = 10045 V rms Vcn = 100-195 V rms Real Vca = 173.2-165 V rms Vbn = 100- 75 V rms Vbc = 173.2- 45 V rms Problem 10.2 A positive-sequence three-phase balanced wye voltage source has a phase voltage of Van = 10020 V . Determine the line voltages of the source. Suggested Solution Vab leads Van by 30 . Therefore, Vab = 100 350 = 173.250 V rms , Vbc = 173.2 - 70 V rms , and Vca = 173.2 - 190 V rms . Problem 10.3 Sketch a phasor representation of an abc-sequence balanced three-phase -connected source, including Vab , Vbc , and Vca if Vab = 2080 V rms . Suggested Solution Vab = 2080 V rms Vbc = 208 - 120 V rms Vca = 208120 V rms Imaginary Vca = 208120 V rms Vab = 2080 V rms Real Vbc = 208-120 V rms Problem 10.4 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Van = 12060 V rms . Label all magnitudes and assume an abc-phase sequence. Suggested Solution Van = 12060 V rms Vbn = 120 - 60 V rms Vcn = 120180 V rms Vab = 20890 V rms Vbc = 208 - 30 V rms Vca = 208 - 150 V rms Imaginary Vab = 20890 V rms Van = 12060 V rms Vcn = 120180 V rms Real Vbc = 208- 30 V rms Vca = 208-150 V rms Vbn = 120- 60 V rms Problem 10.5 A positive-sequence three-phase balanced wye voltage source Van = 24090 V rms . Determine the line voltages of the source. has a phase voltage of Suggested Solution Vab = 3 Van Van + 30 = 415.7120 V rms Vbc = 415.70 V rms Vca = 415.7 - 120 V rms ( ) Problem 10.6 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Vab = 20845 V rms . Label all phasors and assume an abc-phase sequence. Suggested Solution Vab = 20845 V rms Vbc = 208 - 75 V rms Vca = 208165 V rms Van = Vab 3 Vab - 30 = 12015 V rms ( ) Vbn = 120 - 105 V rms Vcn = 120135 V rms Imaginary Vab Vcn Vca Van Real Vbn Vbc Problem 10.7 A positive-sequence balanced three-phase wye-connected source with a phase voltage of 100 V supplies power to a balanced wye-connected load. The per phase load impedance is 40 + j10 . Determine the line currents in the circuit if Van = 0 . Suggested Solution I an = Van 100 + j 0 = = 2.43 - 14 A rms Z 40 + j10 I bn = 2.43 - 134 A rms I cn = 2.43 - 254 A rms Problem 10.8 A positive-sequence balanced three-phase wye-connected source supplies power to a balanced wyeconnected load. The magnitude of the line voltages is 150 V. If the load impedance per phase is 36 + j12 , determine the line currents if Van = 0 . Suggested Solution I an 36 Van j12 Van = 150 3 0 = 86.60 V rms I an = 86.6 + j 0 = 2.28 - 18.43 A rms 36 + j12 I bn = 2.28 - 138.43 A rms I cn = 2.28 - 258.43 A rms Problem 10.9 An abc-sequence balanced three-phase wye-connected source supplies power to a balanced wye-connected load. The line impedance per phase is 1 + j 0 , and the load impedance per phase is 20 + j 20 . If the source line voltage Vab is 1000 V rms , find the line currents. Suggested Solution Van = 100 3 ( 0 - 30 ) = 57.74 - 30 V rms I aA = 57.74 - 30 57.74 - 30 = = 1.99 - 73.6 A rms 21 + j 20 ( 20 + j 20 ) + (1 + j 0 ) I bB = 1.99 - 193.6 A rms I cC = 1.99 - 313.6 A rms Problem 10.10 In a three-phase balanced wye-wye system, the source is an abc-sequence set of voltages with Van = 12060 V rms . The per phase impedance of the load is 12 + j16 . If the line impedance per phase is 0.8 + j1.4 , find the line currents and the load voltages. Suggested Solution Line currents: I AN = Van 12060 12060 = = = 5.566.3 A rms Z + Z L (12 + j16 ) + ( 0.8 + j1.4 ) 21.653.7 I BN = 5.56 - 113.7 A rms I CN = 5.56126.3 A rms Load voltages: VAN = Z Z + Z L Van = 12 + j16 (12060 ) = 111.159.4 V rms 12.8 + j17.4 VBN = 111.1 - 60.6 V rms VCN = 111.1179.4 V rms Problem 10.11 An abc-sequence set of voltages feeds a balanced three-phase wye-wye system. The line and load impedances are 0.6 + j1 and 18 + j14 , respectively. If the load voltage on the a phase is VAN = 114.4718.99 V rms , determine the voltages at the line input. Suggested Solution Ia = VAN = 5.02 - 18.88 A rms 18 + j14 Van = I a ( Z line + Zload ) = I a Z line + VAN = ( 5.02 - 18.88 )( 0.6 + j1) + 114.4718.99 = 12020 V rms Vbn = 120 - 100 V rms Vcn = 120 - 220 V rms Problem 10.12 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages. The load voltage on the a phase is VAN = 108.5879.81 V rms , Z line = 1 + j1.4 , and Z load = 10 + j13 . Determine the input sequence of voltages. Suggested Solution Ia = VAN = 6.6227.38 A rms Zload Van = I a Z line + VAN = I a ( Zline + Zload ) = 12080 V rms Vbn = 120 - 40 V rms Vcn = 120 - 160 V rms Problem 10.13 A balanced abc-sequence of voltages feeds a balanced three-phase wye-wye system. The line and load impedances are 0.6 + j 0.9 and 8 + j12 , respectively. The load voltage on the a phase is VAN = 116.6310 V rms . Find the line voltage Vab . Suggested Solution Van = VAN Zline + Zload 8.6 + j12.9 = (116.6310 ) = 125.510 V rms 8 + j12 Zload Vab = 3 Van Van + 30 = 217.440 V rms ( ) Problem 10.14 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages. Zline = 1 + j1.8 , Zload = 14 + j12 , and the load voltage on the a phase is VAN = 398.117.99 V rms . Find the line voltage Vab . Suggested Solution Ia = VAN 398.117.99 = = 21.59 - 22.61 A rms 14 + j12 Zload Van = I a ( Z line + Zload ) = ( 21.59 - 22.61 )(15 + j13.8 ) = 44020 V rms Vab = 440 350 = 762.150 V rms Problem 10.15 An abc-phase sequence balanced three-phase source feeds a balanced load. The system is connected wyewye and Van = 0 . The line impedance is 0.5 + j 0.2 , the load impedance is 16 + j10 , and the total power absorbed by the load is 1836.54 W. Determine the magnitude of the source voltage Van . Suggested Solution P ,load = 1836.54 = 612.18 W 3 Z L = RL + jX L = 16 + j10 2 I aA = 612.18 612.18 = = 38.26 W 16 RL I aA = 6.19 A rms The total impedance is 16.5 + j10.2 = 19.431.72 Therefore, since Van = 0 , I aA = 6.19 - 31.72 A rms and Van = 6.19 19.4 = 120 V rms Problem 10.16 In a balanced three-phase wye-wye system, the total power loss in the lines is 272.57 W. Van = 105.2831.65 V rms and the power factor of the load is 0.77 lagging. If the line impedance is 2 + j1 , determine the load impedance. Suggested Solution If the total line loss is 272.57 W, then the loss per phase is 272.57 = 90.86 W 3 a I aA 2+j ZL A Van n N 2 2 P ,loss = I aA Re {Zline } = 2 I aA = 90.86 W I aA = 6.74 A rms Then, ZL = 105.28 = 15.62 6.74 Z = cos -1 ( 0.77 ) = 39.8 L (lagging pf) So, Z L = 15.6239.8 = 12 + j10 Problem 10.17 In a balanced three-phase wye-wye system the load impedance is 8 + j 4 . The source has phase sequence abc and Van = 1200 V rms . If the load voltage is VAN = 111.62 - 1.33 V rms , determine the line impedance. Suggested Solution I aA = 111.62 - 1.33 = 12.48 - 27.9 A rms 8 + j4 Vline = 1200 - 111.62 - 1.33 = 8.817.12 V rms Zline = 8.817.12 = 0.7145.02 = 0.5 + j 0.5 12.48 - 27.9 Problem 10.18 In a balanced three-phase wye-wye system the load impedance is 10 + j1 . The source has phase sequence abc and the line voltage Vab = 22030 V rms. If the load voltage VAN = 1200 V rms, determine the line impedance. Suggested Solution Z Y = 10 + j1 Van = Vab 3 Vab - 30 = ( ) 220 3 ( 30 - 30 ) = 127.020 V rms Per phase Y circuit: a Rline jXline 10 Van j1 n N A ZY VAN = Van Z Y + Zline V 127.020 - 1 = 0.595.71 Zline = Z Y an - 1 = (10 + j1) VAN 1200 Problem 10.19 In a balanced three-phase wye-wye system, the load impedance is 20 + j12 . The source has an abcphase sequence and Van = 1200 V rms. If the load voltage is VAN = 111.49 - 0.2 V rms, determine the magnitude of the line current if the load is suddenly short circuited. Suggested Solution I aA = 111.49 - 0.2 = 4.78 - 31.16 A rms 20 + j12 Vline = 1200 - 111.49 - 0.2 = 8.522.62 V rms Zline = Vline 8.522.62 = = 1.7833.78 4.78 - 31.16 I aA I aASC = 120 = 67.42 A rms 1.78 Problem 10.20 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages and Van = 12050 V rms. The load voltage on the a phase is 110.6529.03 V rms and the load impedance is 16 + j 20 . Find the line impedance. Suggested Solution a I aA Zline ZL A Van n N Van = 12050 V rms Z L = 16 + j 20 = 25.651.34 I aA = VAN = 110.6529.03 V rms VAN 110.6529.03 = = 4.32 - 22.31 A rms 25.651.34 ZL Then, Zline = Vline 12050 - 110.6529.03 = = 9.94139.4 4.32 - 22.31 I aA Problem 10.21 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages and Van = 12040 V rms. If the a-phase line current and line impedance are known to be 7.10 - 10.28 A rms and 0.8 + j1 , respectively, find the load impedance. Suggested Solution ZT = 12040 = 16.950.28 = 10.8 + j13 7.10 - 10.28 Z L = ZT - Zline = (10.8 + j13) - ( 0.8 + j1) = 10 + j12 Problem 10.22 An abc-sequence set of voltages feeds a balanced three-phase wye-wye system. If Van = 44030 V rms, VAN = 413.2829.78 V rms, and Z line = 1.2 + j1.5 , find the load impedance. Suggested Solution Ia = Van - VAN 44030 - 413.2829.78 = = 13.94 - 17.93 A rms 1.2 + j1.5 Zline VAN 413.2829.78 = = 19.95 + j 21.93 13.94 - 17.93 Ia Zload = Problem 10.23 In a three-phase balanced positive-sequence system a delta-connected source supplies power to a wyeconnected load. If the line impedance is 0.2 + j 0.4 , the load impedance 6 + j 4 , and the source phase voltage Vab = 20840 V rms, find the magnitude of the line voltage at the load. Suggested Solution Vab = 20840 V rms Van = 12010 V rms Then, VAN = 6 + j4 (12010 ) = 113.88.33 V rms 6.2 + j 4.4 VAB = 3 VAN = 3 (113.8 ) = 197.28 V rms Problem 10.24 Given the network shown, compute the line currents and the magnitude of the phase voltage at the load. a 0.4 j0.8 A 8 208-220 V rms 20820 V rms 8 0.4 0.4 j0.8 j0.8 j6 B 208-100 V rms j6 N 8 j6 C c b Suggested Solution Vab = 20820 V rms Zline = 0.4 + j 0.8 ZY = 8 + j6 Per phase Y circuit: a Zline I aA A Van n ZY N Van = Vab 3 Vab - 30 = 120 - 10 V rms ( ) I aA = Van 120 - 10 = = 11.10 - 49.00 A rms 8.4 + j 6.8 Zline + Z Y I bB = 11.10 - 169 A rms I cC = 11.1071 A rms VAN = I aA Z Y = 11.10 ( 8 + j 6 ) = 11.10 10 = 111 V rms VBN = VCN = VAN = 111 V rms Problem 10.25 In a balanced three-phase delta-wye system the source had an abc-phase sequence. The line and load impedances are 0.6 + j 0.3 and 12 + j 7 , respectively. If the line current I aA = 9.6 - 20 A rms, determine the phase voltages of the source. Suggested Solution Van = ( 9.6 - 20 )(12.6 + j 7.3) = 139.7810.09 V rms Vab = 139.78 3 (10.09 + 30 ) = 242.1140.09 V rms Vbc = 242.11 ( 40.09 - 120 ) = 242.11 - 79.91 V rms Vca = 242.11 ( 40.09 + 120 ) = 242.11160.09 V rms Problem 10.26 An abc-phase-sequence three-phase balanced wye-connected 60-Hz source supplies a balanced deltaconnected load. The phase impedance in the load consists of a 20 - resistor in series with a 50-mH inductor, and the phase voltage at the source is Van = 12020 V rms. If the line impedance is zero, find the line currents in the system. Suggested Solution Z = 20 + j 377 ( 0.05 ) Zline = 0 Van = 12020 V rms Y: Van ZY ZY = VAN VAB 3 Z = = = 6.67 + j 6.21 I aA 3 I AB 3 I aA = 13.10 - 23.28 A rms I aA = I bB = 13.10 - 143.28 A rms I cC = 13.1096.72 A rms Problem 10.27 An abc-phase-sequence three-phase balanced wye-connected source supplies a balanced delta-connected load. The impedance per phase in the delta load is 12 + j 9 . The line voltage at the source is Vab = 120 340 V rms. If the line impedance is zero, find the line currents in the balanced wye-delta system. Suggested Solution Zline = 0 Vab = VAB I AB = 120 340 = 13.863.13 A rms 12 + j 9 I aA = 13.86 3 ( 3.13 - 30 ) = 24.01 - 26.87 A rms , I bB = 24.01 - 146.87 A rms , and I cC = 24.0193.13 A rms Problem 10.28 An abc-phase-sequence three-phase balanced wye-connected source supplies power to a balanced deltaconnected load. The impedance per phase in the load is 14 + j12 . If the source voltage for the a phase is Van = 12080 V rms, and the line impedance is zero, find the phase currents in the wye-connected source. Suggested Solution Van = 12080 V rms Vab = 120 3110 V rms Zline = 0 Vab = VAB I AB = VAB 120 3110 = = 11.2769.4 A rms , Zload 14 + j12 I bB = 19.52 - 80.6 A rms and I cC = 19.52159.4 A rms Problem 10.29 An abc-phase-sequence three-phase balanced wye-connected source supplies a balanced delta-connected load. The impedance per phase of the delta load is 10 + j8 . If the line impedance is zero and the line current in the a phase is known to be I aA = 28.10 - 28.66 A rms, find the load voltage VAB . Suggested Solution If I aA = 28.1 - 28.66 A rms , then I AB = 28.1 3 ( -28.66 + 30 ) = 16.221.34 A rms VAB = I AB Z = (16.221.34 )(10 + j8 ) = 207.840 V rms Problem 10.30 An abc-phase-sequence three-phase balanced wye-connected source supplies power to a balanced deltaconnected load. The impedance per phase of the delta load is 14 + j11 . If the line impedance is zero and the line current in the a phase is I aA = 20.2231.84 A rms, find the voltages of the balanced source. Suggested Solution ZY = Z 14 11 = + j = 5.9338.2 3 3 3 Van = I aA Z Y = ( 20.2231.84 )( 5.9338.2 ) = 12070 V rms Vbn = 120 - 50 V rms Vcn = 120190 V rms Problem 10.31 In a balanced three-phase wye-delta system, the source has an abc-phase sequence and Van = 1200 V rms. If the line current is I aA = 4.820 A rms, find the load impedance per phase in the delta. Suggested Solution If Van = 1200 , then VAB = 120 330 V rms and if I aA = 4.820 , then I AB = 4.8 3 50 A rms Zload = VAB = 70.48 - j 25.65 I AB Problem 10.32 In a balanced three-phase wye-delta system, the source has an abc-phase sequence and Van = 12040 V rms. The line and load impedance are 0.5 + j 0.4 and 24 + j18 , respectively. Find the delta currents in the load. Suggested Solution 1 Using Y conversion, Z Y = Z = 8 + j 6 3 I aA = 12040 = 11.283.02 A rms 8.5 + j 6.4 Then, I AB = 11.28 3 ( 3.02 + 30 ) = 6.5133.02 A rms , I BC = 6.51 - 86.98 A rms , and I CA = 6.51153.02 A rms Problem 10.33 In a three-phase balanced delta-delta system, the source has an abc-phase sequence. The line and load impedances are 0.3 + j 0.2 and 9 + j 6 , respectively. If the load current in the delta is I AB = 1540 A rms, find the phase voltages of the source. Suggested Solution Converting to wye sources, Z Y = Z 3 I aA = 15 3 ( 40 - 30 ) = 2610 A rms Van = ( I an )( Zline + Z Y ) = ( 2610 )( 3.3 + j 2.2 ) = 10343.7 V rms Then, Vab = 103 373.7 = 178.573.7 V rms, Vbc = 178.5 - 46.3 V rms, and Vca = 178.5193.7 V rms Problem 10.34 In a balanced three-phase delta-delta system, the source has an abc-phase sequence. The phase angle for the source voltage is Vab = 40 and I ab = 415 A rms. If the total power absorbed by the load is 1400 W, find the load impedance. Suggested Solution PLtotal = 1400 W P = 1400 = 466.67 W 3 Since Vab = 40 and I ab = 415 A rms P = VAB I AB cos ( 40 - 15 ) No line impedance VAB = 466.67 = 128.73 V rms 4 cos 25 ZL = VAB 128.7340 = = 32.1825 I AB 415 Problem 10.35 A three-phase load impedance consists of a balanced wye in parallel with a balanced delta. What is the equivalent wye load and what is the equivalent delta load if the phase impedance of the wye and delta are 6 + j 3 and 15 + j12 , respectively? Suggested Solution Equivalent Y: Z 'Y = Z ' 3 Ztotal = ( 6 + j3)( 5 + j 4 ) 18 + j39 = = 2.77 + j1.78 ( 6 + j3) + ( 5 + j 4 ) 11 + j 7 Equivalent : Z = 3Z Y = 3 ( 2.77 + j1.78 ) = 8.31 + j 5.34 Problem 10.36 In a balanced three-phase system, the abc-phase-sequence source is wye connected and Van = 12020 V rms. The load consists of two balanced wyes with phase impedances of 8 + j 6 and 12 + j8 . If the line impedance is zero, find the line currents and the phase currents in each load. Suggested Solution a I aA A 8 Van j6 12 I1 j8 I2 n N I1 = 12020 = 12 - 16.87 A rms 8 + j6 12020 = 8.32 - 13.69 A rms 12 + j8 I2 = I aA = I1 + I 2 = 20.3 - 15.57 A rms The other currents are shifted by -120 and -240 Problem 10.37 In a balanced three-phase system, the source is a balanced wye with an abc-phase sequence and Vab = 20860 V rms. The load consists of a balanced wye with a phase impedance of 8 + j 5 in parallel with a balanced delta with a phase impedance of 21 + j12 . If the line impedance is 1.2 + j1 , find the phase currents in the balanced wye load. Suggested Solution Converting Z to Z : Y Z = 7 + j 4 Y Then, Z L = ( 8 + j 5 ) ( 7 + j 4 ) = 4.3530.8 Vab = 20860 V rms Van = 208 3 ( 60 - 30 ) = 12030 V rms I aA = Van 12030 = = 20.3 - 3.2 A rms Z L + Zline 5.933.2 VAN = I aA Z L = ( 20.33 - 3.2 )( 4.3530.8 ) = 88.427.6 V rms I AN = VAN 88.427.6 = = 9.37 - 4.4 A rms ZY 8 + j5 I BN = 9.37115.6 A rms I CN = 9.37 - 124.4 A rms Problem 10.38 In a balanced three-phase system, the source is a balanced wye with an abc-phase sequence and Vab = 20850 V rms. The load is a balanced wye in parallel with a balanced delta. The phase impedance of the wye is 5 + j 3 and the phase impedance of the delta is 18 + j12 . If the line impedance is 1 + j 0.8 , find the line currents and the phase currents in the loads. Suggested Solution Z = Y Z = 6 + j4 3 a 1 I aA j0.8 5 Van j3 j4 6 A n N Van = 208 3 ( 50 - 30 ) = 12020 V rms ZL = ( 5 + j3)( 6 + j 4 ) = 3.2232.18 ( 5 + j 3) + ( 6 + j 4 ) 12020 12020 = = 26.67 - 13.94 A rms Zline + Z L 4.533.94 I aA = VAN = I aA Zload = 85.8818.24 V rms Yload : I AN = VAN = 14.73 - 12.72 A rms 5 + j3 load : VAB = 85.88 348.24 V rms I AB = VAB = 6.8814.55 A rms 18 + j12 The other currents are shifted by -120 and -240 Problem 10.39 In a balanced three-phase system the source, which has an abc-phase sequence, is connected in delta and Vab = 20855 V rms. There are two loads connected in parallel. Load 1 is connected in wye and the phase impedance is 4 + j 3 . Load 2 is connected in wye and the phase impedance is 8 + j 6 . Compute the delta currents in the source if the line impedance connecting the source to the loads is 0.2 + j 0.1 . Suggested Solution a 0.8 I aA j0.1 4 Van j3 j6 8 A n N Van = 208 3 ( 55 - 30 ) = 12025 V rms ZL = ( 4 + j3)(8 + j 6 ) = 3.3336.87 = 2.66 + j 2.0 ( 4 + j 3) + ( 8 + j 6 ) 12025 12025 = = 33.8 - 11.29 A rms Zline + Z L 2.86 + j 2.1 33.8 3 ( -11.29 + 30 ) = 19.5118.71 A rms I aA = I ba = I cb = 19.51 (18.71 - 120 ) = 19.51 - 101.29 A rms I ac = 19.51 (18.71 - 240 ) = 19.51 - 221.29 A rms Problem 10.40 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two parallel wye-connected loads. The phase impedance of load 1 and load 2 is 4 + j 4 and 10 + j 4 , respectively. The line impedance connecting the source to the loads is 0.3 + j 0.2 . If the current in the a phase of load 1 is I AN1 = 1020 A rms , find the delta currents in the source. Suggested Solution By current division: I AN1 = I aA ( 4 + j 4 ) + (10 + j 4 ) 10 + j 4 I aA = 1527.9 A rms Then, since a source: I ab = - I aA 3 ( + 30 ) = 15 3 - 122.1 = 8.64 - 122.1 A rms I bc = 8.64 - 242.1 A rms I ca = 8.64 - 2.1 A rms a I aA I ab I ca c I bc I cC b I bB Problem 10.41 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two loads connected in parallel. The line connecting the source to the loads has an impedance of 0.2 + j 0.1 . Load 1 is connected in wye and the phase impedance is 4 + j 2 . Load 2 is connected in delta, and the phase impedance is 12 + j 9 . The current I AB in the delta load is 1645 A rms . Find the phase voltages of the source. Suggested Solution The Y equivalent circuit is: a 0.2 I aA j0.1 4 Van j2 j3 4 A 16 315 A rms n N Using current division: I aA ( 4 + j 2 ) 8 + j5 = 16 315 I aA = 58.4320.44 A rms Van = I aA ( 0.2 + j 0.1) + 16 315 ( 4 + j 3) = 151.3651.46 V rms ( ) Vab = 262.1681.46 V rms , Vbc = 262.16 - 38.54 V rms , and Vca = 262.16 - 158.54 V rms Problem 10.42 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two loads connected in parallel. Load 1 is connected in wye and has a phase impedance of 6 + j 2 . Load 2 is connected in delta and has a phase impedance of 9 + j 3 . The line impedance is 0.4 + j 0.3 . Determine the phase voltages of the source if the current in the a phase of load 1 is I AN1 = 1230 A rms . Suggested Solution The Y equivalent circuit is: a 0.4 I aA j0.3 3 Van j1 j2 6 A I AN1 n N Using current division: I aA ( 3 + j1) 9 + j3 = 1230 A rms I aA = 3630 A rms Van = ( I aA )( 0.4 + j 0.3) + (1230 )( 6 + j 2 ) = 93.1451.93 V rms Vab = 93.14 3 ( 51.93 + 30 ) = 161.3281.93 V rms Vbc = 161.32 ( 81.93 - 120 ) = 161.32 - 38.07 V rms Vca = 161.32 ( 81.93 - 240 ) = 161.32 - 158.07 V rms Problem 10.43 A balanced three-phase delta-connected source supplies power to a load consisting of a balanced delta in parallel with a balanced wye. The phase impedance of the delta is 24 + j12 , and the phase impedance of the wye is 12 + j8 . The abc-phase-sequence source voltages are Vab = 44060 V rms , Vbc = 440 - 60 V rms and Vca = 440 - 180 V rms , and the line impedance per phase is 1 + j 0.8 . Find the line currents and the power absorbed by the wye-connected load. Suggested Solution Vab = 44060 V rms Van = 440 3 ( 60 - 30 ) = 25430 V rms Equivalent single-phase (a-phase) diagram: a 1 I aA j0.8 12 Van j8 j4 8 A n N ZL = (12 + j8)(8 + j 4 ) 20 + j12 = 5.5429.3 ZT = Zline + Z L = 6.830.9 I aA = 25430 = 37.35 - 1 A rms 6.830.9 I bB = 37.35 - 121 A rms and I cC = 37.35119 A rms VAN = I aA Z L = 206.9328.3 V rms I AN - Y = VAN 206.9328.3 = = 14.37 - 5.4 A rms ZY 12 + j8 2 PY,load = 3 (14.37 ) (12 ) = 7.434 kW Problem 10.44 An abc-sequence wye-connected source having a phase-a voltage of 1200 V rms is attached to a wyeconnected load having an impedance of 8070 . If the line impedance is 420 , determine the total complex power produced by the voltage source and the real and reactive power dissipated by the load. Suggested Solution I aA = 1200 1200 = = 1.45 - 67.88 A rms 8070 + 420 31.12 + j 76.55 VL = I aA Z Y = (1.45 - 67.88 )( 8070 ) = 116.182.12 V rms S , S = VI = (1200 )(1.4567.88 ) = 17467.88 VA ST , S = 3S , S = 52267.88 VA S , L = (116.182.12 )(1.4562.88 ) = 168.4670 = 57.62 + j158.30 VA S L = 3S , L = 505.3870 = 172.86 + j 474.9 VA PL = 172.86 W and QL = 474.9 VAR Problem 10.45 The magnitude of the complex power (apparent power) supplied by a three-phase balanced wye-wye system is 3600 VA. The line voltage is 208 V rms. If the line impedance is negligible and the power factor angle of the load is 25 , determine the load impedance. Suggested Solution S = 3 VL I L IL = 3600 208 3 = 9.99 A rms ZY = 208 3 120 25 = 25 = 12.0125 = 10.88 + j 5.08 9.99 9.99 Problem 10.46 A balanced three-phase wye-wye system has two parallel loads. Load 1 is rated at 3000 VA, 0.7 pf lagging, and load 2 is rated at 2000 VA, 0.75 pf leading. If the line voltage is 208 V rms, find the magnitude of the line current. Suggested Solution S1 = 3000 cos -1 ( 0.7 ) = 300045.57 = 2100 + j 2142.43 VA S 2 = 2000 - cos -1 ( 0.75 ) = 2000 - 41.41 = 1500 - j1322.88 VA ST = S1 + S 2 = 3600 + j819.55 = 3692.1112.82 VA IL = 3692.11 208 3 = 10.25 A rms Problem 10.47 Two industrial plants represent balanced three-phase loads. The plants receive their power from a balanced three-phase source with a line voltage of 4.6 kV rms. Plant 1 is rated at 300 kVA, 0.8 pf lagging and plant 2 is rated at 350 kVA, 0.84 pf lagging. Determine the power line current. Suggested Solution S1 = 300 cos -1 ( 0.8 ) = 30036.87 = 240 + j180 kVA S 2 = 350 cos -1 ( 0.84 ) = 35032.86 = 294 + j189.9 kVA ST = S1 + S 2 = 534 + j 369.9 = 649.634.71 kVA IL = 649.6 4.6 3 = 81.53 A rms Problem 10.48 A cluster of loads is served by a balanced three-phase source with a line voltage of 4160 V rms. Load 1 is 240 kVA at 0.8 pf lagging and load 2 is 160 kVA at 0.92 pf lagging. A third load is unknown except that it has a power factor of unity. If the line current is measured and found to be 62 A rms, find the complex power of the unknown load. Suggested Solution ST = 3 VL I L = 3 ( 4160 )( 62 ) = 446730.54 VA S1 = 240 cos -1 ( 0.8 ) = 192000 + j144000 VA S 2 = 160 cos -1 ( 0.92 ) = 147200 + j 62707 VA (192000 + 147200 + P3 ) + j (144000 + 62707 + 0 ) = ST ( 339200 + P3 ) + ( 206707 ) 2 2 = 446730.54 P3 = 56831 W S3 = P3 + j 0 = 568310 VA Problem 10.49 A balanced three-phase source serves two loads: Load 1: 36 kVA at 0.8 pf lagging Load 2: 18 kVA at 0.6 pf lagging The line voltage at the load is 208 V rms at 60 Hz. Find the line current and the combined power factor at the load. Suggested Solution Vab = 208 V rms S1 = 36 cos -1 ( 0.8 ) = 3636.87 = 28.8 + j 21.6 kVA S 2 = 18 cos -1 ( 0.6 ) = 1853.13 = 10.8 + j14.4 kVA S L = S1 + S 2 = 39.6 + j 36.0 = 53.5242.27 kVA S L = 3 Vab I aA I aA = 53.52 103 208 3 = 148.56 A rms pf L = cos = cos ( 42.27 ) = 0.74 lagging Problem 10.50 A balanced three-phase source serves the following loads: Load 1: 48 kVA at 0.9 pf lagging Load 2: 24 kVA at 0.75 pf lagging The line voltage at the load is 208 V rms at 60 Hz. Determine the line current and the combined power factor at the load. Suggested Solution Vab = 208 V rms S1 = 48 cos -1 ( 0.9 ) = 4825.84 = 43.2 + j 20.92 kVA S 2 = 24 cos -1 ( 0.75) = 2441.41 = 18 + j15.87 kVA S L = S1 + S 2 = 61.2 + j 36.79 = 71.4131.02 kVA S L = 3 Vab I aA I aA = 71.41 103 208 3 = 198.21 A rms pf L = cos = cos ( 31.02 ) = 0.86 lagging Problem 10.51 A small shopping center contains three stores that represent three balanced three-phase loads. The power lines to the shopping center represent a three-phase source with a line voltage of 13.8 kV rms. The three loads are: Load 1: 500 kVA at 0.8 pf lagging Load 2: 400 kVA at 0.85 pf lagging Load 3: 300 kVA at 0.90 pf lagging Find the magnitude of the power line current. Suggested Solution Vab = 13.8 kV rms S1 = 500 cos -1 ( 0.8 ) = 50036.87 = 400 + j 300 kVA S 2 = 400 cos -1 ( 0.85 ) = 40031.79 = 340 + j 210.72 kVA S3 = 300 cos -1 ( 0.9 ) = 30025.84 = 270 + j130.76 kVA ST = S1 + S 2 + S3 = 1010 + j 641.48 = 1196.532.42 kVA ST = 3 Vab I aA I aA = 1196.5 13.8 3 = 50.1 A rms Problem 10.52 The following loads are served by a balanced three-phase source: Load 1: 18 kVA at 0.8 pf lagging Load 2: 8 kVA at 0.8 pf leading Load 3: 12 kVA at 0.75 pf lagging The load voltage is 208 V rms at 60 Hz. If the line impedance is negligible, find the power factor of the source. Suggested Solution S1 = 18 cos -1 ( 0.8 ) = 1836.87 = 14.40 + j10.80 kVA S 2 = 8 - cos -1 ( 0.8 ) = 8 - 36.87 = 6.40 - j 4.80 kVA S3 = 12 cos -1 ( 0.75 ) = 1241.41 = 9.00 + j 7.94 kVA ST = S1 + S 2 + S3 = 29.80 + j13.94 = 32.9025.07 kVA pf = cos = cos ( 25.07 ) = 0.91 lagging Problem 10.53 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 30 kVA at 0.8 pf lagging Load 2: 24 kW at 0.6 pf leading Load 3: unknown If the line voltage and total complex power at the load are 208 V rms and 600 kVA , respectively, find the unknown load. Suggested Solution VAB = 208 V rms pf at source = 1.0 ST = 3 VAB I aA cos -1 (1.0 ) = 60.00 kVA I aA = 60 103 208 3 = 166.8 A rms S1 = 30 cos -1 ( 0.8 ) = 3036.87 = 24 + j18 kVA S2 = 24 - cos -1 ( 0.6 ) = 40 - 53.13 = 24 - j 32 kVA 0.6 P3 = 60.0 - 24 - 24 = 12.0 kW Q3 = 0 - 18 + 32 = 14 kVAR PT = P + P2 + P3 1 QT = Q1 + Q2 + Q3 S3 = 12.0 + j14 kVA = 18.44 kVA at 0.65 pf lagging Problem 10.54 A balanced three-phase source serves the following loads: Load 1: 18 kVA at 0.8 pf lagging Load 2: 10 kVA at 0.7 pf leading Load 3: 12 kW at unity pf Load 4: 16 kVA at 0.6 pf lagging The magnitude of the line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.02 + j 0.04 . Find the magnitude of the line voltage and power factor at the source. Suggested Solution VAB = 208 V rms Zline = 0.02 + j 0.04 S1 = 18 cos -1 ( 0.8 ) = 1836.87 = 14.4 + j10.8 kVA S 2 = 10 - cos -1 ( 0.7 ) = 10 - 45.57 = 7.0 - j 7.14 kVA S3 = 12 0 = 120 = 12 + j 0 kVA 1.0 S 4 = 16 cos -1 ( 0.6 ) = 1653.13 = 9.6 + j12.8 kVA S L = S1 + S 2 + S3 + S 4 = 43 + j16.46 = 46.0420.95 kVA V2 S L = 3 AN Z L V2 Z L = 3 AN SL VAN = VAB 3 = 208 3 = 120 V rms ; Let VAN = 0 (VAN 0 )2 V 2 1202 = 3 AN = 3 Then, Z L = 3 = 0.9420.95 = 0.88 + j 0.34 SL S L 46.04 - 20.95 Per-phase Y circuit: a I aA A Zline I aA = VAN 1200 = = 127.66 - 20.95 A rms ZL 0.9420.95 Van n ZL Van = I aA [ Z line + Z L ] = 124.721.94 V rms Vab = 3 Van = 216.02 V rms N pf = cos Van - IaA = cos ( 22.89 ) = 0.92 lagging ( ) Problem 10.55 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 18 kW at 0.8 pf lagging Load 2: 10 kVA at 0.6 pf leading Load 3: unknown If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 41.93 kVA and the combined power factor at the load is 0.86 lagging, find the unknown load. Suggested Solution VAB = 208 V rms S L = 3 VAB I aA = 41.93 kVA pf L = 0.86 lagging I aA = 41.93 103 208 3 = 116.39 A rms S = cos -1 ( 0.86 ) = 30.68 L S L = 41.9330.68 = 36.06 + j 21.40 kVA = S1 + S 2 + S3 S1 = 18 cos -1 ( 0.8 ) = 22.536.87 = 18 + j13.5 kVA 0.8 S 2 = 10 - cos -1 ( 0.6 ) = 10 - 53.13 = 6 - j8 kVA S3 = S L - S1 - S 2 = ( 36.06 + j 21.40 ) - (18 + j13.5 ) - ( 6 - j8 ) = 12.06 + j15.9 kVA = 19.96 kVA at 0.60 pf lagging Problem 10.56 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 20 kVA at 0.6 pf lagging Load 2: 12 kW at 0.75 pf lagging Load 3: unknown If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 35.52 kVA and the combined power factor at the load is 0.88 lagging, find the unknown load. Suggested Solution VAB = 208 V rms S L = 3 VAB I aA = 35.52 kVA pf L = 0.88 lagging I aA = 35.52 103 208 3 = 98.6 A rms S = cos -1 ( 0.88) = 28.36 L S L = 35.5228.36 = 31.26 + j16.87 kVA = S1 + S 2 + S3 S1 = 20 cos -1 ( 0.6 ) = 2053.13 = 12 + j16 kVA S2 = 12 cos -1 ( 0.75) = 1641.41 = 12 + j10.58 kVA 0.75 S3 = S L - S1 - S 2 = ( 31.26 + j16.87 ) - (12 + j16 ) - (12 + j10.58 ) = 7.26 - j 9.71 kVA = 12.13 kVA at 0.60 pf leading Problem 10.57 A standard practice for utility companies is to divide its customers into single-phase users and three-phase users. The utility must provide three-phase users, typically industries, with all three phases. However, single-phase users, residential and light commercial, are connected to only one phase. To reduce cable costs, all single-phase users in a neighborhood are connected together. This means that even if the threephase users present perfectly balanced loads to the power grid, the single-phase loads will never be in balance, resulting in current flow in the neutral connection. Consider the 60-Hz, abc-sequence network shown. With a line voltage of 41630 V rms , phase a supplies the single-phase users on A Street, phase b supplies B Street and phase c supplies C Street. Furthermore, the three-phase industrial load, which is connected in delta, is balanced. Find the neutral current. a b c I AN I BN B Street 30 kW pf = 1 A B C I CN C Street 60 kW pf = 1 Three-phase 36 kW pf = 0.5 lagging 2400 V rms n 240 - 120 V rms 240120 V rms I nN A Street 48 kW pf = 1 N Suggested Solution PA = 48, 000 = Van I AN I AN = PA = 2000 A rms Van PB = 125 - 120 A rms VBN PC = 250 - 240 A rms VCN I BN = I CN = I Nn = I AN + I BN + I CN = 108.9783.41 A rms I nN = 108.97 - 96.59 A rms Problem 10.58 A three-phase abc-sequence wye-connected source with Van = 2200 V rms supplies power to a wyeconnected load that consumes 150 kW of power at a pf of 0.8 lagging. Three capacitors are found that each have an impedance of - j 2.0 , and they are connected in parallel with the previous load in a wye configuration. Determine the power factor of the combined load as seen by the source. Suggested Solution Van = 2200 V rms P = 50 kW 1 pf = 0.8 lagging ZC = - j 2 Original situation per phase: S old = 50 cos -1 ( 0.8 ) 0.8 = 62.536.87 = 50 + j 37.5 kVA Corrected situation: Pnew = Pold = 50 kW Qnew = Qold + QC = ( 37.5 103 ) + QC QC = - Van ZC 2 = -24.2 kVAR Qnew = 13.3 kVAR S new = 51.7414.90 kVA pf new = cos new = cos (14.90 ) = 0.97 lagging Problem 10.59 A three-phase abc-sequence wye-connected source with Van = 2200 V rms supplies power to a wyeconnected load that consumes 150 kW of power at a pf of 0.8 lagging. Three capacitors are found that each have an impedance of - j 2.0 , and they are connected in parallel with the previous load in a delta configuration. Determine the power factor of the combined load as seen by the source. Suggested Solution Van = 2200 V rms P = 50 kW 1 pf = 0.8 lagging ZC = - j 2 Z CY = ZC 2 =-j 3 3 Original situation per phase: S old = 50 cos -1 ( 0.8 ) 0.8 = 62.536.87 = 50 + j 37.5 kVA Corrected situation: Pnew = Pold = 50 kW Qnew = Qold + QC = ( 37.5 103 ) + QC QC = - Van 2 Z CY = -72.6 kVAR Qnew = -35.1 kVAR S new = 61.09 - 35.07 kVA pf new = cos new = cos ( -35.07 ) = 0.82 leading Problem 10.60 Find C in the network shown such that the total load has a power factor of 0.9 lagging. + 4.6 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 6 MVA 0.8 pf lagging Suggested Solution Vab = 4.6 kV rms pf new = 0.9 lagging S L1 = S L 3 3 S L 3 = 6 MVA f = 60 Hz pf old = 0.8 lagging = 2 cos -1 0.8 = 1.60 + j1.20 MVA Old situation per phase: S old = 1.60 + j1.20 MVA New situation: S new = 1.60 cos -1 ( 0.9 ) 0.9 = 1.60 + j 0.77 kVA = Pold + j ( Qold + QC ) QC = 0.77 - 1.20 = -0.43 MVAR But, 2 QC = - CYVan = - 2 CYVab 3 CY = 161.7 F Problem 10.61 Find C in the network shown such that the total load has a power factor of 0.9 leading. + 4.6 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 6 MVA 0.8 pf lagging Suggested Solution pf new = 0.9 leading f = 60 Hz Original complex power per phase: S old = 236.87 = 1.6 + j1.2 MVA Corrected complex power per phase: S new = 1.6 - cos -1 ( 0.9 ) 0.9 = 1.6 - j 0.77 MVA Qold + QC = Qnew QC = -0.77 - 1.20 = -1.97 MVAR But, 2 QC = - CYVan = - 2 CYVab 3 CY = 740.9 F Problem 10.62 Find C in the network shown such that the total load has a power factor of 0.92 leading. + 34.5 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 20 MVA 0.707 pf lagging Suggested Solution Vab = 34.5 kV rms f = 60 Hz S3 = 20 MVA pf old = 0.707 lagging pf new = 0.92 leading Old complex power per phase: S old = 20 cos -1 ( 0.707 ) = 4.71 + j 4.71 MVA 3 New complex power per phase: S new = 4.71 - cos -1 ( 0.92 ) = 4.71 - j 2.01 MVA 0.92 2 QC = - CVab = Qnew - Qold = -6.72 MVAR C = 15.0 F Problem 10.63 Find C in the network shown such that the total load has a power factor of 0.92 lagging. + 34.5 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 20 MVA 0.707 pf lagging Suggested Solution Vab = 34.5 kV rms f = 60 Hz S3 = 20 MVA pf old = 0.707 lagging pf new = 0.92 lagging Old complex power per phase: S old = 20 cos -1 ( 0.707 ) = 4.71 + j 4.71 MVA 3 New complex power per phase: S new = 4.71 cos -1 ( 0.92 ) = 4.71 + j 2.01 MVA 0.92 2 QC = - CVab = Qnew - Qold = -2.70 MVAR C = 6.0 F Problem 10FE-1 A wye-connected load consists of a series RL impedance. Measurements indicate that the rms voltage across each element is 84.85 V. If the rms line current is 6 A, find the total complex power for the threephase load configuration. Suggested Solution VL = 84.85 + j84.85 = 12045 V 2 84.85 2 84.85 ST = 3 ( 6 ) + j ( 6) 6 6 = 1527.3 + j1527.3 = 216045 VA or ST = 3 (12045 )( 6 ) = 216045 VA Problem 10FE-2 A balanced three-phase delta-connected load consists of an impedance of 12 + j12 . If the line voltage at the load is measured to be 230 V rms, find the magnitude of the line current and the total real power absorbed by the three-phase configuration. Suggested Solution I = 230 = 13.55 A rms 12 + j12 I L = 3 I = 23.47 A rms Ptotal = 3 VL I L cos 45 = 6.61 kW or Ptotal = 3 I R = 3 (13.55 ) (12 ) = 6.61 kW 2 2 Problem 10FE-3 Two balanced three-phase loads are connected in parallel. One load with a phase impedance of 24 + j18 is connected in delta, and the other load has a phase impedance of 6 + j 4 and is connected in wye. If the line-to-line voltage is 208 V rms, determine the line current. Suggested Solution Assume 2080 V rms For : I = IL = 2080 = 6.933 - 36.87 A rms 24 + j18 ( 3 ) ( - 30)( 6.933 - 36.87) = 12 - 66.87 A rms For Y: 208 - 30 3 IL = = 16.64 - 63.69 A rms 6 + j4 Total line current: I L = 12 - 66.87 + 16.64 - 63.69 = 12.09 - j 25.96 = 28.64 - 65 A rms or Convert Y 8 + j6 Ztotal = ( 6 + j 4 )(8 + j 6 ) 24 + j 68 = = 4.1935.06 ( 6 + j 4 ) + (8 + j 6 ) 14 + j10 Then 208 - 30 3 IL = = 28.66 - 65.06 A rms 4.1935.06 Problem 10FE-4 The total complex power at the load of a three-phase balanced system is 2430 kVA . Find the real power per phase. Suggested Solution S = 24, 00030 VA = 20, 785 + j12, 000 VA P = 20, 785 = 6.928 kW 3 Problem 11.1 Determine the driving point impedance at the input terminals of the network shown in fig P11.1 as a function of s. R L C V0(t) Vi(t) Suggested Solution L 1 SL S 2 LCR + SL + R C Z ( s) = R + ( SL) = R + = R+ = 1 SC 1 + S 2 LC S 2 LC + 1 + SL SC Problem 11.2 Determine the voltage transfer function Vo(s)/Vc(s) as a function of s for the network shown in fig p11.1. Suggested Solution L C L 1 1 + SL SL) ( VO SL SC C = SC = = 2 (S ) = L R L S LCR + SL + R 1 Vi R+( SL) + RSL + SC C SC C R+ 1 + SL SC SL S LCR + SL + R 2 Problem 11.3 Determine the voltage transfer function Vo(s)/Vc(s) as a function of s for the network shown in fig p11.3. R1 R2 C Vi(t) V0(t) L VO (S ) = Vi ( R2 + SL) ( R2 + SL) + R1 SC 1 SC ( R2 + SL)( SCR1 + 1) ( R2 + SL)( SCR1 + 1) = = 2 2 R2 + SCR1 R2 + S LCR1 + SL + R1 S LCR1 + S ( L + CR1 R2 ) + R1 + R2 R1 + R2 1 )( S + ) L CR1 = 1 R R + R2 S2 + + 2S + 1 LCR1 R1C L (S + R2 1 )( S + ) L CR1 1 R R + R2 + 2S + 1 S2 + LCR1 R1C L (S + Suggested Solution ( R2 + SL) = R1 ( R2 + SL) + SCR1 + 1 Problem 11.4 Find the transfer impedance Vo(s)/Is(s) for the network shown in fig 11.4. 1F iS (t) 2 + 2H 4 V0 _ Suggested Solution 2 1F iS (t) I0 + 2H 4 V0 _ 2S + 2 IO = I S 1 2S + 2 + 4 + S 2 IO 2S + 2S V 8S ( S + 1) = 2 , VO = 4 I O , SO O = 2 I S 2S + 6S + 1 IS 2 S + 6 S +1 8S ( S + 1) 2S 2 + 6S +1 Problem 11.5 Find the driving point impedance at the input terminal of the circuit in fig 11.5. 1F 3 Z 3 3 3 1H Suggested Solution 1 1 1 Z 3 1H Zi ( s) = 4(1 + S ) 1 5( S + 1) 2 +1+ = 5+ S S S ( S + 5) Problem 11.6 Draw the bode plot for the network function. Suggested Solution H ( jw) = jw5 + 1 jw20 + 1 Problem 11.7 Draw the bode plot for the network function. H ( jw) = jw2 + 1 jw10 + 1 Suggested Solution H ( jw) = jw2 + 1 jw10 + 1 Problem 11.8 Draw the bode plot for the network function. H ( jw) = 10(10 jw + 1) (100 jw + 1)( jw + 1) Suggested Solution H ( jw) = Zeros: Poles: H(jo) 10(10 jw + 1) (100 jw + 1)( jw + 1) r/s & 1 r/s 10 = 20dB 0.1 0.01 = Problem 11.9 Draw the bode plot for the network function. H ( jw) = jw ( jw + 1)(0.1 jw + 1) Suggested Solution H ( jw) = Zero :dc Poles: 1& 10 r/s jw ( jw + 1)(0.1 jw + 1) Problem 11.10 Draw the bode plot for the network function. H ( jw) = 10 jw + 1 ( jw)(0.1 jw + 1) Suggested Solution H ( jw) = 10 jw + 1 ( jw)(0.1 jw + 1) zeros : 0.1 r/s poles : dc & 10 r/s Problem 11.11 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( jw) = 10 ( jw)(0.1 jw + 1) Suggested Solution Problem 11.12 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( jw) = 20(0.1 jw + 1) ( jw)(0.1 jw + 1)(0.01 jw + 1) Suggested Solution Problem 11.13 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( jw) = 100( jw) ( jw + 1)( jw + 10)( jw + 50) Suggested Solution 1 ( jw) S H ( jw) = jw jw ( jw + 1)( + 1)( +1) 10 50 Problem 11.14 Draw the bode plot for the network function. H ( jw) = 16 ( jw) ( jw2 + 1) 2 Suggested Solution H ( jw) = zeros: none poles: also: 2@dc, 10 r/s 16=24.1dB 16 ( jw) ( jw2 + 1) 2 Problem 11.15 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( jw) = 640( jw + 1)(0.01 jw + 1) ( jw) 2 ( jw + 10) Suggested Solution 640( jw + 1)(0.01 jw + 1) 64( jw + 1)(0.01 jw + 1) = ( jw) 2 ( jw + 10) ( jw) 2 (0.1 jw + 1) H ( jw) = poles: 2@dc, 10 r/s zeroe at : 1 &100 r/s also: 64=36.1dB Problem 11.16 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( jw) = 105 (5 jw + 1) 2 ( jw) 2 ( jw + 10)( jw + 100) 2 Suggested Solution 105 (5 jw + 1) 2 H ( jw) = ( jw) 2 ( jw + 10)( jw + 100) 2 (5 jw + 1) 2 H ( jw) = ( jw) 2 (0.1 jw + 1)(0.01 jw + 1) 2 poles: 2@dc, 10 r/s & 2@ 100r/s zeroe at : 2@ 0.2r/s also: 64=36.1dB Problem 11.17 Sketch the magnitude characteristic of the bode plot for the transfer function. G ( j ) = 10 j ( j + 1)( j + 10 ) 2 Suggested Solution G ( j ) = 10 j ( j + 1)( j + 10 ) 2 = 0.1 j ( j + 1)( 0.1 j + 1) 2 poles at : 1 and 2@ 10 r/s zeros at : dc also : 0.1 = - 20 dB Problem 11.18 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( j ) = 100 ( j ) 2 ( j + 1)( j + 10 ) ( j + 50 ) 2 Suggested Solution H ( j ) = 100 ( j ) 2 ( j + 1)( j + 10 ) ( j + 50 ) 2 (1/ 50 )( j ) H ( j ) = 2 ( j + 1)( 0.1 j + 1) ( 0.02 j + 1) 2 poles :1, 50 and 2@ 10 r/s zeros: 2 @ dc also: 1/50 = - 34 dB Problem 11.19 Sketch the magnitude characteristic of the bode plot for the transfer function. G ( j ) = - 2104 ( j + 1) ( j + 10 )( j + 100 ) 2 2 Suggested Solution G ( j ) = - 2104 ( j + 1) ( j + 10 )( j + 100 ) 2 2 = - (1/10 ) 2 ( j + 1) ( 0.1 j + 1)( 0.01 j + 1) 2 2 poles : 2@ 1 r/s, 1@ 10 r/s, 2@ 100 r/s zeros: 2 @ dc also: 0.1 = - 20 dB Problem 11.20 Sketch the magnitude characteristic of the bode plot for the transfer function. 64 ( j + 1) G ( j ) = - j 3 ( 0.1 j + 1) 2 Suggested Solution 64 ( j + 1) G ( j ) = - j 3 ( 0.1 j + 1) 2 poles : 3@dc, 1@ 10 r/s zeros: 2 @ 1 r/s also: 64 = 36.1 dB Problem 11.21 Sketch the magnitude characteristic of the bode plot for the transfer function. G ( j ) = - 2 ( j + 1) 3 Suggested Solution G ( j ) = - 2 ( j + 1) 3 poles: 3 @ 1 r/s zeros: 2@ dc Problem 11.22 Draw the Bode plot for the network function. H ( j ) = 72 ( j + 2 ) 2 j ( j ) + 2.4 j + 144 Suggested Solution H ( j ) = H ( j ) = 72 ( j + 2 ) 2 j ( j ) + 2.4 j + 144 ( 0.5 j + 1) 2 j ( j /12 ) + ( j / 60 ) + 1 zeros : 1@ 2 r/s simple pole : 1 @ dc complex poles T = (1/12)s $ 2=1/60 so Wo = 12 r/s = 0.1 Problem 11.23 Sketch the magnitude characteristics of the Bode plot for the transfer function. G ( j ) = 104 ( j + 1) - 2 + 6 j + 225 j ( j + 450 )( j + 50 ) 2 Suggested Solution G ( j ) = 104 ( j + 1) - 2 + 6 j + 225 j ( j + 450 )( j + 50 ) 2 2 2 ( j + 1) ( j /15 ) + ( 2 / 75 ) j + 1 G ( j ) = 2 ( j )( j / 450 + 1)( 0.02 j + 1) poles : dc 2@ 50 and 450 r/s simple pole : 1 r/s complex poles T = (1/15)s $ 2=2/75 so Wo = 15 r/s = 0.2 also: 2=6dB Problem 11.24 Sketch the magnitude characteristic of the Bode plot for the transfer function. H ( j ) = 81( jw + 0.1) j - 2 + 3.6 j + 81 Suggested Solution j 0.1 + 1 0.1 H ( j ) = - ( )2 3.6 j j + 1 + 81 81 Problem 11.25 Sketch the magnitude characteristic of the bode plot for the transfer function. H ( j ) = 6.4( j ) 2 ( j + 1)(1 - + 8 j + 64) Suggested Solution H ( j ) = 0.1( j ) ( j + 1)(- 2 64 + 1 j + 1) 8 POLES : 0.5r/s & 2@10r/s ZEROS: 0.1 r/s Problem 11.26 Determine H(jw) if the amplitude characteristic for H(jw) is shown in fig 26. Suggested Solution H ( jo) = -40dB = 0.01 H ( j ) = 10 j + 1 100(2 j + 1)(0.1 j + 1) 2 Poles: 10r/s & 2@ 80r/s Zeroes:1&120r/s Problem 11.27 Find H(jw) if its magnitude characteristic is shown in fig 27. Suggested Solution H ( jo) = 40dB = 100 j 100( j + 1)( + 1) 120 H ( j ) = j (0.1 j + 1)( + 1) 2 80 poles: 10 r/s $ 2 @ 80 r/s zeroes : 0.1 r/s H(jO) = 40 dB = 100 Problem 11.28 Find H(jw) if its magnitude characteristic is shown in fig 28. Suggested Solution There is a zero at w =10r/s, A pole at w=0 and A double pole at w=20r/s. Since the initial slope of 20dB /dec cuts the odB line at w=0, the gain is 10. .: j + 1) 10 H ( j ) = j + 1) 2 j ( 20 10( Problem 11.29 Find H(jw) if its amplitude characteristic is shown in fig 29. Suggested Solution The initial slope of 20dB/dec will cut the odB line at w=40r/s. Therefore the gain is 40. The zeros are at w=50r/s and w=1000r/s. The pole are ar w=0 and there is a double pole at w=400.: H ( j ) = 40( j j + 1)( + 1) 50 1000 j + 1) 2 j ( 100 Problem 11.30 Find H(jw) if its magnitude characteristics is shown in fig 11.30. Suggested Solution Poles: 0.8,100 $ 2@700 r/s Zeros: dc , 12 r/s H ( j 0.05) = -20dB = 0.1 jw K ( jw) + 1 12 H ( jw) = 2 jw jw + 1 + 1 (1.25 jw + 1) 100 100 H ( j 0.05) = K (0.05)(1) = 0.1 K = 2 (1) jw 2( jw) + 1 12 H ( jw) = 2 jw jw (1.25 jw + 1) + 1 + 1 100 100 Problem 11.31 Find H(jw) if its amplitude characteristics is shown in fig 11.31. Suggested Solution The initial slope indicates that the gain is 1,there are zeroes at w=0 and w=30. The poles are at w=1, w=100 and a double pole w=8/5 jw 1( jw) + 1 30 H ( jw) = 2 jw jw ( jw + 1) + 1 + 1 100 8 Problem 11.32 Given the magnitude characteristics in fig 11.32, find G(jw). Suggested Solution Poles : 2@ dc, 1@200r/s Zeros: 1.5 $ 10 r/s H ( j 0.02) = 80dB = 104 2 jw K (0.1 jw + 1) + 1 3 H ( jw) = jw jw ( jw) 2 + 1 + 1 80 200 4 H ( j 0.02) = 10 k = 400 2 jw + 1 400(0.1 jw + 1) 3 H ( jw) = jw jw + 1 + 1 ( jw) 2 80 200 Problem 11.33 Find G(jw) if the amplitude characteristic fir this function is shown in fig 11.33. Suggested Solution Zeros: 2@100 r/s Simple poles: dc $ 900 r/s Complex poles : =1/20s $ =0.1 So Wo=20 r/s $ 2=0.01 Also, H ( j 0.8) = 20dB = 10 H ( jw) = k (0.1 jw + 1) 2 2 jw jw jw + 1 + 1 ( jw) + 900 20 100 H ( j 0.8) = 10 K = 8 H ( jw) = 0.8(0.1 jw + 1) 2 2 jw jw jw + 1 + + 1 ( jw) 900 20 100 Problem 11.34 A series of RLC circuit resonates at 2000 rad/s. If C= 20F and it is known that the impedance at resonance is 2.4 ohm, compute the value of L, the Q of the circuit and the bandwidth. Suggested Solution r 1 L = 12.5mH WO = 2000 = s LC W L THEN Q= O = 10.42 R W 2000 r = 192 BW = O = Q 10.42 s Problem 11.35 A series of resonant circuit has a Q of 120 and a resonant frequency of 60,000 rad/s. Determine the halfpower frequencies and the bandwidth of the circuit. Suggested Solution BW = WO 60 K r = = 500 120 Q s 2 1 1 +1 WLO , HI = WO 2Q 2Q WHI = 60.25Kr / s ALSO BW=WHI = WLO WLO = 59.75 Kr / s Problem 11.36 Given the series RLC circuit in fig 11.36 if R=10ohm , find the values of L and C such that the network will have a resonant frequency of 100 kHz and a bandwidth of 1kHz. R L Vi(t) C Suggested Solution WO = 1 Kr = 200 s LC W L (100 K )(2 ) L Q= O = = 100 L = 1.59mH R 10 1 WO = C = 1.59nF LC C = 1.59nF Problem 11.37 Given the network in fig 11.37, find Wo ,Q, Wmax and Vo(max). 1 2mH + 6 cos wt V 10F V0 _ Suggested Solution 1 R vs(t) L V0 _ L + WO = 1 r = 7071 s LC Q= WO L = 14.14 R Wmax = WO (1 - Q VS 1 r ) = 7062 2 2Q s = 84.89V VO max = 1 1- 4Q 2 r s WO = 7071 Q=14.14 Wmax = 7062 r VO s max = 84.89V Problem 11.38 Repeat the problem 11.37 if the value of R is changed to 0.1ohm. Suggested Solution 1 R vs(t) L V0 _ L + WO = 1 LC Q= WO L R 1 ) 2Q 2 WO = 7071 r s Q=141 Wmax = 7071 r VO s Wmax = WO (1 - max = 846V Problem 11.39 A series RLC circuit is driven by a signal generator. The resonant frequency of the network is known to be 1600 rad/s and at that frequency the impedance seen by the signal generator is 5ohm. If C=20F, find L,Q and the bandwidth. Suggested Solution WO = 1 L = 19.5mH LC BW = R / L = 256r / s W Q = O = 6.25 BW L = 19.5mH BW = 256r / s Q = 6.25 Problem 11.40 A variable frequency voltage source drives the network in fig 11.40. Determine the resonant frequency ,Q,BW and the average power dissipated by the network at resonance. 4 R 5F 50mH 12 cos t V Suggested Solution WO = 1 kr 2 s LC Q= WO L = 25 R WO = 80r / s Q P = 18W BW = Problem 11.41 In the network in fig 11.41 the inductor value is 30mH, and the circuit is driven by a variable frequency source. If the magnitude of the current at resonance is 12A,Wo=1000rad/s and L=10mH, find C, Q and the bandwidth of the circuit. R 36 cos (t + 45O) V L C Suggested Solution 36 R I O = R = 3 IO = WO = 1 kr 1 C = 10nF s LC Q= WO L = 3.33 R WO = 300r / s Q BW = Problem 11.42 Given the network in fig 11.42, find Vo(max). 0.5 4mH + R 12 cos t V L C 25F V0 _ Suggested Solution WO = BW = 1 kr 3162 s LC WO = 25.98 Q = Q VS 1 1- 4Q 2 = 305.1V VO max Problem 11.43 A parallel RLC resonant circuit with a resonant frequency of 20,000rad/s has admittance at resonance of 1ms. If the capacitance of the network is 5F, find the values of R and L. Suggested Solution 1 = 1K 0.001 1 L = 2 = 500 H WO C R= Problem 11.44 A parallel RLC resonant circuit with a resistance of 200ohm. If it is known that the bandwidth is 80rad/s, find the values of the parameters L and C. Suggested Solution BW = 1 1 80 = C = 62.5 F RC 200C WHE = BW + WLO = 80r / s WO = 1 L = 22.73mH LC L = 22.73mH Problem 11.45 A parallel RLC circuit, which is driven by a variable frequency 2-a current source, has the following values: R=1Kohm,L=100mH, and C=10F. Find the bandwidth of the network, the half-power frequencies , and the voltage across the network at the half-power frequencies. Suggested Solution BW = 2 1 100r / s RC WO = 1 1kr / s LC WO = WHIWLO and WHI - WLO = BW WLO 2 W = O WHI 2 WHI - BWWHI - WO = 0 WHI = WLO BW + BW 2 + 4WO 1051.25r / s 2 = 951.25r / s 2 2 AT RESONANCE VO = 2 KV AND POWER FREQUENCIES VO = 2 KV 1.41KV 2 Problem 11.46 A parallel RLC circuit, which is driven by a variable-frequency 10-A source, has the following parameters: R=500ohm, L=0.5 mH, and C=20F. Find the resonant frequency, the BW, and the average power dissipated at the half-power frequencies. WO = 1 kr 10 s LC Suggested Solution 1 BW = 100r / s RC Q= WO = 100 BW W = WO P = 25KW PLC = PHI = 12.5KW Problem 11.47 Consider the network in fig 11.47. if R=2Kohm, L=20mH, C=50F, and Rs=infinity, determine the resonant frequency Wo, the Q of the network, and the bandwidth of the network. What impact does an Rs of 10Kohm have on the quantities determined. Rs R L C Suggested Solution WO = 1 r = 1000 s LC BW = 1 10r / s RC Q= WO = 100 BW RS = 10 K RS R = 1.67 K BW = 12r / s Q = 83.3 Problem 11.48 The source in the network in fig 11.48 is i(t) = cos1000t + cos1500t A.R=200ohm and C=500F. If Wo =1000rad/s, find L, Q, and the BW. Compute the o/p voltage Vo(t) and discuss the magnitude of the output voltage at the two input frequencies. + R L _ C V0 Suggested Solution L= 1 WO C 2 = 2mH BW = 1 = 10r / s RC Q= WC = 100 BW Z EQ = R L/C j (WL - 1 ) WC RL C 2 Z EQ = R L + WCR - WC C 2 Z EQ = 2.4 -89.30 Z EQ = 2.4 VO 2 = 2.4 Problem 11.49 Determine the parameters of a parallel resonant circuit which has the following properties: Wo = 2Mrad/sec, BW = 20 krad/sec, and an impedance of 2000ohm. WO = 1 C = 25nF LC Suggested Solution 1 BW = L = 10 H RC Problem 11.50 Determine the value of C in the network shown in fig 11.50 in order for the circuit to be in resonance. C 4 cos 2t V 6 4 4H Suggested Solution C vs(t) iC(t) R2 R1 L AT RESONANCE Vs(t) AND Is(t) ARE IN PHASE. SO, THE IMPEDANCE SEEN BY THE SOURCE, Zs, IS PURELY RESISTIVE. 1 1 Z S = R1 + R2 + = REQ + jo jwc jwL 1 ZS = 4 + ( 6 + j8 ) j 2c 4 3 24 + + j (32 - ) C C = N ( jw) = R ZS = eq 1 D( jw) 10 + j (8 - ) 2C IF Zs IS RESISTIVE, THEN THE PHASE ANGLES OF N(jw) AND D(jw) MUST BE EQUAL. 32 - 3 / C 64C 2 - 25C + 1 = 0 C = 45.2mF ,345.4mF 24 + 4 / C Problem 11.51 Determine the equation for the nonzero resonant frequency of the impedance shown in fig 11.51. Z C L R Suggested Solution N N 1 jwLR + L / C = Z = jwL R + = jwc R + j (WL - I / LC ) D D FOR RESONANCE, Z=Req +jo , OR, IF N = D , N = D WO LR WO - 1/ WO C 1 = L/C R WO CR WO = 2 1 WO = L RC ( - RC ) R 1 r/s LC - ( RC ) 2 Problem 11.52 Determine the new parameters of the network shown in fig 11.52 if Znew = 10000 Zold. L = 0.5H C = 1/8 F 2 Suggested Solution L = 0.5H C = 1/ 8F R = 2 Z NEW = 10000 Z OLD KM = 10000 Problem 11.53 Determine the new parameters of the network shown in fig 11.52 if Wnew = 10000 Wold. Suggested Solution L = 0.5H C = 1/ 8F R = 2 WNEW = 10000WOLD KF = 10000 Problem 11.54 Given the network in fig 11.54 sketch the magnitude characteristic of the transfer function. + + V1(t) _ V0(t) _ Suggested Solution Z EQ = jwL GV ( jw) = ( R) = jwL 1 + jwL / R jwL 1 [ jwL / R ] jWC Z EQ VO = = V1 Z EQ + jwL / C jwL + ( jw) 2 GV ( jw) = ( jw) 2 + 10 jw + 100 FIND FORM, GV ( jw) = 0.01( jw) 2 jw jw +1 ( )2 + 10 10 NETWORK IS A HIGHPASS FILTER. Problem 11.55 Given the network in fig 11.55, sketch the magnitude characteristic of the transfer function. GV ( jw) = VO ( jw) V1 10H + + V1(t) _ 1mF 100 V0(t) _ Suggested Solution LET Z EQ = ( R ) R 1 + jwRC Z EQ VO R = = Vi Z EQ + jwL R + jwL(1 + jwRC ) jwL = VO 1 = Vi ( jw) 2 LC + jwL + 1 R VO 100 = 2 Vi ( jw) + jw(10) + 1 VO 1 = jw 2 jw Vi ( ) + +1 10 10 NETWORK IS A LOW-PASS FILTER. Problem 11.56 Determine what type of filter the network shown in fig 11.56 represents by determining the voltage transfer function. + R1 V1(t) _ R2 C V0(t) _ + Suggested Solution + R1 V1(t) _ R2 C Z EQ = R1 1 R1 = jwc 1 + jwR1c (1 + jwR1c) R2 VO R2 = = Vi R2 + Z EQ R1 + R2 + jwR1 R2 c VO R2 (1 + jwR1c) = WHERE Vi R1 + R2 (1 + jwRc) R = R1 R2 < R1 1 RC 1 WP = RC WP > WZ WZ = NETWORK IS A HIGHER FILTER. Problem 11.57 Determine what type of filter the network shown in fig 11.57 represents by determining the voltage transfer function. + R1 V1(t) _ R2 C V0(t) _ + Suggested Solution + R1 V1(t) _ R2 C V0(t) _ + Z EQ = R1 GV = 1 jwLR1 = jwL R1 + jwL VO R2 ( R1 + jwL) R2 = = Vi R2 + Z EQ R1 R2 + jwLR1 + jwLR1 jwL R1 GV = jwL 1+ R R = R1 R2 < R1 1+ R1 L R WP = L WP < WZ WZ = NETWORK IS A LOW-PASS FILTER. Problem 11.58 Given the lattice network shown in fig 11.58 determine what type of filter this network represents by determining the voltage transfer function. R1 + C + Vi(t) _ R1 V0(t) _ R2 Suggested Solution + R1 Vi(t) _ + V0 1/jwc R1 R2 R 1 R1 jwcR2 1 1 - jwcR2 1 - VO = Vi = Vi - = Vi R1 + R1 R + 1 2 1 + jwcR2 2 1 + jwcR2 2 jwc THEREFORE 2 VO 1 1 + ( wcR2 ) = = 0.5 2 1 + ( wcR2 ) 2 Vi AN ALL PASS FILTER. Problem 11.59 Given the network shown in fig 11.59 and employing the voltage follower analyzed in chapter 3 determine the voltage transfer function and its magnitude characteristic. What type of filter does the network represent? + + VS (t) _ _ V0(t) Suggested Solution + + + VS (t) _ V1 V0(t) _ - R = 1 C = 1F V1 1/ jw 1 = = Vi (1/ jw) + 1 1 + jw VO 1 = V1 1 + jw 2 SECOND ORDER LOW-PASS Problem 11.60 Determine the voltage transfer function and its magnitude characteristic for the network shown in fig 11.60 and identify the filter properties. C2 R2 + R1 + C1 Vi(t) _ V0(t) _ Suggested Solution - Z 2 ( jw) VO ( jw) = Vi Z1 ( jw) Z 2 ( jw) = R2 / jwc2 1 R1 + jwc2 1 jwc1 Z1 ( jw) = R1 + VO jwc1 R2 ( jw) = Vi ( jwc1 R1 + 1)( jwc2 R2 + 1) THIS IS A 2ND ORDER BAND PADD FILTER. Problem 11.61 Given the network in fig 11.61 find the transfer function and determine what type of filter the network represents. VO ( jw) Vi + + Vi(t) R2 _ V0(t) _ Suggested Solution VO Z ( jw) = 1 + 2 Vi Z1 Z 2 = R2 Z1 = R1 + 1 jwc VO ( jwc( R1 + R2 ) + 1) jwcR2 ( jw) = 1 + = ( jwc1 R1 + 1) ( jwc1 R1 + 1) Vi this is a high pass filter Problem 11.62 Repeat problem 11.54 for the network shown in fig 11.62. + + Vi(t) R1 _ R2 C V0(t) _ Suggested Solution + + Vi(t) R1 _ R2 C V0(t) _ Z eq = R1 1 R1 = jwc 1 + jwcR1 USING STANDARD NON-INVERTING GAIN EQUATION, GV = Z VO R + R2 + jwcR1 R2 = 1 + eq = 1 Vi R2 R2 (1 + jwcR1 ) GV = (1 + R1 (1 + jwcR) ) R2 (1 + jwcR1 ) R = R1 R2 1 CR 1 WP = CR1 WZ = SINCE R<R1, WZ > WP NETWORK IS A LOW-PASS FILTER. Problem 11.63 In all OTA problem, the specifications are gm - Iabc sensitivity = 20, maximum gm = 1mS with range of 4 decades. For the circuit in figure 11.50, find gm and Iabc values required for a simulated resistance of 10Kohm. Suggested Solution io = g m (-vin ) & io = -iin vin = - Req = io i = in gm gm 1 vin = iin g m g m = 100 s and IARC = 5 A Problem 11.64 In all OTA problem, the specifications are gm - Iabc sensitivity = 20, maximum gm = 1mS with range of 4 decades. For the circuit in figure 11.53, find gm and Iabc values required for a simulated resistance of 10Kohm. Repeat for 750Kohm. Suggested Solution io1 = -i1 = - g m1v1- Req = 1 g m1 1 gm2 - io 2 = -i2 = - g m 2 v2 Req = For 10k 1 = 100 s and I ARC = 5 A 10k For 750k g m1 = g m 2 = 1.33 s and I ARC = 66 A g m1 = g m 2 = Problem 11.65 Use the summing circuit in figure 11.51 to design a circuit that realizes the following function. Vo = 7V1 + 3V2 Suggested Solution io1 + io 2 = io = io 3 = g3vo g1v1 + g 2 v2 = g3vo vo = g1 g v1 + 2 v2 g3 g3 g1 g = 7, 2 = 3 g3 g3 arbitrarily select g3 = 100 s g 2 = 700 s g1 = 300 s and I ARC1 = 5 A, I ARC2 = 35 A and IARC3 = 15 A Problem 11.66 Prove that the circuit in figure p11.66 is a simulated inductor. Find the inductance in terms of C, gm1 and gm2. gm1 + + Vi(t) _ gm2 + C Suggested Solution gm1 + + Vi(t) _ I02 gm2 + C VC _ I01 + dvi dt io 2 = - g m 2 vi = iin dv c dio 2 c diin g m1vin = c i = - = dt g m 2 dt g m 2 dt c diin c vin = or i eq = g m1 g m 2 dt g m1 g m 2 io1 = g m1vin = c Problem 11.67 In the Tow-Thomas biquad in figure 11.57, C1=20pF, C2 =10 pF, gm = 10S , gm2 = 80S, gm3 =10S . find the low pass filter transfer function for the Vi1 V02 i/p- o/p pair. Plot the corresponding Bode plot. Suggested Solution wo = g1 g 2 = 2 Mr / s c1c2 wo g3 = = 1Mr / s Q c2 vo 2 wo 4*1012 = = vi1 s 2 + wo s + w 2 s 2 + 106 s + 4*1012 o Q 2 Problem 11.68 Find the transfer function of the OTA filter in figure 11.68. Express Wo and Q in terms of the capacitances and transconductances. What kind of filter is it. Vin + + gm2 + Vo gm1 gm3 Suggested Solution - C2 Vo Vin + + C1 gm2 + I 03 gm1 gm3 I o1 = g1 (Vin - Vx ) I o 2 = - g 2Vo I o 3 = - g3Vo I o1 + I o 2 + (Vo - Vx ) jwc2 = jwc1Vx I o 3 = (Vo - Vx ) jwc2 making substitutions. g1 (Vin - Vx ) - g 2Vo + jwc2Vo - jwc2Vx = jwc1Vx - g3Vo = jwc2Vo - jwc2Vx from the last eq jwc2Vx = g3Vo + jwc2Vo Vx = (1 + then g1Vin - g1 (1 + g3 g g )Vo - g 2Vo + jwc2Vo - jwc2 (1 + 3 )Vo = jwc1 (1 + 3 )Vo jwc2 jwc2 jwc2 g3 )Vo jwc2 jwc2 ( g1 + g 2 + g3 ) + g1 g3 - w2c1c2 + jwg1 g3 g1Vin = jwc1 2 jwc1 g1Vin = - w c1c2 + jwc2 ( g1 + g 2 + g3 ) + g3c1 Vo Vo jwc1 g1 = Vin - w2 + jw[ ( g1 + g 2 + g3 ) + g3c1 / c2 ] + g1 g3 c1 c1c2 wo = Q= g1 g3 c1c2 g1 g 2 c1c2 c2 ( g1 + g 2 + g3 ) + g3c1 This is a band pass filter. Problem 11.69 Find the transfer function of the OTA filter in fig 11.69. Express Wo and Q in terms of the capacitances and transconductances. What kind of filters is it? V0 Vin + - gm2 + + gm1 gm3 Suggested Solution V0 Vin + - gm2 + I03 + gm1 gm3 I o1 = - g1Vx I o 2 = - g 2Vo (Vin - Vx ) jwc1 + I o1 + I o 2 + (Vo - Vx ) jwc2 = 0 jwc1Vin - jwc1Vx - g1Vx - g 2Vo + jwc2Vo - jwc2Vx = 0 jwc1Vin - ( jwc1 + g1 + jwc2 )Vx + ( jwc2 - g 2 )Vo = 0 in addition (Vo - Vx ) jwc2 = - g3Vo or jwc2Vx = ( jwc2 + g3 )Vo jwc2 + g3 g3 Vx = Vo = 1 + Vo jwc2 jwc2 then, g jwc1Vin = ( jwc1 + g1 + jwc2 ) 1 + 3 - jwc2 + g 2 Vo jwc2 - w2 c1c2 + jwc2 (( g1 + g 2 + g3 ) + g3c1 / c2 ) + g1 g3 jwc1Vin = Vo jwc2 - w2 c1c2 Vo = Vin - w2 c1c2 + jwc2 ( g1 + g 2 + g3 + g3 (c1 / c2 )) + g1 g3 Vo - w2 = Vin - w2 + jw( ( g1 + g 2 + g3 + g3 (c1 / c2 )) ) + g1 g3 c1 c1c2 Therefore, wo = Q= g1 g3 c1c2 g1 g 2 c1c2 c2 ( g1 + g 2 + g3 ) + g3c1 A high pass filter. Problem 11.70 Refer to the ac/dc converter low-pass filter application of example 11.25. if we put the converter to use powering a calculator, the load current can be modeled by a resistor as shown in fig 11.25. The load resistor will affect both the magnitude of the dc component of Vof and the pole frequency. Plot both the pole frequency and the ratio of the 60-Hz component of the o/p voltage to the dc component of Vof versus Rl for 100ohm less than equal to Rl, which is less than equal to 100 kohm. Comment on the advisable limitations on Rl if (a)the dc component of Vof is to remain within 20 % of its 9-v ideal value: (b) the 60 Hz component of Vof remains less than 15% of the dc component. Requivalent 500 AC/DC Convertor 53.05F RL Suggested Solution Req Vdc C Voc RL Vac THE CIRCUIT BELOW MODELS THE SITUATION DESCRIBED IN THE TEXT. H= VO RL RL = = VS RL + Req + jwcRL RL + Req RL 0.8 RL + Req 1 1 + jwcR WHERE R=RL Req a).H ( jo) = RL 2 K b). H ( j 377) < 0.15 H ( jo) OR, R > 330 Since, R = Req RL RL > 970.6 Requirement a) is more stringent. Problem 11.71 Referring to Ex 11.28 design a notch filter for the tape deck for use in Europe, where power utilities generate at 50 Hz. Suggested Solution Rtape = 50 Ramp = 1K WZ = 1 = 2 (50) LC CHOOSE YEILDS, C=100 F L=101mH Problem 11FE-1 Determine the resonant frequency of the circuit in fig 11PFE-1 and find the voltage Vo at resonance. + 12 cos t V 2 1mH 10F V0 _ Suggested Solution 1 = 10kr / s LC 12 00 I= = 6 00 2 6 -900 I vc = = -4 -5 = 60 -900 v jwc 10 10 wL also vc = Qvs = o (12) = 60v R wo = Problem 11FE-2 Given the series circuit in fig 11PFE-2, determine the resonant frequency and find the value of R so that the BW of the network about the resonant frequency is 200 r/s. R 20mH 10F Suggested Solution 1 = 1kr / s LC wL w R R Q = o and BW= o BW = and 200= R = 4 R Q L 0.00020 wo = Problem 11FE-3 Given the low-pass filter circuit in fig 11PFE-3, determine the frequency in Hz at which the output is down from the dc, or very low frequency, output. + 5K Vi(t) _ 1F V0(t) _ + Suggested Solution vo 1/ jwc 1 = = 3dBpo int vin R + 1/ jwc 1 + jwRc w= f3dB 1 = 200r / s RC = 31.83Hz Problem 11FE-4 Given the band-pass shown in fig 11PFE-4, find the components L and R necessary to provide a resonant frequency of 1000 r/s and a BW of 100 r/s. + 10F Vi(t) _ L R V0(t) _ + Suggested Solution w= 1 L = 100mH LC R BW = R = 10 L Problem 11FE-5 Given the low-pass shown in fig 11PFE-5, find the half-power frequency and the gain of this circuit, if the source frequency is 8Hz. + 2K 10F V0(t) _ Suggested Solution w= 1 = 50r / s RC 50 f = = 7.96 Hz 2 The dc gain is unity and the gain at the Half Power Frequency 7.98 is approximate to 8 Hz is by definition 0.707 of the dc gain. Problem 12.1 Find the Laplace Transform of the function f (t ) = te - t (t - 1) Suggested Solution F ( s ) = te - at (t - 1)e - st dt 0 let g (t ) = te - at F ( s ) = g (t ) (t - 1)e - st dt = g (1)e- s (1) = e- ( s + a ) 0 Problem 12.2 Find the Laplace transform of the function f (t ) = te - a (t -1) (t - 1) Suggested Solution f (t ) = te - a (t -1) (t - 1) let g (t ) = te - ( a -1)t F ( s ) = g (t ) (t - 1)e - st dt = g (1)e- s (1) = e- s 0 F ( s ) = e- s Problem 12.3 If f (t ) = e - at cos( t ) show s+a F (s) = ( s + a ) 2 + ( ) 2 Suggested Solution F ( s ) = e - at e - st cos( t )dt = e- ( a + s ) [ 0 0 e - j t + e - j t ]dt 2 1 F ( s ) = [ e - ( s + a - j ) t dt + e- ( s + a + j )t dt ] 0 2 0 1 -1 1 e - ( s + a - j )t | - e- ( s + a + j )t | ] = [ 0 0 s + a + j 2 s + a - j 1 1 -1 = [ + ] 2 s + a - j s + a + j 1 s + a + j + s + a - j ] = [ 2 ( s + a - j )( s + a + j ) s+a F (s) = ( s + a ) 2 + ( ) 2 F (s) = s+a ( s + a ) 2 + ( ) 2 Problem 12.4 Find F(s) if f (t ) = e - at sin( t )u (t - 1) Suggested Solution f (t ) = e - at sin( t )u (t - 1) F ( s ) = e - s L[e - a (t +1) sin( )(t + 1)] = e - ( s + a ) L[e - at sin (t + 1)] = e - ( s + a ) L[e - at (sin t cos + cos t sin )] = e-( s+a ) [ cos ( s + a ) sin + ] 2 2 ( s + a ) + ( ) ( s + a ) 2 + ( ) 2 cos ( s + a ) sin + F ( s) = e- ( s + a ) 2 2 2 2 ( s + a ) + ( ) ( s + a ) + ( ) Problem 12.5 If f (t ) = t cos( t )u (t - 1) find F ( s ) Suggested Solution f (t ) = t cos( t )u (t - 1), F ( s ) = e - s L[(t + 1) cos (t + 1)] = e - s L[(t + 1)(cos t cos - sin t sin )] = e - s L[t cos t cos + cos t cos - t sin t sin - sin t sin ] L[cos t cos ] = s cos s2 + 2 s sin L[sin t sin ] = 2 s +2 -d s cos ( s 2 + 2 ) cos - s cos (2 s ) ( 2 ) = -[ L[t cos t cos ] = ] ds s + 2 ( s 2 + 2 )2 sin (2 s) -d sin L[t sin t sin ] = ( 2 ) = -[ 2 ] 2 ds s + ( s + 2 )2 F (s) = e- s [ 2 s 2 cos - ( s 2 + 2 ) cos 2 s sin s cos sin + 2 + + ] (s 2 + 2 )2 (s + 2 )2 s 2 + 2 s 2 + 2 2 s 2 cos - ( s 2 + 2 ) cos 2 s sin s cos sin + 2 + + ] (s 2 + 2 )2 (s + 2 )2 s 2 + 2 s 2 + 2 F (s) = e- s [ Problem 12.6 Find F(s) if f (t ) = te - at u (t - 4) Suggested Solution f (t ) = te - at u (t - 4) F ( s ) = e -4 s L[(t + 4)e - a ( t + 4) ] = e -4( s + a ) L[(t + 4)e- at ] = e -4( s + a ) L[te- at + 4e - at ] = e -4( s + a ) [ 1 4 + ] 2 ( s + a) ( s + a) F ( s ) = e -4( s + a ) [ 1 4 + ] 2 ( s + a) ( s + a) Problem 12.7 Use the shifting Theorem to determine L { f ( t )} where f (t ) = [e - (t - 2) - e-2(t - 2) ]u (t - 2) Suggested Solution f (t ) = [e - (t - 2) - e-2(t - 2) ]u (t - 2) The shifting Theorem states L[ f (t - t0 )u (t - t0 )] = e -t0 s F ( s) L[ f (t - t0 )u (t - t0 )] = e -t0 s F ( s) let g (t ) = (e - e )u (t ) so 1 1 G ( s) = - s +1 s + 2 F ( s ) = e -2 s G ( s ) -t -2 t F ( s ) = e -2 s [ 1 1 - ] s +1 s + 2 e -2 s F (s) = ( s + 1)( s + 2) Problem 12.8 Use the shifting Theorem to determine L{f(t)} where f (t ) = [e - ( t - 2) - e - ( t -1) ]u (t - 1) Suggested Solution f (t ) = [e - ( t - 2) - e - ( t -1) ]u (t - 1) let g (t ) = (t + e - t )u (t - 1) so G ( s) = 1 1 + 2 s +1 s F (s) = e- sG(s) F ( s ) = e -2 s [ 1 1 + 2] s +1 s F ( s ) = e -2 s [ 1 1 + 2] s +1 s Problem 12.9 Use Property Number 5 to find L{f(t)} if f (t ) = e - at u (t - 1) Suggested Solution f (t ) = e - at u (t - 1) let g (t ) = e - at Thm _ 5 : L[ g (t )u (t - t0 )] = e - t0 s L[ g (t + t0 )] so L[e - at u (t - 1)] = e - sL[e - a (t +1) ] = e- ( s + a ) L[e- at ] sin ce 1 L[e - at ] = s+a -( s+a ) e F (s) = s +1 F ( s) = e- ( s + a ) s +1 Problem 12.10 Use Property Number 6 to find L{f(t)} if f (t ) = te - at u (t - 1) Suggested Solution f (t ) = te - at u (t - 1) let g (t ) = tu (t - 1) Thm _ 6 : L[e - at g (t )] = G ( s + a ) 1 1 + ) = G(s) s2 s 1 1 F ( s) = G ( s + a) = e- ( s + a ) ( + ) 2 ( s + a) ( s + a) L[ g (t )] = L[tu (t - 1)] = e - s L[t - 1] = e - s ( F ( s) = G ( s + a) = e- ( s + a ) ( 1 1 + ) 2 ( s + a) ( s + a) Problem 12.11 Given the following functions F(s), find f(t), 4 ( s + 1) + ( s + 2) 10 s F (s) = ( s + 1) + ( s + 4) F (s) = Suggested Solution A. F (s) = for s = -1 4 ( s + 1) + ( s + 2) 4 = 4 = k1 s+2 for s = -2 4 = -4 = k2 s +1 so F (s) = 4 4 - s + 2 s +1 f (t ) = (4e - t - 4e -2t )u (t ) f (t ) = (4e - t - 4e -2t )u (t ) B. F (s) = 10 s ( s + 1) + ( s + 4) for s = -1 10 s -10 = = k1 3 s+4 for s = -4 10 s 40 = = k2 s +1 3 so F (s) = -10 / 3 40 / 3 - s +1 s+4 -10 -t 40 -4t f (t ) = ( e + e )u (t ) 3 3 f (t ) = ( -10 -t 40 -4t e + e )u (t ) 3 3 Problem 12.12 Given the following functions F(s), find f(t) s + 10 ( s + 4)( s + 6) 24 F (s) = ( s + 2)( s + 8) F (s) = Suggested Solution A. F (s) = for s = -4 s + 10 ( s + 4)( s + 6) s + 10 = 3 = k1 s+6 for s = -6 s + 10 = -2 = k2 s+4 so F (s) = 3 2 - s+4 s+6 f (t ) = (3e -4t - 2e -6t )u (t ) f (t ) = (3e -4t - 2e -6t )u (t ) B. F (s) = 24 ( s + 2)( s + 8) for s = -2 24 = 6 = k1 s +8 for s = -8 24 = -4 = k2 s+2 so F (s) = 4 4 - s + 2 s +8 f (t ) = (4e -2t - 4e -8t )u (t ) f (t ) = (4e -2t - 4e -8t )u (t ) Problem 12.13 Given the following functions F(s), find f(t) if F (s) = s +1 s ( s + 2)( s + 3) s2 + s + 1 F (s) = s ( s + 2)( s + 1) Suggested Solution A. F (s) = s +1 s ( s + 2)( s + 3) s +1 = 1/ 6 = k1 ( s + 2)( s + 3) s +1 1 for s = -2, = = k2 s ( s + 3) 2 s +1 -2 for s = -3, = = k3 s ( s + 2) 3 1/ 6 1/ 2 2 / 3 so F ( s ) = + - s s+2 s+3 2 1 1 and f (t ) = + e -2t - e-3t u (t ) 3 6 2 for s = 0, B. 2 1 1 f (t ) = + e -2t - e-3t u (t ) 3 6 2 F (s) = s2 + s + 1 s ( s + 2)( s + 1) s2 + s + 1 = 1/ 2 = k1 for s = 0, ( s + 2)( s + 3) s2 + s +1 = -1 = k2 for s = -1, s ( s + 2) s2 + s + 1 3 = = k3 for s = -2, s ( s + 2) 2 1/ 2 -1 3 / 2 + + so F ( s ) = s s +1 s + 2 1 3 and f (t ) = ( + e - t + e -2t )u (t ) 2 2 1 3 f (t ) = ( + e - t + e -2t )u (t ) 2 2 Problem 12.14 Given the following functions F(s), find f(t). s 2 + 5s + 4 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 6) F (s) = s ( s 2 + 8s + 12) F (s) = Suggested Solution A. s 2 + 5s + 4 ( s + 2)( s + 4)( s + 6) ( s + 1)( s + 4) ( s + 1) A B F (s) = = = + ( s + 2)( s + 4)( s + 6) ( s + 2)( s + 6) s + 2 s + 6 -1 A = F ( s ) ( s + 2) s =-2 = 4 5 B = F ( s ) ( s + 6) s =-6 = 4 F (s) = F (s) = 5 / 4 1/ 4 - s+6 s+2 5 1 f (t ) = ( + e -6t - e-2t )u (t ) 4 4 5 1 f (t ) = ( + e -6t - e-2t )u (t ) 4 4 B. F (s) = ( s + 3)( s + 6) s ( s 2 + 8s + 12) ( s + 3)( s + 6) ( s + 3) A B = = + s ( s + 2)( s + 6) s ( s + 2) s s + 2 3 A = F ( s) s s =0 = 2 -1 B = F ( s ) ( s + 2) s =-2 = 2 F (s) = F (s) = 3 / 2 1/ 2 3 / 2 - + s s+2 s+2 3 1 f (t ) = ( - e -2t )u (t ) 2 2 3 1 f (t ) = ( - e -2t )u (t ) 2 2 Problem 12.15 Given the following functions F(s), find f(t). s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 6) F (s) = s ( s 2 + 10s + 24) F (s) = Suggested Solution A. s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 4) ( s + 3) A B F (s) = = = + ( s + 2)( s + 4)( s + 6) ( s + 6)( s + 2) s + 6 s + 2 F (s) = F (s) = 1/ 4 3 / 4 + s+2 s+6 1 3 f (t ) = ( e -2t + e-6t )u (t ) 4 4 1 3 f (t ) = ( e -2t + e-6t )u (t ) 4 4 B. F (s) = F (s) = ( s + 3)( s + 6) s ( s 2 + 10s + 24) A B ( s + 3)( s + 6) ( s + 3) = = + ( s )( s + 4)( s + 6) ( s + 6)( s + 2) s s + 4 3 1 f (t ) = ( + e -4t )u (t ) 4 4 F (s) = 3 / 4 1/ 4 + s s+4 3 1 f (t ) = ( + e -4t )u (t ) 4 4 Problem 12.16 Given the following functions F(s) find f(t) s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = Suggested Solution A. F (s) = s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) Matlab code EDU syms s t EDU ilaplace((s^2+7*s+12)/(s+2)/(s+4)/(s+6)) ans = 3/4*exp(-6*t)+1/4*exp(-2*t) or 3 1 f (t ) = ( e -6t + e-2t ) 4 4 3 1 f (t ) = ( e -6t + e-2t ) 4 4 B. F ( s) = 10( s + 2) ( s 2 + 4s + 5) Matlab code ilaplace((s+3)*(s+6)/(s*(s^2+10*s+24))) ans = 1/4*exp(-4*t)+3/4 or 3 1 f (t ) = ( + e -4t ) 4 4 3 1 f (t ) = ( + e -4t ) 4 4 Problem 12.17 Given the following functions F(s) find f(t) 10 ( s + 2 s + 2) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = 2 Suggested Solution A. F (s) = 10 ( s + 2s + 2) k1 k1 10 F (s) = 2 = + s + 2s + 2 s + 1 - s + 1 + 2 -5 5 + s +1- s +1+ f ( ) = 10e - t cos(t - 90o )u (t ) f ( ) = f ( ) = 10e - t cos(t - 90o )u (t ) B. F ( s) = F (s) = so 10( s + 2) ( s 2 + 4s + 5) k1 k1 10( s + 2) = + 2 s + 4s + 5 s + 2 - s + 2 + 5 5 + s + 2 - s + 2 + f ( ) = 10e -2t cos(t )u (t ) f ( ) = f ( ) = 10e -2t cos(t )u (t ) Problem 12.18 Given the following functions F(s) find inverse Laplace functions. 10 ( s + 2 s + 2) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = 2 Suggested Solution A F (s) = 10 ( s + 2s + 2) 2 F (s) = k1 k1 10( s + 2) = + 2 s + 4s + 5 s + 2 - s + 2 + for s = -1 + 10( s + 1) = 5 = k1 s +1+ so 5 5 F (s) = + s +1- s +1+ f (t ) = 10e -6 cos tu (t ) B f (t ) = 10e -6 cos tu (t ) F ( s) = F (s) = 10( s + 2) ( s 2 + 4s + 5) k3 s +1 k k2 = 1+ + s ( s + 4s + 5) s s + 2 - s + 2 + 2 for s=0 s +1 = 1/ 5 = k1 2 s + 4s + 5 for s = -2 + s +1 = 0.31 - 108.43o = k2 s( s + 2 + ) F (s) = 1/ 5 0.31 - 108.43o 0.31108.43o + + s s + 2 + s + 2 + 1 f (t ) = ( + 0.62e -2t cos(t - 108.43o ))u (t ) 5 1 f (t ) = ( + 0.62e -2t cos(t - 108.43o ))u (t ) 5 Problem 12.19 Given the following functions F(s), find f(t) s ( s + 6) ( s + 3)( s 2 + 6s + 18) ( s + 4)( s + 8) F (s) = s ( s 2 + 8s + 32) F (s) = Suggested Solution A. F (s) = s ( s + 6) ( s + 3)( s 2 + 6s + 18) s ( s + 6) s ( s + 6) = 2 ( s + 3)( s + 6s + 18) ( s + 3)( s + 3 + 3 )( s + 3 - 3 ) k k2 k3 = 1 + + s + 3 s + 3 + 3 s + 3 - 3 at F (s) = s = -3 s ( s + 6) = -1 = k1 ( s + 3)( s 2 + 6s + 18) at s = -3 + 3 s ( s + 6) = 1 = k2 ( s + 3)( s 2 + 6s + 18) so F (s) = 1 1 -1 + + s + 3 s + 3 + 3 s + 3 - 3 f (t ) = (e -3t + 2e -3t cos 3t )u (t )) f (t ) = (e -3t + 2e -3t cos 3t )u (t )) B. F (s) = ( s + 4)( s + 8) s ( s 2 + 8s + 32) ( s + 4)( s + 8) s ( s + 6) = 2 ( s )( s + 8s + 32) ( s + 3)( s + 3 + 3 )( s + 3 - 3 ) k k2 k2 = 1+ + s s + 4 + 4 s + 4 - 4 at F (s) = s=0 ( s + 4)( s + 8) = 1 = k1 ( s )( s 2 + 8s + 32) at s = -4 + 4 ( s + 4)( s + 8) -1 = = k2 2 ( s )( s + 8s + 32) 2 so /2 1 1/ 2 F (s) = + + s s + 4 + 4 s + 4 - 4 f (t ) = (1 + e -4t cos(4t - 90o )u (t )) f (t ) = (1 + e -4t cos(4t - 90o )u (t )) Problem 12.20 Given the following functions F(s) find f(t) 6 s + 12 ( s + 4 s + 5)( s 2 + 4s + 8) s ( s + 2) F (s) = 2 s + 2s + 2 F (s) = 2 Suggested Solution 6s + 12 (( s + 2) + 12 )(( s + 2) 2 + 22 ) k k k k = 1 1 + 1 1 + 2 2 + 2 2 s + 2 - j1 s + 2 + j1 s + 2 - j 2 s + 2 + j 2 F (s) = 2 k11 = F ( s )( s + 2 - j1) |s =-2+ j1 = 10o k11 = F ( s )( s + 2 - j 2) |s =-2- j 2 = -10o f (t ) = [2e -2t cos(t ) - 2e-2t cos(2t )]u (t ) f (t ) = [2e -2t cos(t ) - 2e -2t cos(2t )]u (t ) s 2 + 2s + 2 - 2 2 = 1- 2 2 ( s + 2 s + 2) ( s + 2s + 2) 2 F (s) = 1 - ( s + 1) 2 + (1) 2 F (s) = f (t ) = [ (t ) - 2e -t sin(t )]u (t ) f (t ) = [ (t ) - 2e - t sin(t )]u (t ) Problem 12.21 Use Matlab to solve Problem 12.19 Suggested Solution F (s) = s ( s + 6) ( s + 3)( s 2 + 6s + 18) Matlab Code >> syms s t >> ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18))) ans = -exp(-3*t)+2*exp(-3*t)*cos(3*t) >> f (t ) = 2e -3t cos(3t ) - e -3t f (t ) = 2e -3t cos(3t ) - e -3t F (s) = ( s + 3)( s + 8) s ( s 2 + 8s + 32) Matlab Code >> syms s t >> ilaplace((s+4)*(s+8)/(s*(s^2+8*s+32))) ans = 1+exp(-4*t)*sin(4*t) >> f (t ) = 1 + e -4t sin(4t ) f (t ) = 1 + e -4t sin(4t ) Problem 12.22 Given following functions F(s), find f(t) s +1 F (s) = 2 s ( s + 2) s+3 F (s) = ( s + 1) 2 ( s + 4) Suggested Solution A. F (s) = at s=0 s +1 1 = k1 = s+2 2 d s +1 1 [ ] = k2 = ds ( s + 2) 4 at s = -2 s +1 -1 = k3 = 2 4 (s) so F (s) = 1/ 2 1/ 4 1/ 4 + - s2 s s+4 1 1 1 -2t f (t ) = ( + t - e )u (t ) 4 2 4 k k k s +1 = 1+ 2+ 3 2 s ( s + 4) s s s+2 2 1 1 1 f (t ) = ( + t - e -2t )u (t ) 4 2 4 B. F (s) = ( s + 4)( s + 8) s ( s 2 + 8s + 32) f (t ) = 1 + e -4t sin(4t ) F (s) = at s = -1 s+3 2 = k1 = s+4 3 d s+3 1 [ ] = k2 = ds ( s + 4) 9 at s = -4 s+3 1 = k3 = 2 ( s + 1) 9 so F (s) = 2/3 1/ 9 1/ 9 + - 2 s +1 s + 4 ( s + 1) 1 2 1 f (t ) = ( e - t + te - t - e -4t )u (t ) 9 3 9 k k1 k s+3 = + 2 + 3 2 2 s +1 s + 4 ( s + 1) ( s + 4) ( s + 1) 1 2 1 f (t ) = ( e- t + te- t - e-4t )u (t ) 9 3 9 Problem 12.23 Given the following functions F(s), find f(t) s +8 F (s) = 2 s ( s + 6) 1 F (s) = 2 s ( s + 1) 2 Suggested Solution A. F (s) = at s=0 s +8 4 = k1 = s+6 3 d s +8 -1 [ ] = k2 = ds ( s + 6) 18 at s = -6 s +8 1 = k3 = 2 18 ( s) so F (s) = 4 / 3 1/18 1/18 - + s2 s s+6 4 1 1 -6t f (t ) = ( t - + e )u (t ) 3 18 18 k k k s +8 = 1+ 2+ 3 2 s s+6 s ( s + 6) s 2 4 1 1 f (t ) = ( t - + e -6t )u (t ) 3 18 18 B. F (s) = k3 k k k s +8 = 1+ 2+ + 4 2 2 2 s ( s + 1) s +1 s ( s + 1) s 2 at s=0 1 = k1 = 1 ( s + 1) 2 d 1 [ ] = k2 = -2 ds ( s + 1) 2 at s = -1 1 1 = k3 = 18 ( s)2 d 1 [ ] = k2 = 1 ds ( s ) 2 so 1 2 1 1 F (s) = 2 - - + s s ( s + 1) 2 s + 1 f (t ) = (t - 2 + te - t + 2e- t )u (t ) f (t ) = (t - 2 + te- t + 2e - t )u (t ) Problem 12.24 Given the following functions F(s), find f(t) F (s) = s+4 ( s + 2) 2 s+6 F (s) = s ( s + 1) 2 Suggested Solution F (s) = k1 k s+4 = + 2 2 2 s+2 ( s + 2) ( s + 2) at s = -2 s + 2 = 2 = k1 d [ s + 2] = k2 = 1 ds so F (s) = 2 1 + ( s + 2) 2 s + 2 f (t ) = (2te -2t + e -2t )u (t ) f (t ) = (2te -2t + e -2t )u (t ) B. F (s) = at s=0 s+6 = k1 = 6 ( s + 1) 2 at s = -1 s+6 = k2 = -5 s d s+6 [ ] = k3 = -6 ds s so F (s) = 6 5 6 - - 2 s ( s + 1) s +1 f (t ) = (6 - 5te - t - te - t )u (t ) k k k2 s+6 = 1+ + 3 2 2 s ( s + 1) s +1 s ( s + 1) f (t ) = (6 - 5te- t - te - t )u (t ) Problem 12.25 Given the following functions F(s) find f(t) F (s) = s2 ( s + 1) 2 ( s + 2) s 2 + 9 s - 20 s ( s + 4)3 ( s + 5) F (s) = Suggested Solution A. F (s) = at s = -1 s2 = k1 = 1 ( s + 2) at s = -2 s2 = k3 = 4 ( s + 1) 2 d s2 [ ] = k2 = -3 ds ( s + 1) 2 so F (s) = 1 3 4 - + ( s + 1) 2 s + 1 s + 2 k k1 k s2 = + 2 + 3 2 2 s +1 s + 2 ( s + 1) ( s + 2) ( s + 1) f (t ) = (te - t - 3e - t + 4te-2t )u (t ) f (t ) = (te - t - 3e- t + 4te-2t )u (t ) B. F (s) = at s=0 k k k2 s 2 + 9s - 20 1 = = 1+ + 3 3 2 2 s ( s + 4) ( s + 5) s ( s + 4) s ( s + 4) s+4 1 = k1 = 1/10 ( s + 4) 2 at s = -4 1 = k2 = -1/ 4 s d 1 [ ] = k3 = -1/16 ds s so F (s) = 1/16 1/ 4 1/16 - - 2 s s+4 ( s + 4) 1 t -4 t 1 - 4 t f (t ) = ( - e - e )u (t ) 16 4 16 1 t 1 f (t ) = ( - e-4t - e -4t )u (t ) 16 4 16 Problem 12.26 Find f(t) if F(s) is given by expression F (s) = s ( s + 1) ( s + 2)3 ( s + 3) Suggested Solution F (s) = s ( s + 1) A B C D = + + + ( s + 2)3 ( s + 3) s + 3 ( s + 3)3 ( s + 3) 2 s + 2 A = F ( s )( s + 3) |s =-3 = -6 B = F ( s )( s + 2) 2 |s =-2 = 2 -6 2 C D + + + C + 2D = 7 3 8 4 2 -6 + 2+C + D C + D =1 F (-1) = 0 = 2 D = 6, C = -5 -6 2 5 6 + - + F (s) = 3 2 s + 3 ( s + 3) ( s + 3) s+2 F (0) = 0 = f (t ) = (t 2e -2t - 5te -2t + 6e -2t - 6e -3t )u (t ) f (t ) = (t 2e-2t - 5te -2t + 6e-2t - 6e-3t )u (t ) Problem 12.27 Find f(t) if F(s) is given by F (s) = 12( s + 2) s 2 ( s + 1)( s 2 + 4s + 8) Suggested Solution F (s) = at s=0 12( s + 2) = 3 = k1 ( s + 1)( s 2 + 4s + 8) d 12( s + 2) [ ] = k2 = -3 ds ( s + 1)( s 2 + 4 s + 8) at s = -1 12( s + 2) 12 = k3 = ( s 2 )( s 2 + 4 s + 8) 5 at s = -2 + 2 12( s + 2) 1 = k4 = - 26.56o 3 s 2 ( s + 1) so 1 1 - 26.56o 26.56o 3 3 12 / 5 3 3 F (s) = 2 - + + + s s +1 s + 2 - 2 s + 2 + 2 s 12 - t 2 -2t f (t ) = [-3 + 3t + e + e cos(2t - 26.56o )]u (t ) 5 3 k k1 k2 k4 k4 + + 3 + + s 2 s s + 1 s + 2 - 2 s + 2 + 2 f (t ) = [-3 + 3t + 12 - t 2 -2t e + e cos(2t - 26.56o )]u (t ) 5 3 Problem 12.28 Use Matlab to solve Problem 12.25 Suggested Solution s2 ( s + 1) 2 ( s + 2) Matlab Code F (s) = >> syms s t >> >> ilaplace(s^2/((s+1)^2*(s+2))) ans = (-3+t)*exp(-t)+4*exp(-2*t) >> f (t ) = te- t - 3e - t + 4e-2t f (t ) = te- t - 3e - t + 4e-2t s 2 + 9s + 20 s ( s + 4)3 ( s + 5) Matlab Code >> syms s t >> ilaplace((s^2+9*s+20)/(s*(s+4)^3*(s+5))) F (s) = ans = -1/4*t*exp(-4*t)-1/16*exp(-4*t)+1/16 >> f (t ) = 0.0625{1 - e -4t - 4te-4t } f (t ) = 0.0625{1 - e -4t - 4te-4t } Problem 12.29 Find the inverse Laplace transform of the following functions e- s F (s) = s +1 F (s) = 1 - e -2 s s 1 - e- s s+2 F (s) = Suggested Solution a. F (s) = let G( s) = 1 g (t ) = e - t u (t ) s +1 f (t ) = g (t - 1) f (t ) = e - (t -1) u (t - 1) e- s s +1 b. F (s) = let G( s) = 1 g (t ) = u (t ) s f (t ) = g (t ) - g (t - 2) f (t ) = u (t ) - u (t - 2) 1 - e -2 s s c. F (s) = let G( s) = 1 g (t ) = e -2t u (t ) s+2 f (t ) = g (t ) - g (t - 1) f (t ) = e -2t u (t ) - e-2(t -1) u (t - 1) 1 - e- s s+2 f (t ) = e - (t -1) u (t - 1) f (t ) = u (t ) - u (t - 2) f (t ) = e-2t u (t ) - e -2(t -1) u (t - 1) Problem 12.30 Find the inverse Laplace transform of the following function ( s + 1)e - s F (s) = s ( s + 2) F (s) = 10e -2 s ( s + 1)( s + 3) Suggested Solution F (s) = at s=0 s +1 = k1 = 1/ 2 s+2 at s = -2 s +1 = 1/ 2 = k2 s 1/ 2 1/ 2 F (s) = e- s [ ] + s s+2 1 1 f (t ) = u (t - 1) + e -2(t -1) u (t - 1) 2 2 F (s) = k k 10e -2 s = e -2 s [ 1 + 2 ] ( s + 1)( s + 3) s +1 s + 3 k k ( s + 1)e - s = e- s [ 1 + 2 ] s ( s + 2) s s+2 at s = -1 10 = k1 = 5 s+3 at s = -3 10 = k2 = -5 s +1 5 5 F ( s ) = e -2 s [ ] - s +1 s + 3 f (t ) = [5e- (t - 2) - 5e-3(t - 2) ]u (t - 2) f (t ) = 1 1 u (t - 1) + e-2(t -1) u (t - 1) 2 2 f (t ) = [5e - (t - 2) - 5e -3(t - 2) ]u (t - 2) Problem 12.31 Find the inverse Laplace transform f(t) if F(s) is se - s F (s) = ( s + 1)( s + 2) Suggested Solution F (s) = let G( s) = s A B = + ( s + 1)( s + 2) s + 1 s + 2 A = G ( s )( s + 1) |s =-1 = -1 2 1 - g (t ) = [2e -2t - e- t ]u (t ) s + 2 s +1 f (t ) = g (t - 1) f (t ) = [2e -2(t -1) - e - (t -1) ]u (t - 1) se - s ( s + 1)( s + 2) f (t ) = [2e -2(t -1) - e - (t -1) ]u (t - 1) B = G ( s )( s + 2) |s =-2 = 2 G( s) = Problem 12.32 Find f(t) if F(s) is given by the following functions s 2 + 2s + 3 ( s + 2)e -4 s F ( s ) = e -2 s [ ] F (s) = 2 s ( s + 1)( s + 2) s ( s + 1) Suggested Solution A. F ( s ) = e -2 s [ at s=0 s 2 + 2s + 3 3 = k1 = ( s + 1)( s + 2) 2 at s = -1 s 2 + 2s + 3 = k2 = -2 s ( s + 2) at s = -2 3 s 2 + 2s + 3 = k3 = 2 s ( s + 1) so 3/ 2 3/ 2 -2 ] F ( s ) = e -2 s [ + + ( s + 1) ( s + 2) s 3 3 f (t ) = [ - 2e- (t - 2) + e -2(t - 2) ]u (t - 2) 2 2 B. F (s) = k k k ( s + 2)e -4 s = e -4 s [ 1 + 2 + 3 ] s 2 ( s + 1) s 2 s ( s + 1) k3 k k s 2 + 2s + 3 ] = e -2 s [ 1 + 2 + ] s ( s + 1)( s + 2) s ( s + 1) ( s + 2) at s=0 s+2 = k1 = 2 s +1 d s+2 [ ] = -1 = k 2 ds s + 1 at s = -1 s+2 = k3 = 1 s +1 so F ( s ) = e -4 s [ 2 1 1 ] - + 2 s s ( s + 1) f (t ) = [2(t - 4) - 1 + e- (t - 4) ]u (t - 4) 3 3 f (t ) = [ - 2e- (t - 2) + e -2(t - 2) ]u (t - 2) 2 2 f (t ) = [2(t - 4) - 1 + e- (t - 4) ]u (t - 4) Problem 12.33 Find f(t) if F(s) is given by following functions F (s) = e- s [ 2( s + 1) ] ( s + 3)( s + 2) F (s) = 10( s + 2)e -2 s ( s + 4)( s + 1) Suggested Solution A. F (s) = e- s [ at s = -2 2( s + 1) = k1 = -2 ( s + 3) at s = -3 2( s + 1) = k2 = 4 ( s + 2) so F ( s ) = e -2 s [ -2 4 + ] s+2 s+3 f (t ) = [-2e - (t -1) + 4e -3(t -1) ]u (t - 1) k k 2( s + 1) ] = e- s [ 1 + 2 ] ( s + 3)( s + 2) s+2 s+3 B F (s) = at s = -1 10( s + 2) 10 = k1 = ( s + 4) 3 at s = -4 10( s + 2) = k2 = 4 ( s + 1) so 10 / 3 4 + F ( s ) = e -2 s [ ] s +1 s + 4 10 20 f (t ) = [ e- (t - 2) + e -4(t - 2) ]u (t - 2) 3 3 k k 10( s + 2)e -2 s = e -2 s [ 1 + 2 ] ( s + 4)( s + 1) s +1 s + 4 f (t ) = [-2e- (t -1) + 4e-3(t -1) ]u (t - 1) 10 20 f (t ) = [ e- (t - 2) + e -4(t - 2) ]u (t - 2) 3 3 Problem 12.34 Solve the following differential equations using Laplace Transform dx(t ) + 3x(t ) = e -2t dt dx(t ) + 3 x(t ) = 2u (t ) dt Suggested Solution A. dx(t ) + 3x(t ) = e -2t , x(0) = 1 dt dx(t ) 1 1 s+3 sX ( s ) - x(0) = X ( s )( s + 3) = +1 = dt s+2 s+2 s+2 1 X (s) = x(t ) = e -2t u (t ) s+2 B. dx(t ) + 3x(t ) = 2u (t ), x(0) = 2 dt dx(t ) 2 2( s + 1) sX ( s ) - 2 + 3 X ( s ) X ( s )( s + 3) = + 2 = dt s s 2 x(t ) = [1 + 2e -3t ]u (t ) 3 x(t ) = e -2t u (t ) 2 x(t ) = [1 + 2e -3t ]u (t ) 3 Problem 12.35 Solve the following differential equations using Laplace transform d 2 y (t ) 2dy (t ) + + y (t ) = e-2t , y (0) = y '(0) = 0 A. dt dt 2 d 2 y (t ) 4dy (t ) B. + + 4 y (t ) = u (t ), y (0) = 0; y '(0) = 1 dt dt 2 Suggested Solution A. d 2 y (t ) 2dy (t ) + + y (t ) = e-2t , y (0) = y '(0) = 0 2 dt dt 1 s+2 1 A B C Y (s) = ; C = Y ( s )( s + 2) |s =-2 = 1 = + + ( s + 1) 2 ( s + 2) ( s + 1) 2 ( s + 1) ( s + 2) s 2Y ( s ) + 2sY ( s ) + Y ( s ) = Y ( s )[ s 2 + 2s + 1] = A = Y ( s )( s + 1) 2 |s =-1 = 1 1 1 = 1 + B + B = -1 2 2 1 1 1 Y (s) = - + ( s + 1) 2 ( s + 1) ( s + 2) Y (0) = y (t ) = (te - t - e - t + e -2t )u (t ) B. d 2 y (t ) 4dy (t ) + + 4 y (t ) = u (t ), y (0) = 0; y '(0) = 1 dt 2 dt s 2Y ( s ) - sy '(0) - y (0) + 4 sY ( s ) - 4 y (0) + 4Y ( s ) = 1 s2 + 1 +s= s s 2 s +1 A B C Y (s) = = + + s ( s + 2) 2 s ( s + 2) 2 s + 2 1 A = Y ( s ) s |s = 0 = 4 B = Y ( s )( s + 2) 2 |s =-2 = -2.5 Y ( s )[ s 2 + 4s + 4] = 1 5 3 Y (-1) = -2 = - - + C C = 4 2 4 1 1 10 3 + ] Y (s) = [ - 4 s ( s + 2) 2 s + 2 1 y (t ) = [1 - 10te -2t + 3e-2t ]u (t ) 4 1 s y (t ) = (te - t - e - t + e-2t )u (t ) 1 y (t ) = [1 - 10te -2t + 3e -2t ]u (t ) 4 Problem 12.36 Use Laplace transform to find y(t) if dy (t ) + 5 y (t ) + 4 y ( x)dx = u (t ), y (0) = 0, t > 0 dt 0 t Suggested Solution In Laplace terms 4 1 sY ( s ) + 5Y ( s ) + Y ( s ) = 5 s 2 Y ( s )[ s + 5s + 4] = 1 so Y (s) = so Y (s) = 1/ 3 1/ 3 - s +1 s + 4 1 1 y (t ) = [ e - t - e -4t ]u (t ) 3 3 k k 1 1 = = 1 + 2 k1 = 1/ 3; k2 = -1/ 3 s + 5s + 4 ( s + 1)( s + 4) s + 1 s + 4 2 1 1 y (t ) = [ e- t - e -4t ]u (t ) 3 3 Problem 12.37 Solve the integrodifferential equations using Laplace transforms. t dy (t ) + 2 y (t ) + y ( )d = 1 - e -2t , y (0) = 0, t > 0 dt 0 Suggested Solution In Laplace terms 1 1 1 sY ( s ) + 2Y ( s ) + Y ( s ) = - s s s+2 1 1 1 Y ( s )[ s + 2 + ] = - s s s+2 so Y (s) = at s = -1 2 = 2 = k1 s+2 d 2 [ ] = -2 = k2 ds s + 2 at s = -2 2 = 2 = k3 ( s + 1) 2 so F (s) = 2 2 2 - + 2 ( s + 1) s +1 s + 2 k k1 k + 2 + 3 2 ( s + 1) s +1 s + 2 f (t ) = [2te- t - 2e - t + 2e -2t ]u (t ) f (t ) = [2te- t - 2e- t + 2e -2t ]u (t ) Problem 12.38 Determine the y(t) in the following equation if all initial conditions are zero. d 3 y (t ) d 2 y (t ) dy (t ) +4 +3 = 10e -2t 3 dt dt dt 2 Suggested Solution d 3 y (t ) d 2 y (t ) dy (t ) +4 +3 = 10e -2t 3 dt dt 2 dt All intial conditions are zero. 10 s+2 A B C D 10 Y (s) = = + + + s ( s + 2)( s + 1)( s + 3) s s + 2 s + 1 s + 3 A = Y ( s ) s |s = 0 = 5 / 3 s 3Y ( s ) + 4 s 2Y ( s ) + 3sY ( s ) = B = Y ( s )( s + 2) |s =-2 = 5 C = Y ( s )( s + 1) |s =-1 = -5 D = Y ( s )( s + 3) |s =-3 = -5 / 3 Y (s) = 5/3 5 5 5/3 + - - s s + 2 s +1 s + 3 5 5 y (t ) = ( + 5e -2t - 5e - t - e -3t )u (t ) 3 3 5 5 y (t ) = ( + 5e -2t - 5e - t - e -3t )u (t ) 3 3 Problem 12.39 Find f(t) using convolution if F(s) is 1 F (s) = ( s + 1)( s + 2) Suggested Solution F (s) = let 1 s +1 1 F2 ( s ) = s+2 f1 (t ) = e- t u (t ) F1 ( s ) = f 2 (t ) = e -2t u (t ) f (t ) = e - (t - ) e-2 d = e - t e - d =e- t [e - ]t0 0 0 t t 1 ( s + 1)( s + 2) f (t ) = e- t [1 - e - t ] f (t ) = (e- t - e-2t )u (t ) f (t ) = (e - t - e -2t )u (t ) Problem 12.40 Use convolution to find f(t) if 1 F (s) = ( s + 1)( s + 2) 2 Suggested Solution F (s) = let 1 ( s + 2) 2 1 F2 ( s ) = s +1 f1 (t ) = te -2t u (t ) F1 ( s ) = f 2 (t ) = e - t u (t ) f (t ) = e - (t - ) e -2 d = e - t e- d =e - t [-e - - e - ]t0 0 0 t t 1 ( s + 1)( s + 2) 2 f (t ) = e [-e - te + 1] f (t ) = (e - t - te -2t - e-2t )u (t ) -t -t -t f (t ) = (e - t - te -2t - e-2t )u (t ) Problem 12.41 Determine the intial and final value of F(s) given by the expression F (s) = 2( s + 2) s ( s + 1) F (s) = F (s) = 2( s 2 + 2s + 6) s ( s + 1) 2s 2 s ( s + 1)( s 2 + 2 s + 2) Suggested Solution A. F (s) = for 2( s + 2) ]= 2 s ( s + 1) 2( s + 2) ]= 4 t , lim s 0 sf ( s ) = lim s 0 [ s ( s + 1) B. t = 0+ , lim s sf ( s ) = lim s [ F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2( s 2 + 2 s + 6) ]= 2 s ( s + 1) 2( s 2 + 2 s + 6) s ( s + 1) 2( s + 2) s ( s + 1) 2( s 2 + 2 s + 6) ]= 0 s ( s + 1) C. F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2s 2 ]= 2 s ( s + 1)( s 2 + 2s + 2) 2s 2 s ( s + 1)( s 2 + 2 s + 2) 2s 2 ]= 0 s ( s + 1)( s 2 + 2s + 2) Problem 12.42 Find the initial and final value of the time function f(t) if F(s) is given as F (s) = 10( s + 2) ( s + 1)( s + 3) F (s) = F (s) = ( s 2 + 2s + 4) ( s + 1)( s 3 + 4s 2 + 8s + 10) 2s ( s 2 + 2 s + 2) Suggested Solution A. F (s) = for 10( s + 2) ] = 10 ( s + 1)( s + 3) 10( s + 2) ]= 0 t , lim s 0 sf ( s ) = lim s 0 [ ( s + 1)( s + 3) B. t = 0+ , lim s sf ( s ) = lim s [ F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ ( s 2 + 2s + 4) ]= 0 ( s + 1)( s 3 + 4 s 2 + 8s + 10) ( s 2 + 2s + 4) ( s + 1)( s 3 + 4s 2 + 8s + 10) 10( s + 2) ( s + 1)( s + 3) ( s 2 + 2s + 4) ]= 0 ( s + 1)( s 3 + 4s 2 + 8s + 10) C. F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2s 2 ]= 2 s ( s + 1)( s 2 + 2s + 2) 2s ( s + 2s + 2) 2 2s 2 ]= 0 s ( s + 1)( s 2 + 2s + 2) Problem 12.43 Find the final values of the time function f(t) if F(s) is given as F (s) = 10( s + 1) ( s + 2)( s + 3) F (s) = 10 s + 4s + 4 2 Suggested Solution A. F (s) = for 10( s + 1) ] = 10 ( s + 2)( s + 3) 10( s + 1) t , lim s 0 sf ( s ) = lim s 0 [ ]= 0 ( s + 2)( s + 3) B. 10 F (s) = 2 s + 4s + 4 for 10 t = 0+ , lim s sf ( s ) = lim s [ 2 ]= 0 s + 4s + 4 10 t , lim s 0 sf ( s ) = lim s 0 [ 2 ]= 0 s + 4s + 4 t = 0+ , lim s sf ( s ) = lim s [ 10( s + 1) ( s + 2)( s + 3) Problem 12.44 In the network in fig, the switch opens at t =0. Use laplace transform to find I(t) for t>0. t=0 3H 3H 12V 6 i(t) 12 Suggested Solution t=0 + Vl 3H 12V 6 i(t) 12 12 6 I(s) 3H for t<0 VL = 0V and i (0- ) = i (0+ ) = 1A 18I ( s ) + 3sI ( s ) - 3i (0) = 0 I ( s )[3s + 18] = 3 1 I (s) = s+6 i (t ) = e -6t u (t ) i (t ) = e -6t u (t ) Problem 12.45 The switch in the circuit opens at t=0. Find I(t) for t>0 using Laplace transforms. 2 4 t=0 3 12V 2H Suggested Solution 2 R1 4 t=0 3 R3 12V R2 + VL 2 4 R4 t=0 3 12V iL(o) 6 3 4 i (t) 2H 2 di (t ) =0 dt 9 I ( s ) + 2 I ( s ) - 2i (0+ ) = 0 9 I ( s) = [ s + ] = 4 2 4 I ( s) = 9 s+ 2 3i (t ) + 4i (t ) + 2i (t ) + 2 i (t ) = 4e 9 - t 2 u (t ) A i (t ) = 4e 9 - t 2 u (t ) A Problem 12.46 In the circuit in fig, the switch moves from position 1 to 2 at t =0. Use Laplace transforms to find v(t) for t>0. 1 t=0 2 12V + v(t) 100F 3K 6K Suggested Solution 1 t=0 2 12V R1 + v(t) 100F R2 R1 + V(o) 12V - R2 R2 + V(t) - V (t ) Cdv(t ) + =0 R dt V (s) + CV ( s ) - Cv (0) = 0 R 1 4 V ( s )[ s + ] = v (0) V ( s ) = RC s+5 v(t ) = 4e-5t u (t )V v(t ) = 4e -5t u (t )V Problem 12.47 In the network the switch closes at t=0. Use Laplace transforms to find Vc(t) for t>0. 1 12V t=0 1 2H Vc(t) 0.5F + Suggested Solution 1 12V t=0 1 2H iL 0.5F + t 12 - Vc(t ) = 1 dVc(t ) Vc( x)dx + C L dt 0 s 12 1 - Vc( s ) = Vc( s ) + Vc( s) Vc(t) s 2s 2 12 1 s = Vc( s )[1 + + ] s 2s 2 12 2s + 1 + s 2 ] = Vc( s )[ s 2s 24 24 Vc( s ) = = 2s + 1 + s 2 ( s + 1) 2 Vc(t ) = 24te- t u (t )V Vc(t ) = 24te - t u (t )V Problem 12.48 In the network the switch opens at t=0. Use Laplace transforms to find iL(t) for t>0. 3 iL(t) t=0 1A 1H 0.5F Suggested Solution 3 iL(t) t=0 1A 0.5F 1H - ic + Vc(t) For _ t > 0 L t diL 1 = 3iC + iC (t )dt dt C0 and iL + iC = 1 iC = 1 - iL now, 1 - I L ( s) ) sI L ( s ) = 3 - 3I L ( s ) + 2( s 2 2 I L ( s )[ s + ] = 3 + s s 3s + 2 A B = + A = 4, B = -1 I L ( s) = 2 s + 3s + 2 s + 2 s + 1 4 1 - I L ( s) = s + 2 s +1 iL (t ) = (4e -2t - e- t )u (t ) iL (t ) = (4e -2t - e- t )u (t ) Problem 12.49 In the network the switch opens at t=0. Use Laplace transforms to find V0(t) for t>0. 3K t=0 4K + Vc(t) 2K + V(t) - 12V - Suggested Solution 3K t=0 4K + + Vc(t) 12V 4K + V(t) 0.1mF 2K - 2K - V(t) Vc(0- ) = 12( 2+4 ) = 8v = Vc(0+ ) 2+4+3 8 v0 (0+ ) = vc (0+ ) = v 3 1 i (t )dt - vc (0) = 0 0.1m 0 t 6ki (t ) + I ( s ) 0.8m - =0 s s 0.8 4/3 I ( s) = mA = 1 0.6s + 1 s+ 0.6 4 - t / 0.6 i (t ) = e u (t )mA 3 8 V0 (t ) = 2ki (t ) = e- t / 0.6V 3 0.6 I ( s ) + i (t ) = 4 - t / 0.6 e u (t )mA 3 Problem 12FE-1 The output function of a network is expressed using Laplace transforms in the following form. V0 ( s ) = 12 s ( s + 1)( s + 2) Find the output as a function of time v0(t). Suggested Solution V0 ( s ) = 12 A B C = + + s ( s + 1)( s + 2) s s + 1 s + 2 12 |s = 0 = A = 6 ( s + 1)( s + 2) 12 |s =-1 = B = -12 s ( s + 2) 12 |s =-2 = C = 6 s ( s + 1) V0 (t ) = (6 - 12e- t + 6e-2t )u (t )V V0 (t ) = (6 - 12e- t + 6e-2t )u (t )V Problem 12FE-2 The Laplace transform function representing the output voltage of a network is expressed as V0 ( s ) = 120 s ( s + 10)( s + 20) Determine the time domain function and the value of the v0(t) at t=100mSec. Suggested Solution V0 ( s ) = 120 A B C = + + s ( s + 10)( s + 20) s s + 10 s + 20 120 |s = 0 = A = 0.6 ( s + 10)( s + 20) 120 |s =-10 = B = -1.2 s ( s + 20) 120 |s =-20 = C = 0.6 s ( s + 10) V0 (t ) = (0.6 - 1.2e -10t + 0.6e-20t )u (t )V V0 (t ) |t = 0.1sec = 0.24V V0 (t ) |t = 0.1sec = 0.24V Problem 12FE-3 The Laplace transform function for the output voltage of a network is expressed in the following form V0 ( s ) = 12( s + 2) s ( s + 1)( s + 3)( s + 4) Determine the final value i.e. vo ( t ) as t , of this voltage. Suggested Solution V0 ( s ) = and V0 () = sV0 ( s ) V0 () = 12( s + 2) |s = 0 = 2v s ( s + 1)( s + 3)( s + 4) 12( s + 2) s ( s + 1)( s + 3)( s + 4) V0 = 2v Problem 13.1 Find the input impedance Z(s) of the network in fig 3.1. 2 1 1 1F Z(s) 2H 2 Suggested Solution Z1 Z(s) Z2 1 1( ) S = 1 Z1 ( S ) = 1 1+ ( ) S +1 S 6S + 8 Z2 (S ) = 2S + 3 6S + 8 ZZ Z (S ) = 1 2 = 2 Z1 + Z 2 6 S + 16S + 11 Problem 13.2 Find the input impedance Z(s) of the network in fig 3.2 (a). when the terminals B-B' are open circuited and (b). when the terminals B-B' are closed circuited. 1F A B 1 Z(s) 2 2H A B Suggested Solution 1/S 1 Z(s) 2 S Z (S ) = 2S + 1 S +1 1/S 1 Z(s) 2 S Z (S ) = 2S + 1 S +1 Problem 13.3 Use lap lace transforms to find v(t) for t>0 in the network shown in fig. Assume zero initial conditions. 1 2 5u(t)V 1F 10u(t)V Suggested Solution 1 2 5/S 1/S 10/S 1 + 5/S 1/S _ Voc ZTH Z ZTH + V - 5 1/ S 5 1 = S 1 + 1/ S S 1 + S 1 ZTH = 1 (1/ S ) = 1+ S 10 2 10 2 S + 3 S + 1 VS = VOC + = S 2 + ZTH S 1 + S 3 + 2S VOC = VS = 10 V (t ) = 10u (t )V Problem 13.4 Use lap lace transforms and node analysis to find i1(t) for t>0 in the network shown in fig. Assume zero initial conditions. 2 i1(t) 6 u(t) V 12 u(t) V 1H Suggested Solution 2 i1(t) 2/S 6/S 12/S S ZTH = S (2 / S ) = 2S 2+ S2 VOC = 6 S 6 = S 1 + 2 / S 2 + S 2 12 6 12 + + VOC 2 S 2+ S2 = 9 S + 12 S I1 = = 2S Z + ZTH S ( S 2 + S + 2) 2+ 2 + S2 A K K - + I1 = + S 1 7 1 7 S- - j S- + j 2 2 2 2 i1 (t ) = 6 + 16.12e - t / 2COS ( 7t / 2 + 26.370 ) u (t )V 2 i1(t) ZTH 12/S S 2 i1(t) ZTH 2/S VOC 12/S Problem 13.5 For the network shown in fig 13.5 find io(t),t>0. 1 1F 1H 1 4u(t) V 2 e-tu(t) V i0(t) Suggested Solution 1 1/S S 1 4/S 2 1/(S+1) i0(t) 1 1 4 1 V ( + + 1) = 2 - S 2 S S +1 2 2(- S + 4 S + 4) V= S ( S + 1)(3S + 2) I O = 0.5V = (- S 2 + 4 S + 4) S ( S + 1)(3S + 2) iO (t ) = (2 - 16.12e -t - 4 / 3e-2t / 3 )u (t )V Problem 13.6 For the network shown in fig 13.5 find Vo(t),t>0. e-2tu(t) V 1 1F 1 2u(t) A 1H 1 Suggested Solution 1/(S+2) 1 1/S 1 2/S S 1 V0 KCL at the Supernode is: 2 V1 V V + 2+ 2 = 1 + 1/ S S 2 3 1 V1 + = V2 S +2 SOLVING FOR V2 YIELDS: V2 = 2(3S 2 + 6 S + 4) S 2 + 2S + 4 / 3 VO = ( S + 2)(3S 2 + 6 S + 2) ( S + 2)( S + 0.5 - j.646)( S + 0.5 + j.646) K K2 K2 = 1 + + S + 2 ( S + 0.5 - j.646) ( S + 0.5 + j.646) K1 = 0.5, K 2 = 0.316 -37.760 VO (t ) = ( 0.5e -2t + 0.632e -2t COS (0.646t - 37.760 ) ) u (t )V Problem 13.7 For the network shown in fig 13.5 find Vo(t),t>0. 1 1 1H 10u(t) mA 1 1F 9u(t) V Suggested Solution 1 1 S 0.01/S 1 1/S 9/S 1 0.01/S 1 0.01/S 1 + VOC 1/S 9/S _ 1 Z0G ZTH + VOC 0.01/S _ 9 1/ S 9 = S 1 + 1/ S S ( S + 1) 1 ZTH = 1 1/ S = S +1 0.005( S - 899) Z EQ 10-2 - VOC V = = ( S + 1) 2 S Z EQ + ZTH + 1 A B + V= 2 ( S + 1) S +1 A = -4.5 A B + B = 5*10-3 V (899) = 0 = 2 (900) 900 VOC = V (t ) = (5e - t - 4500te- t )u (t )mV Problem 13.8 Find Vo(t),t>0 in the network shown in fig using node equations. 1 4u(t) V 1H 2u(t) A + 1F e-tu(t) A 1 V0 _ Suggested Solution 2/S + 1/S 1/(S+1) 1 V0 _ V 2 1 + = O + SVO = VO ( S + 1) S S +1 1 A B C 2 + 3S VO = = + + 2 2 S ( S + 1) S ( S + 1) S + 1 A=2 B =1 LET S=-2, VO (-2)=2=-0.5A+B-C C=-2 VO (t ) = 2 + 2e -t + te- t u (t )V Problem 13.9 Find Vo(t),t>0 in the network shown in fig using node equations. 12u(t) V 1H 4e u(t) V -t 1 1F + 6u(t) V 1 1 V0 _ Suggested Solution 12/S S 4/(S+1) + 6/S 1 1/S 1 V0 _ 1 KCL : V3 - VO + (V2 - VO ) S = VO VO = 6 4 2S + 6 - = S S + 1 S ( S + 1) 12 14 S + 18 V3 - V2 + = S S ( S + 1) 2( S + 9) VO = S ( S + 2) A B VO = + S ( S + 2) V2 = V3 + SV2 S +2 A=9 B = -7 VO (t ) = 9 - 7e -2t u (t )V Problem 13.10 For the network shown in fig find Vo(t),t>0 using mesh equations. 0.5F 2 1H + 2u(t) A 1 1 1 V0 _ 4u(t) V Suggested Solution 2/S 0.5F 2 1H 1S + 2u(t) A 2/S I1 1 I2 4u(t) V 4/S I3 _ 1 1 V0 BY MESH ANALYSIS, -I1 +4I 2 -I3 =4/S -I 2 + I 3 ( S + 2) = 0 GIVEN 1.5 = VO = I3 S(S+7/4) 1.5 K K2 VO ( S ) = = 1+ S(S+7/4) S (S+7/4) I1 = 2 / S , I 3 SOLVES TO K1 = 6 / 7, K 2 = -6 / 7 6 1 1 VO ( S ) = - 7 S (S+7/4) 6 VO (t ) = (1 - e -7 / 4t ) u (t )V 7 Problem 13.11 For the network shown in fig find Vo(t),t>0 using mesh equations. i0(t) 1 1F 2 1 2H e-tu(t) A 4u(t) V Suggested Solution i0(t) 1 1/S 2 1 2S e-tu(t) A 1/(S+1) 4u(t) V 4/S 4 - V1 = V1S + lO S V -V 1 lO = V2 + = 1 2 (S+1) 2 YIELDS : 1 2 lO ( S ) = - S (3S+4) 2 lO (t ) = (1 - e -4 / 3t ) u (t ) A 3 Problem 13.12 Use loop equations to find Vo(t),t>0 in the network shown in fig. 2u(t) A 1 1 2 1F e-tu(t) A 1H Suggested Solution 2/S I1 1 1 1/S 1/(S+1) I2 1S I3 I1 = 2 1 , I3 = - (S+1) S 1 2 I 2 + ( I 2 - I1 ) + ( S + )( I 2 - I 3 ) = 0 S 1 2 S 1 I 2 (2 + 1 + S + ) = + + S S (S+1) S(S+1) I 2 (S ) = S 2 + 2S + 3 (S+1)(S 2 + 3S + 1) 2 2 1.28 1.72 -( S 3 + 6S 2 + 8S + 2) =- - + + S(S+1)(S+2.62)(S+0.381) S (S+1) S + 2.62 S + 0.38 I O = I 2 - I1 = lO (t ) = (-2 - 2e - t + 1.28e-2.62t + 1.72e -0.38t )u (t ) A Problem 13.13 Use loop equations to find Vo(t),t>0 in the network shown in fig. 1 1H 1 1F + 2u(t) A 4u(t) V 1 V0 _ Suggested Solution 1 1S 1 1/S I2 + 2/S 4/S I1 2 V0 _ Mesh eq's: 4 = ( S + 4) I1 - 3I 2 S 2 I2 = S 10 I1 = S ( S + 4) VO = ( I1 - I 2 )2 10 -2S + 2 -4 S + 4 2 - 2 = VO = 2= S ( S + 4) S ( S + 4) S ( S + 4) S B A + VO = ( S + 4) S A =1 B = -5 VO (t ) = (1 + 5e -4t )u (t )V Problem 13.14 Use mesh equations to find io(t),t>0 in the network shown in fig. 1 4u(t) V 1F 1 1 2H e-tu(t) A Suggested Solution 1 4/S I1 1 I2 1/S 1 2S I3 1/(S+1) IO = I 2 I3 = 1 S +1 4 = (1/ S + 1) I1 - I 2 / S S 0 = -(1/ S ) I1 + I 2 (3 + 1/ S ) - I 3 SOLVING 1 FOR I1 & PUT INTO 2. I1 = SO I2 + 4 S S +4 1 4 I 2 3S + 1 - = ( S + 1) + S + 1 = S + 1 S + 1 ( S + 1) S +4 (4 + S ) / 3 B A IO = S + 1 = = + 2 3S + 4 S S ( S + 4 / 3) ( S + 4 / 3) S S +1 A =1 B=- 2 3 2 iO (t ) = (1 - e -4t / 3 )u (t )V 3 Problem 13.15 Use loop equations to find Vo(t),t>0 in the network shown in fig. + 1H 1 2u(t) A 1 V0 (t) _ 4u(t) V Suggested Solution + S I1 4/S 2/S 1 I2 1 V0 (t) _ MESH EQUATIONS: VO = I 2 4 2 2 = SI1 + I 2 (2 + ) & I 2 - I1 = S S S SO 2 I1 = I 2 - S AND 4 2 2S + 4 K K - = -2 + I 2 (2 + + S ) I 2 = 2 = + S S S + 2 S + 2 S + 1 - j1 S + 1 + j1 2(-1 + j1) = 2 -450 K = j2 VO (t ) = 2 2e - t COS (t - 450 )u (t )V Problem 13.16 Use loop analysis to find Vo(t),t>0 in the network shown in fig. 4e-tu(t) A 1F 6u(t) A 1 1H + 1 2u(t) A 1 V0 _ Suggested Solution 4/(S+1) 1F I3 6/S I1 1 S + I2 1 I4 2u(t) A 1 V0 _ 4 S +1 2 I2 = S -6 I3 = S I1 = KVL FOR LOOP 4 IS 1( I 2 - I1 ) + 1( I 2 - I1 + I 4 ) + S (- I 3 + I 4 ) + I 4 = 0 -(6 S 2 + 6 S + 4) S ( S + 1)( S + 3) 1( I 2 - I1 ) + 1( I 2 - I1 + I 4 ) + S (- I 3 + I 4 ) + I 4 = 0 I4 = I4 = 2 20 / 3 -(6 S 2 + 6 S + 4) -4 / 3 = + - S ( S + 1)( S + 3) S S +1 S + 3 SINCE VO (t ) = (1) I 4 4 20 VO (t ) = (- + 2e - t - e -3t )u (t )V 3 3 Problem 13.17 Use mesh analysis to find Vo(t),t>0 in the network shown in fig. 4u(t) A iX(t) + 1F (iX(t))/2 1H 1 _ 1 V0 Suggested Solution 4/S I1 + 1F IX/2 I2 1H 1 I3 1 _ V0 4 S I I2 = X 2 I1 = I X = I3 - 4 S THEN KVL FOR THE MESH 3: S [ I 3 - 0.5( I 3 - 4 / S ) ] + 1[ I 3 - 4 / S ] + I 3 = 0 I3 = 2 -4( S - 2) -6 & VO = I 3 (1) = + S ( S + 4) 3 S +4 VO (t ) = (2 - 6e -4t - 20 -3t e )u (t )V 3 Problem 13.18 Use loop equations to find Vo(t),t>0 in the network shown in fig. i1(t) 1F 1 1H + 2i1(t) 2 V0 2vA(t) 2u(t) A _ Suggested Solution I1 + VA 1/S 1 S + 2I1 I2 2 V0 2VA 2/S _ 2 1 + 2 I1 = I1 (1) + I 2 (2 + ) S S OR 2 = - SI1 + I 2 (2 S + 1) ALSO I1 - I 2 = 2VA = 2 I1 (1) I1 = - I 2 NOW , 2 = I 2 [ S + 2 S + 1] I 2 = VO = 2 I1 = 2/3 S + 1/ 3 2 2/3 = 3S + 1 S + 1/ 3 4 VO (t ) = e - t / 3u (t )V 3 Problem 13.19 Use superposition to find Vo(t),t>0 in the network shown in fig. + 1H 4u(t) V 1 2u(t) A 0.5F 1 V0 _ Suggested Solution + 1S 4/S 2/S 1 2/S 1 V0 _ FOR THE 4V SOURCE, VO ( S ) = 4 1 4 = 2 S S + 2 + 2 / S S + 2S + 2 FOR THE 2A SOURCE: VO ( S ) = 2 2S .S (2 + 2 / S ) = 2 S S + 2S + 2 4 2S 2S + 4 + 2 = 2 S + 2S + 2 S + 2S + 2 S + 2S + 2 2 FINALLY, VO ( S ) = AS IN P 16.5, VO ( S ) = YIELDS 2S + 4 S + 2S + 2 2 VO (t ) = 2 2e -t COS (+450 )u (t )V Problem 13.20 Use source transformation to solve problem 13.19. + 1H 4u(t) V 1 2u(t) A 0.5F 1 V0 _ Suggested Solution + 1S 4/S 2/S 1 2/S 1 V0 _ 1 4 2 1 VI ( S ) + = 2+ S S 2+ 2/ S S S (2S + 2)(2 S + 4) VI ( S ) = 2 2 S ( S + 2S + 2) 1 VO ( S ) = VI ( S ) 2+ 2/ S (2 S + 4) K1 K1 VO ( S ) = 2 = + ( S + 2S + 2) S + 1 - r S + 1 + r K1 = K1 = (2 S + 4) = 2 -450 ( S + 1 - r ) S =-1+ r (2 S + 4) = 2 -450 ( S + 1 - r ) S =-1+ r 2 -450 2 -450 + (S + 1 - r ) (S + 1 + r ) VO ( S ) = VO (t ) = 2 2e - t COS (t - 450 )u (t )V Problem 13.21 Use Thevenin's theorem to solve problem 13.19. + 1H 4u(t) V 1 2u(t) A 0.5F 1 V0 _ Suggested Solution + 1S 4/S 2/S 1 2/S 1 V0 _ WITH THE OUTPUT RESISTOR AS A LOAD, VOC = ZTH 4 2 2S + 4 + S= S S S 2 S2 + S + 2 = +1+ S = S S BY VOLTAGE DIVISION, 2S + 4 1 2S + 4 VO ( S ) = 2 S = 2 S +S+2 S +S +2 1 S 2S + 4 FVO ( S ) = 2 S +S +2 VO (t ) = 2 2e -t COS (t - 450 )u (t ) Problem 13.22 Use Thevenin's theorem to solve problem 13.11. i0(t) 1 1F 2 1 2H e-tu(t) A 4u(t) V Suggested Solution i0(t) 1 1/S 2 1 2S e-tu(t) A 1/(S+1) 4u(t) V 4/S VOC = 4 1/ S 1 S +4 . + = S 1 + 1/ S S + 1 S ( S + 1) 1/ S S +2 ZTH = 1 + 1 1/ S = +1 = 1 + 1/ S S +1 VOC S + 4 S +1 S +4 = = IO (S ) = ZTH + 2 S ( S + 1) 3S + 4 S (3S + 4) 2 iO (t ) = [1 - e -4 / 3t ]u (t ) 3 Problem 13.23 Use Thevenin's theorem to find Vo(t),t>0 in the network. 1H 1 1 1F 2u(t) A + 1u(t) V 1 V0 _ Suggested Solution 1S 1 1 1/S 2/S + 1/S ZTH V0 _ Using superposition VOC = ZTH 1 2 (4 s + 1) - (2s + 1) = - S S s = 2s + 2 1 -(4 S + 1) (2 S + 1 / 2) =- VO = VOC = S ( S + 3 / 2) ZTH + 1 (2 S + 3) S A B VO = + S S + 3/ 2 A = -1/ 3 B = -5 / 3 1 S VO (t ) = [- - e -3/ 2t ]u (t )V 3 3 Problem 13.24 Use Thevenin's theorem to find Io(t),t>0 in the circuit shown. 2u(t) A 1 1 1 1F e-tu(t) A 1 1H i0(t) Suggested Solution 2/S 1 1 1 1/S 1/(S+1) ZTH i0(t) 1 ZTH S VOC -(2 S + 1) 1 1 2 VOC = - (1) = S ( S + 1) S +1 S S 1 S +1 ZTH = 1 + = S S (2S + 1) (2 S + 1) -VOC S ( S + 1) IO = = = 1 + S + ZTH 1 + S + S + 1 ( S + 1)( S + 1 + S ( S + 1)) S (2 S + 1) A B C IO = = + + 3 3 2 ( S + 1) ( S + 1) ( S + 1) ( S + 1) A = -1 LET , S = 0, I O (O) = 1 = A + B + C B + C = 2 LET , S = -0.5, I O (-0.5) = 0 = 8 A + 4 B + 2C 4 B + 2C = 8 SOLUTIONS ARE B=Z AND C=0 SO, I O = - iO (t ) = (2te - t - t 2 -t e )u (t ) A 2 1 2 + 3 ( S + 1) ( S + 1) 2 Problem 13.25 Use Thevenin's theorem to find Vo(t),t>0 in the network. + 4u(t) V 2H _ 0.5F e u(t) A -2t 1 V0 2 2u(t) A Suggested Solution + ZTH 2S VOC - 2/S 1/(S+2) 2 2/S 4/S ZTH 1 USING SUPERPOSITION, 2 1 2 2S 4 2 S + 4 S + 8 VOC = 2S + = + = S ( S + 2) S+2 S S+2 S ZTH = 2 S + 2 2 VO = VOC + 4 / S 2S 2 + 8S + 16 = 1 + ZTH S ( S + 2)(2S + 3) 2 S 2 + 8S + 16 A B C VO = = + + S ( S + 2)(2 S + 3) S S + 2 C + 1.5 A = 8/3 B=4 C = -17 / 3 8 17 VO (t ) = ( + 4e -2t - e -3t / 2 )u (t )V 3 3 Problem 13.26 Use Thevenin's theorem to find Vo(t),t>0 in the network. 2u(t) A 2 4u(t) V 1 0.5F Suggested Solution 2/S 2 4/S 2/S VOC ZTH S ZTH + V0 _ USING SUPERPOSITION, VOC = VOC = 4 S + 2/ S 2 S S + 2+ 2/ S + S S + 2+ 2/ S 2 S 8S2 +8 S ( S + 1 - j1)( S + 1 + j1) VO = VO = VOC (1) 8S+8/S = 2 1+ZTH S + 2 S + 2 + 2S2 + 4 (8 / 3)(S2 + 1) S (S2 + (2 / 3) S + 2) A = 4/3 K = 1.27310.050 4 VO (t ) = ( + 2.546e - t / 3COS ( 17t + 10.050 ))V 3 Problem 13.27 Use Thevenin's theorem to find Vo(t),t>0 in the network. 1H 1 2u(t) A 1 + V0 _ 2vA(t) Suggested Solution + 1 2/S V0 _ 2vA 1 2/S ISC 2vA ZTH S + V0 _ VOC = 3VA 2 / S = VA SO, VOC = 6 / S I SC = 2 / S ZTH = VOC / I SC = 3 VO = VOC 6/ S A B = = + S + 1 + ZTH S + 4 S S + 4 A = 1.5 B = -1.5 VO (t ) = 1.5(1 - e -4t )u (t )V Problem 13.28 Use laplace transform to find Vo(t),t>0 in the network.assume that the circuit has reached steady state at t=0-. + 15V 4 4 2H 4 V0 _ Suggested Solution iL 15V 2H 4 4 V0 _ 2S + + 4 2 4 V0 _ 4 VO = 2.5 2S + 6 5 VO = S +3 VO (t ) = 5(e -3t )u (t )V Problem 13.29 Find Io, (t),t>0 in the network. 1 2 4 1F 2H 3 Suggested Solution 2 4 1/S 2/S 4/3 1/S + 8/S V0 _ SOURCE TRANSFORMATIONS: i2 (0- ) = 0 VC (0- ) = 0 VOC = 8 2S 16S 8S 2S + 4 / 3 + 1/ S = 2 S 2 + 4 S / 3 + 1 = S 2 + 2 S / 3 + 1/ 2 S (8) S 2 ZTH = 2 S (4 / 3 + 1/ S ) = 2 6S + 4S + 3 (48) S 2 2 V 48S 2 I OC = OC = 6S 2+ 4S + 3 = 2 (8) S + 6S 8S + 6S + 24S 2 + 16S + 12 ZTH + +4 6S 2 + 4S + 3 1.5S (48) S = IO = 2 52 S + 22S + 12 S 2 + 11 S + 6 16 16 K K- IO = + 11 263 11 263 S+ - j S+ + j 32 32 32 32 0 K = 0.906 34.15 iO (t ) = 1.812e -0.34t COS (0.51t + 34.150 )u (t ) A Problem 13.30 find Io(t),t>0 in the network. 2 12V 2H t=0 1 1 3 14V Suggested Solution t=0 iL 1 1 3 14V 2 12V 12/S 2 2S 1 I0 FOR t= 0iL = 12 14 1 + = 8A Z 3.5 1 + 1 12 -V V + 16 V S = + 12 - SV = V + 16 + S Z 2S 2 SO, V= -4 2S + 1 V -1 IO = = 2 S + 0.5 iO (t ) = -e -0.5t u (t ) A Problem 13.31 Use Thevenin's theorem to find Vo(t),t>0 in the network. + 10V 1 2 1H 1F V0 _ 1 Suggested Solution + 10V 1 VC _ + 1H 2 1/S 1 V0 _ + Zeq V0 _ for t=0i L =0 20 V 3 1 1 ( S + 2) /( S + 1) Z eq = 1 ( S + 2) = S +2= 1 S S +1 S +2+ S +1 ( S + 2) Z eq = 2 S + 3S + 3 20 ( S + 2) VO = 3 3 3 3 3 S + S - j S + S + j 2 2 2 2 k k - VO = + 3 3 s + 1.5 - j s + 1.5 + j 2 2 VC = 20 3 -1.5 + j 3 +2 2 = 3.85 -300 j 3 3t - 300 )V 2 k = VO (t ) = 7.7e -3t / 2COS ( Problem 13.32 find Io(t),t>0 in the network. 2 12V 6 2H 4 7 5 Suggested Solution iL 2 12V 6 2H 3 2S 6 4 I0 12 iL (0- ) = 2 A THE TRANSFORMED NETWORK FOR t>0 IS IO = 4 12 1.5 = 2S + 9 4 + 12 S + 9 / 2 iO (t ) = 1.5e -9t / 2u (t ) A Problem 13.33 find Vo(t),t>0 in the network. 3 24V 1H 3 2 4 + V0 12V _ Suggested Solution 3 24V 1H 3 I1 2 I2 4 12V 6 I1 - 3I 2 = 12 -3I1 + 9 I 2 = 12 12 -3 12 9 12(9 + 3) I1 = = = 144 / 45 6 -3 54 - 9 -3 9 iL (O - ) = 3.2 A V1 - 6.4 V1 + 12 / S V1 + + =0 2S 3 6 SOLVING FOR V1 YIELDS: 14.4 S+1 9.6 V0 = S+1 VO (t ) = 9.6e - t u (t )V V1 = Problem 13.34 find Vo(t),t>0 in the network. 4 + 2 4H 6A 6 3 V0 _ Suggested Solution 2 6 4 6 iL 3 4 2 I1 I2 3 + 6/S 6 4S I3 V0 _ 24 s 12s = 4s + 6 2s + 3 4/3 S 12 S 4 2 12S 16 S + 6 VO = 3I 2 + ( I1 - I 3 ) = + = 2 S + 3 S S 2 S + 3 S ( S + 1.5) A B VO = + S S + 1.5 A=4 loop analysis:I1 =6/S, I3 =4/S, 9I 2 -2I1 =0 I 2 = B = 12 VO (t ) = (4 + 12e -3t / 2 )u (t )V Problem 13.35 find Vo(t),t>0 in the network. 2K + 12K 12V 2K 4K V0 _ 600F 4V Suggested Solution + 12 12/S 2K + 4K V0 _ V0 _ 4/S + R 12/S 2K 4K V0 _ 1.67K/S 4/S FOR t>0 R=6K 12K=4K V1 =VO +4/S KCL: 12 - V1 V V - 12 / S S = 1 + O 1.67 K 4 K R 2K + S 8 - VO V VO - 8 / S 4S2 VO - 32 S S = + O 8 = 2 SVO + 1.67 K 4 K 5 4K 2K + 2+ S 3S OR, 4S2 VO - 32 S 12SVO - 96 S 8 = 2 SVO + 8 = 2 SVO + 5 6S + 5 2S + 3 A B 6S + 5 / 3 = + VO = S ( S + 5 /12) S S + 5 /12 A=4 B=2 VO (t ) = (4 + 2e -5t /12 )u (t )V Problem 13.36 find Vo(t),t>0 in the network. 2K 4K 2K 2K 3K + 8K V0 _ 4K 4K Suggested Solution 2K 4K 3K + V0C _ 2K 4K 2K 2K + Vc 8K V0 _ 4K 4K VOC = 24 VOC 6K 3K - 12 9K 9 = 12V RTH = 3K 6 K = 2 K V = 12(4 K / 8 K ) = 6V VC = V {(4 / 6) - (4 /12)} VC = 2V NODAL ANALYSIS: 12 - VO V VO - 2 / S S = O + 6K 4 K 4 K + 10 K / S 12 + SVO SVO S 2VO - 2 S = + 6 4 4S + 10 AND 6 S 2VO - 12 S + 3SVO -5SVO + 24 2S + S 6S 2VO - 12 S = -10 S 2VO - 25SVO + 48S + 120 24 - 2SVO = VO [16 S 2 + 25S ] = 60 S VO = A = 24 / 5 B = -21/ 20 VO (t ) = ( 24 21 -25t /16 - e )u (t )V 5 20 A B + S S + 24 /16 Problem 13.37 find Vo(t),t>0 in the network. 1H 1 + 1 4u(t) V 2H 2H 1 V0 _ Suggested Solution S 1 + I1 I2 2S 1 4/S 2S 1 V0 _ 4 = I1 (2 S + 1) - SI1 S 0 = - SI1 + I 2 (2 S + 2) I1 = I 2 ((2 S + 2) / S ) SOLVING THE ABOVE TWO EQUATIONS WE GET: 4 4 (2S + 1)(2 S + 2) = I2 - S I2 = 2 & VO = (1) I 2 S S 3S + 6 S + 2 A B 4 = + VO = 2 3S + 6 S + 2 S + 0.42 S + 1.58 A = 1.15 B = -1.15,VO(t ) = 1.15[e -0.42t - e -1.58t ]u (t )V Problem 13.38 find Vo(t),t>0 in the network. 2H 6 8 + 10u(t) V 2 0.5F 4H 8H 4 V0 _ Suggested Solution 2 + 10/S 2/S V0C _ 2S 6 8 + ZTH VOC I1 4S 8S 4 V0 _ VOC = I1 ( ZTH + 6 + 4 S ) + 2 SI 2 & 0 = 2 SI1 + I 2 (8S + 12) FROM THE SECOND LOOP EQUATION, I1 = VOC = 10 4S + 6 2 = I1[2 S - + 6 + 4 S )] ( S(S+1) S S +1 (-4S+6) I 2 , AND, S 10 = I 2 [2 S 3 + 2S 2 - (4S + 6)(2 + 6S + 6 + 4S 2 + 4S )] -5 / 7 S + 31/ 7 S 2 + 46 / 7 S + 24 VO = 4 I 2 I2 = 3 FROM A CRC MATH HANDBOOK, A K K - VO = + + S+0.513 S + 0.257 - j1/ 2 S + 0.257 + j1/ 2 A = -9.05 K = 5.08 27.200 VO (t ) = -9.05e -0.513t + 10.16e-0.257 t COS (0.5t + 27.20)V Problem 13.39 find Vo(t),t>0 in the network. 1:2 1/8 F 2H 12u(t)V 2 16 Suggested Solution 1:2 8/S 12/S 2S 2 16 I0 8/S + 8S 24/S I0 8 16 V0 _ REFER PRIMARY TO SECONDARY , N=2 12 12 24 N( ) = S S S 2 S + 2 (2 S + 2) N 2 = 8S + 8 24 3 S = 2 IO = 8S + 24 + 8 / S S + 3S + 1 A B + IO = S + 0.38 S + 2.62 A = 1.34 B = -1.34 iO (t ) = 1.34[e -0.38t - e-2.62t ]u (t )V Problem 13.40 find Vo(t),t>0 in the network. 2:1 1F + 12u(t) V 8 2H 1 V0 _ Suggested Solution 2:1 1/S + 12/S 8 2S 1 V0 _ 1/S + 2 6/S 1 V0 _ S/2 REFER PRIMARY TO SECONDARY , N=2 12 12 6 N( ) = S S S 2 S + B (2 S + 8) N 2 = 0.5S + 2 6 12 S = 2 VO = 0.5S + 3 + 1/ S S + 6S + 2 A B + VO = S + 0.35 S + 5.65 A = 2.31 B = -2.31 VO (t ) = 2.31[e -0.35t - e-5.65t ]u (t )V Problem 13.41 find Vo(t),t>0 in the network. 1:2 + 2 12u(t) V 1F 4 1/4F V0 _ 4 Suggested Solution 1:2 + 2 1/S 12/S 4 4/S V0 _ 4 + 24/S 8 4/S 4 4/S 4 V0 _ REFER PRIMARY TO SECONDARY , N=2 12 12 24 N( ) = S S S 1/ S + 2 (1/ S + 2) N 2 = 4 / S + 8 24 4 VO = ( ) S 16 + 8 / S 96 6 + VO = 16S + 8 S + 0.5 VO (t ) = e - t / 2V Problem 13.42 Determine the output voltage Vo(t) in the network in fig if the input is given by the source in fig. 1 1F 2 1 + V0 vi(t) _ Suggested Solution 1 1/S 2 1 + V0 Vi _ 1 1/S 2 + V0C Vi _ ZTH VOC 3 1 VOC Therevin ' s EQ: 2 S +1 VOC =Vi 1 = Vi ( 3S + 1) 3+1/S NEW CIRCUIT: 1 VO =VOC 4+ZTH S +1 3S + 1 ) 3S + 1 12 S + 4 + 2 S + 2 ( S + 1) 6( S + 1) VO = VO = (1 - e - S ) S (14S + 6) (14 S + 6) VO =Vi ( B A -S VO = + (1 - e ) S S + 3/ 7 A =1 B = -4 / 7 4 VO (t ) = 1 - e -3t / 7 u (t )V 7 Problem 13.43 find Vo(t),t>0 in the network in fig if the input is represented by the waveform shown in fig. 1 1 1 + 1 i0(t) 6 2 V0 _ i0(t) 12 t(s) 1 Suggested Solution S + I0 2 6 V0 _ 6 24 6 144 -S VOC =2IO = S + S + 8 = S ( S + 8) (1 - e ) S+8 6S + 12 ZTH = 6 ( S + 2) = S +8 NEW CIRCUIT: VO =VOC VO = 2 ( S + 8)2 = VOC 3+ZTH 9S + 36 32 S(S+4) ADD THE TIMESHIFT TERM: VO = B 32 A -S (1 - e - S ) = + (1 - e ) S(S+4) S S + 4 A=8 B = -8 VO (t ) = 8 - 8e -4( t -1) u (t - 1)V Problem 13.44 Determine the output voltage Vo(t) in the network in fig if the input is given by the source in fig. 1 1 1H + Vi(t) 1 1F 1 V0 _ Vi(t) 1 t(s) 1 Suggested Solution 1 1 1S + Vi 1 1/S 1 V0 _ Vi (t ) = 1[u (t ) - u (t - 1)] Vi ( S ) = 1 (1 - e - S ) S 1 V1 V1 / 3 V1 1 S +1 VO = = = 2 0.5 + 1 + S 2 S + 1 + S + 1 S + 1 S +1 S +1 2 2 1 1 B A -S -S VO = (1 - e ) = + (1 - e ) 3 S ( S + 1) S S + 1 A = 1/ 3 B = -1/ 3 1 1 1 1 VO (t ) = - e - (t ) u (t ) - - e - ( t -1) u (t - 1)V 3 3 3 3 Problem 13.45 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. 1 1F Vi(t) 2 3 1 + V0 _ Suggested Solution 1 1S Vi 2 3 1 + V0 _ ZTH + 3 1 V0 _ VOC 2 S +1 VOC =Vi 1 = Vi ( 3S + 1) 3+1/S 1 2S + 2 ZTH = 2 ( + 1) = S 3S + 1 1 VO =VOC 4+ZTH VO S +1 = Vi 14 S + 6 Problem 13.46 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. + VS (t) R2 V0C _ R1 C Suggested Solution + VS (t) R2 V0C _ R1 1/S Vo - VS VS V Z = S = 1+ Z R2 Vo R2 Z = R1 1 R1 = SC 1 + SCR1 Vo R 1 + SCR = 1 + 1 ; VS R2 1 + SCR1 R = R1 R2 NOTE THAT THE 1+R1/R 2 = dC GAIN. Problem 13.47 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. + rs(t) V0 _ R2 R1 C Suggested Solution + VS V0C _ R2 R1 1/S Vo - VS V V R + 1/ SC = S O = 1+ 1 R1 + 1/ SC R2 VS R2 OR, 1 1 Vo R R = 1 + 1 + = 1 + 1 1 + ; VS R2 SCR2 R2 SCR R = R1 + R2 1 Vo R = 1 + 1 1 + VS R2 SCR NOTE THAT THE 1+R1/R 2 = GAIN AT S Problem 13.48 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. + VS (t) R2 V0 _ R1 C Suggested Solution + VS (t) R2 V0 _ R1 1/S Vo - VS VS V R1 = O = 1+ R1 R2 + 1/ SC VS R2 + 1/ SC OR, SCR1 Vo = 1+ ; VS 1 + SCR2 FINALLY , S + 1/(C ( R + R )) Vo R 1 2 = 1 + 1 1 VS R2 S+ CR2 NOTE THAT THE 1+R 1/R 2 = HIGH FREQUENCY GAIN Problem 13.49 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. R1 R2 C2 C1 vs(t) + V0 _ Suggested Solution KCL AT NODE V1 IS VS - V1 V1 = + SC2V1 + C1S (V1 - VO ) R1 R2 AT THE -VE TERMINAL OF THE OP-AMP SINCE i - = 0 SC2V1 + VO =0 R3 SOLVING THE TWO EQUATIONS YIELDS: -S VO C1 R1 = VS C1 C2 R1 + R2 S2 + S + + C1C2 R3 C1C2 R3 C1C2 R1 R2 R3 Problem 13.50 Determine the transfer function for the network shown in fig. If a step function is applied to the network What type of damping will the network exhibit. 1F 1 vs(t) 1 1 1F V0 Suggested Solution 1/S 1 Vs 1 1 1/S V0 USING KCL: AT V1: VS -V1 = V1 +(V1 -VO )S+SV1 OR, VS =V1 (2S+2)-VO (S) AT (-) INPUT: SV1=-VO V1 =-VO /S NOW, V V S VS = O (2S+2)-SVO = VO (S+2+2/S) - O = 2 S VS S +2S+2 CHARACTERISTIC EQ: S2 +2S+2 ROOTS ARE S=-1 j1 Problem 13.51 Find the transfer function Vo(s)/Vi(s) for the network shown in fig. If the step function is applied to the network, will the response be overdamped underdamped, or critical damped. 1F 1 vs(t) 1 1F V0 Suggested Solution 1/S 1 vs 1 1/S V0 NODAL ANALYSIS, @ V1 : VS - V1 (V -V ) = (V1 -VO )S+ 1 O 1 1 VS = V1 ( S + 2) - VO (S+1) V1 - V0 = (VO )S V1 = VO ( S + 1) 1 VS = VO [ ( S + 1)( S + 2) - 1] = VO ( S + 1)( S + 1) = VO ( S + 1) 2 VO 1 = VS ( S + 1) 2 CRITICALLY DAMPED. @(+) INPUT: Problem 13.52 The voltage response of the network to a unit step input is. V0 ( s ) = 2( s + 1) s ( s + 12 s + 27) 2 Is the response overdamped. Suggested Solution A Network has a step responce of 2( s + 1) V0 ( s ) = 2 s ( s + 12 s + 27) is it overdamped? completing the square s 2 + 12 s + 27 = ( s + 3)( s + 4) s = -3, s = -9 system is overdamped. Problem 13.53 The transfer function of the network is given by the expression G ( s) = 100 s s 2 + 22s + 40 Determine the damping ratio, the underdamped natural frequency and the type of response that will be exhibited by the network. Suggested Solution Asystem has teh transfer function 100s G ( s) = s 2 + 22 s + 40 find , 0 The characteristic equation s 2 + 22s + 40 = 0 with analogy to s 2 + 2 0 s + 0 2 0 = 40,2 0 = 22 = 11 2 10 overdamped Problem 13.54 The current response of a network to a unit step input is I 0 (s) = 10( s + 2) s ( s 2 + 11s + 30) 2 Is the response underdamped? Suggested Solution If step responce is 10( s + 2) I 0 (s) = 2 2 s ( s + 11s + 30) How is it Damped? The characteristic equation is s 2 ( s 2 + 11s + 30) = 0 s 2 ( s + 6)( s + 5) Roots s = -6, s = -5 Being real and unequal it is overdamped. Problem 13.55 The voltage response of a network to a unit step input is V0 ( s ) = 10 s ( s + 8s + 16) 2 Is the response critically damped. Suggested Solution If step responce is 10 V0 ( s ) = 2 s ( s + 8s + 16) The characteristic equation is ( s 2 + 8s + 16) = 0 ( s + 4) 2 = 0 Roots s = -4 Critically Damped. Problem 13.56 For the network in fig choose the value of C for critical damping. + C vs(t) 6H C 1 V0 _ Suggested Solution + C vs 6H C 1 V0 _ Parallel RLC Circuit R = 1, L = 6h, Ceq = C + 1F ; Characteristic Equation is s 1 + =0 s2 + RCeq LCeq s 2 + 2 n s + n = 0 2 For critical damping, = 1 1 1 = n = LCeq 6Ceq RCeq = Ceq = Ceq = Ceq = 6 2 1 2 n = 6 Ceq 2 6 = 1.5 = C + 1 4 C = 0.5F Problem 13.57 For the filter in fig choose the values of c1 and c2 to place poles at s=-2 and s=-5 rad/s. 100K 100K vs(t) C1 C2 Suggested Solution 100K 100K vs C1 C2 + V0 _ Both opamps are connected as unity gain bufffers R = 100k 1/ sc1 V1 = Vs ( ) 1/ sc1 + R also V0 = V1 ( or V0 1 1 [ ] = 2 Vs R c1c2 ( s + 1 )( s + 1 ) Rc1 Rc2 For poles at -2 and -5 the transfer function is Vo 1 1 ] [ = 2 Vs R c1c2 ( s + 2)( s + 5) so Rc1 = 0.5 Rc2 = 0.2 c1 = 5 F c2 = 2 F c1 and c2 interchagable 1/ sc2 ) R + 1/ sc2 Problem 13.58 Find the steady state response Vo(t) for the network in fig. + 1H 1 2 12 cos (t) 1F 2 1 V0 _ Suggested Solution + S 1 2 12 cos (t) 1/S 2 1 V0 _ Using KCL V1 V2 V2 V2 + + + =0 s + 1 2 1/ s 3 V1 = V2 - Vi Vi (t ) = 12 cos(t )V Vz ( 1 1 1 1 1 + + s + ) = Vi ( + ) 3 s +1 2 s +1 2 Solve fo rthe Transfer Function 3s + 9 V2 = 2 Vi 6 s + 11s + 11 V0 V s+3 = 2 = 2 Vi 3 / Vi 6 s + 11s + 11 For Steady State s = j1 V0 Vi = j1 3 + j1 = 0.26 - 47.2o -6 + 11 + j11 Also | Vin |= 12 so V0 = 12(0.26) - 47.2o Steady Staet Phasor V0 (t ) = 3.13cos(t - 47.2o )V Problem 13.59 Find the steady state response Vo(t) for the network in fig. 10 cos(t) A + 2 1 1H 1F 1 V0 _ Suggested Solution Is + 2 1 1S 1/S 1 V0 _ Nodal Analysis V1 V1 V + + V2 ( s ) + 2 = 0 1 s 1 1 V 1(1 + ) = -V 2( s + 1) s or V1 = - sV2 Is + V1 - V2 = V2 s + V2 2 V1 = V2 (2 s + 3) - 2 I s - sV2 = V2 (2 s + 3) - 2 I s V2 = V0 = I s ( 2 ) 3s + 3 Transfer _ Function V0 2 / 3 = Is s +1 For steady state s=j1 and V0 2/3 2 = = - 45o I s 1 + j1 3 | I s |= 10 so, 10 2 - 45o 3 V0 (t ) = 4.7 cos(t - 45o )V V0 = Problem 13.60 Find the steady state response Vo(t) for the network in fig. 4 cos 2t V v1(t) + 1 0.5H 2V1(t) 2 V0 _ Suggested Solution v1 VI + 1 v1 S/2 2 V0 _ Nodal _ Analysis 2 V V1 + 2V1 + V0 ( ) + 0 = 0 s 2 V1 = V0 - VI 3V0 - 3VI + so, 4 6VI = V0 (6 + + 1) s V0 6s = VI 7 s + 4 at Steady State s= j2 since 2V0 V0 + =0 s 2 Vi (t ) = 4 cos(2t ) V0 Vf = j2 j12 = 0.8215.95o 4 + j14 sin ce | Vi |= 4 V0 = 4(0.82)15.95o = 3.315.95o V0 (t ) = 3.3cos(2t + 15.95o )V Problem 13.61 Find the steady state response Io(t) for the network shown the fig. 4 cos 2t V 2i1(t) i1(t) 0.5F 2 1 i0(t) Suggested Solution VI 2i1 i1 2/S S 2 1 i0(t) Loop _ Equations I 0 Loop : 2 I1 = ( s + 2) I 0 - sI1 I1 = I 0 I1 Loop : VI = s ( I1 - I 0 ) + I1 + 2 VI = I 0 (1 + ) s 2 I1 s I0 s = VI s + 2 sin ce Vi (t ) = 4 cos(2t ) at Steady State j = j2 I0 j2 1 = = 45o Vi s + j 2 2 sin ce | VI |= 4 I0 = 4 45o 2 I 0 (t ) = 2 2 cos(2t + 45o ) A Problem 13.62 Find the steady state response Io(t) for the network shown the fig. 0.5F 2 i1(t) 12 cos 2t V 2i1(t) 1 i0(t) Suggested Solution 2/S S 2 i1 VI 2i1 1 i0 KCL V0 s + 2 I1 + (V0 - VI ) = 0 1 2 I 0 = V0 /1, I1 = VI / s I0 + or s s 2 s2 - 4 ) I 0 (1 + ) = VI ( - ) = VI ( 2 2 s 2s s+2 s2 - 4 I0 ( ) = VI ( ) 2 2s Finally I0 s - 2 = VI s 2VI s s + I 0 ( ) - VI ( ) = 0 2 2 s at steady state s=j2 Vi (t ) = 12 cos(2t ) I0 VI = 2 j2 - 2 = 245o j2 and I 0 =| VI | 245o = 12 245o I 0 = 12 2 cos(2t + 45o ) A Problem 13.63 Find the steady state response Io(t) for the network shown the fig. 4 cos tu(t) A 1F 1 + Vx + 1 1H 2Vx(t) 1 V0 _ Suggested Solution + IIN 1 1F Vx I3 + 1 1S I2 2Vx(t) 1 V0 _ Loop Equations I 3 - I IN =0 S 1 1 I 2 ( S + 1) + I 3 (2 + ) - I IN (1 + ) = 0 S S VO I 3 I 3 (2) + I 2 ( S ) + ( I 2 - I IN )(1) + I 2 - I 3 = 2VA = 2( I IN - I 3 ) I IN (2 + 2 1 2 1 - 1 - ) + I3[ S + 1 - 2 + 2 - + ] = 0 S S S 3 1 SI IN (1 + ) = V0 [- S 2 - S + 1] S V0 -( S + 1) = I IN S 2 + S - 1 SINCE I IN (t ) = 4COSt (t > 0) at steady state s j1 V0 -(1 + J 1) = = 0.6371.57O I IN -2 + J 1 SINCE | iIN |= 4 A V0 = 4(0.63)71.57O V0 (t ) = 2.52COS (t + 71.57O )V Problem 13.64 Find the steady state response Io(t) for the network shown the fig. 1H 1F + 8 cos 2tu(t) V V0 _ 4 cos 2tu(t) V Suggested Solution 1S 1/S 2I0 + V0 Va _ Vb VA (t ) = 8cos 2t Va (t ) = 4 cos 2t at steady state VA = 80o V VB = 40o V KCL V1 - VA + (V2 - VA ) s = I1 s I1 + V2 + V0 = 0 also V2 = V0 + VB V1 = 2 I 0 + V2 = 3V2 = 3V0 + 3VB combinig these equations yeilds V0 = VA ( s2 + 1 s2 + s + 3 ) - VB ( 2 ) s 2 + 2s + 3 s + 2s + 3 at steady state 8(-4 + 1) - 4(-4 + 3 + j 2) = 5.2297.77o V0 = -1 + 3 + j 4 V0 (t ) = 5.22 cos(2t + 97.77o )u (t )V Problem 13FE-1 A single loop second order circuit is described by the following differential equation What is the correct form of the total (natural plus forced) response? Suggested Solution The Characteristic Eqn is s 2 + 2s + 2 = 0 or ( s + 1 j1) = 0 with a constant forcing function - the answer is (0) Problem 13FE-2 If all initial conditions are zero in the network in fig, find transfer function Vo(s)/Vs(s) and determine the type of damping exhibited by the network. 1/4 F 2H Vs(t) 2 Suggested Solution V0 2 2s s s ( s) = = = = 4 2s + s 2 + 4 s 2 + s + 2 Vs 1 7 2 + 2s + s+ s j s 2 4 The network is underdamped Problem 13FE-3 The initial conditions in the circuit in fig are zeros. Find the transfer function Io(s)/Is(s) and determine the type of damping exhibited by the circuit. 1/3 F 1H is(t) 4 i0(t) Suggested Solution I0 s+4 s ( s + 4) s ( s + 4) (s) = = = 3 s 2 + 4s + 3 ( s + 1)( s + 3) Is s+4+ s The network is overdamped Problem 14.1 Find the exponential Fourier series for the signal shown. Suggested Solution Cn = 1 T0 - jn t -T0 -e 0 dt + 2 0 1 T0 T0 2 0 e - jn0t dt 1 Cn = -T0 jn0 T jn0 0 - jn0 T20 -1-1+ e 2 E 0 = Cn = 2 T0 1 1 jn - jn 2 - e - e = 2 jn 2 - 2 cos ( n ) 2 jn C0 = 0 0 1 - cos ( n ) = 2 Cn = jn jn n = 2 jn0t f (t ) = e n =- jn n 0 n odd n n even odd Problem 14.2 Find the exponential Fourier series for the periodic pulse train shown. Suggested Solution C0 = 10 ( 0.1) 1 =1 T0 = 1 0 = 2 1 Cn = T0 T0 10 0 10e - jn0t dt - jn jn 10 e - e 10 j2 T0 - jn 10 5 10 e - jn0t 10 = Cn = 1- e 5 = 0 - jn 0T0 jn - jn 0T0 10 - jn 10 Cn = e sin n n 10 f (t ) = 10 1 - jn 10 e sin n e jn0t n n =- 10 Problem 14.3 Find the exponential Fourier series for the periodic signal shown. Suggested Solution T0 = 5 1 Cn = T0 0T0 = 2 use time shift theorem with t0 = 2sec T0 5 0 6e - jn0 t - jn 6 jn 5 5 dt = e - e jn 2 6e 5 Cn = sin n and f (t ) = Cn e - jn0t e jn0t n 5 n =- 6 - jn f (t ) = e sin n e j 0.4 t 5 n =- n - jn Problem 14.4 Compute the exponential Fourier series for the waveform that is the sum of the two waveforms shown by computing the exponential Fourier series of the two waveforms and adding them. Suggested Solution For V1 ( t ) , Cn1 = 2 T0 T0 = 4 0T0 = 2 e - jn0t dt = 2 jn 2 - jn 1- e 2 T0 4 0 ForV2 ( t ) , note V2 ( t ) is half as large and time shifted 1sec, or Cn2 = Cn1 2 e - jn0 (1) T0 4 = Cn1 2 e - jn 2 Cn = Cn1 + Cn2 jn - jn - jn - jn 1 1 e 4 +e 4 - jn Cn = 1- 2 - 2e 2 + e 2 - e = 2 jn 2 jn 3 e - jn 4 Cn = 1 jn 3 - jn 1 - e 4 cos n 4 V (t ) = 3 i - jn jn t 1 - e 4 cos n e 2 V 4 n =- jn Problem 14.5 Find the exponential Fourier series for the signal shown. Suggested Solution Comparing f (t ) to the waveforms in table14.1 notice that f ( t ) matches the even function symmetry triangular wave where A = 1 T0 = 2 & 0 = C0 = A 1 = 2 2 Cn = -2 A -2 n 0 = 2 2 2 2 n n n odd f (t ) = -2 1 + 2 2 e jn t 2 n =- n n 0 n odd Problem 14.6 Given the waveform shown, determine the type of symmetry that exists if the origin is selected at: (a) l1 and (b) l2. Suggested Solution If origin is at l1, then v (t) = -v(- t) odd symmetry If origin is at l2, then v (t) = v (-t) even symmetry Problem 14.7 What type of symmetry is exhibited by the two waveforms shown? (b) Suggested Solution In part a) f1 (t) = -f1 (t) In part b) f2 (t) = -f2 (t- T0/2) odd symmetry half wave symmetry Problem 14.8 Derive the trigonometric Fourier series for the waveform shown. Suggested Solution bn = 2 T0 T0 2 T - 0 2 2 20 40 sin ( n 0 t ) t cos ( n 0 t ) t sin ( n 0 t )dt = 2 - 2 T0 n 0 T0 ( n 0 ) - T0 2 T0 bn = bn = 40 sin ( n ) T0 cos ( n ) sin ( n ) T0 cos ( n ) - + - 2 2 2 2n0 2n 0 T0 ( n0 ) ( n0 ) 40 T0 n +1 20 cos ( n ) = ( -1) 2 n T0 n 0 n +1 n =1 V ( t ) = ( -1) 20 sin ( nt ) n Problem 14.9 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution 1 3 a0 = (1)(1) + (1)(1) = T0 = 2sec 2 4 2 1 2 2 an = t cos ( n 0 t ) dt + cos ( n 0 t ) dt T0 0 T0 1 2 t sin ( n0 t ) cos ( n 0 t ) 2 + + sin ( n 0 t ) 2 ( n0 ) 0 n0T0 n 0 1 sin ( n ) cos ( n ) sin ( 2n ) sin ( 2n ) cos ( n ) - 1 1 + - + - = an = 2 2 2 n n n ( n ) ( n ) ( n ) 2 an = T0 1 bn = 2 T0 1 0 t sin ( n 0 t ) dt + 2 T0 2 1 sin ( n 0 t ) dt 1 sin ( n0 t ) t cos ( n0 t ) cos ( n 0 t ) 1 bn = - + 2 n 0 n 0 ( n0 ) 0 2 sin ( n ) cos ( n ) cos ( n ) cos ( 2n ) 1 bn = - + - =- 2 n n n n ( n ) a0 = 3 4 an = cos ( n ) - 1 ( n ) 2 bn = - 1 n Problem 14.10 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution Even symmetry, T0 = 4 0 = 2 bn = 0 1 1 (1)(1)( 2 ) 1 2 a0 = = 4 4 n an = t cos 0 2 1 cos ( n0 t ) t sin ( n 0 t ) t dt = + 2 n 0 ( n0 ) 0 cos n sin n 2+ 2 - 1 = 4 cos n - 1 + 2 sin n an = 2 2 2 n0 ( n0 ) ( n0 ) ( n ) 2 n 2 a0 = 1 4 an = 4 ( n ) 2 2 cos n 2 - 1 + n sin n 2 Problem 14.11 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution V ( t ) can be expressed as the sum of the two wave forms shown below where V1 ( t ) is the wave form in Problem 14.10 Since V2 ( t ) is of even form a 02 = - 1 2 bn2 = 0 T0 2 0 T0 = 4 2 4 an2 = T0 an2 = V2 (t )cos ( n 0 t ) dt = - cos ( n 0 t ) dt = 1 -1 n 0 sin ( n ) - sin n 2 2 n sin n 2 1 4 2 and an1 = cos n - 1 + sin n 2 4 ( n ) 2 n 2 bn = 0 From problem14.10 : a01 = a0 = a01 + a02 = - an = 4 1 4 ( n ) 2 4 cos n 2 - 1 + n sin n 2 Problem 14.12 Find the trigonometric Fourier series for the waveform shown. Suggested Solution Half wave symmetry a0 = 0 an = bn = 0 an = 4 T0 1 0 T0 2 0 T0 = 4sec for n even 0 = 2 f ( t ) cos ( n0 t ) dt for n odd 2 1 an = cos ( n0 tdt ) + 2cos ( n 0 t dt ) sin n 0 t 2sin n 0 t an = + n 0 0 n 0 1 n sin 2 an = 1 2 n + 2sin ( n ) - 2sin 2 n 2 -2 n sin an = n 2 bn = 4 T0 To 2 0 f ( t ) sin ( n 0 t ) dt = sin ( n 0 t ) dt + 2 sin ( no t ) dt 0 1 0 1 2 bn = cos ( no t ) n 0 + 1 cos ( no t ) no 1 n 1 - cos 2 = 2 n + 2cos 2 n 2 - 2cos ( n ) Since bn holds only for n odd , bn = v1 + 1 = 2 n =1 n odd 6 n 3 1 n sin no t - sin cos no t n n 2 Problem 14.13 Find the trigonometric Fourier series for the waveform shown. Suggested Solution - 2 4 This waveform is similar to that of prob. 14.15. If one multiplis ( t ) in 14.15 by T = 2 sec 2 0 = 1r / s a0 = 2 - 2 = negative , one gets this waveform here. From 14.15. 1 4 -1 a0 = 4 a0 = an = 1 n 2 2 (1 - cos ( n ) ) bn = 2cos ( n ) - 1 n 1 bn = (1 - 2cos ( n ) ) n an = 1 ( cos ( n ) - 1) n2 Problem 14.14 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution T0 = 4 sec an = an = so, an = sin ( n ) n 0 + 1 2 ( n 0 ) 2 0 = 2 0 a0 = 1 1 1 2 ( 2 )( 2 ) - (1)(1) = 4 T0 4 3 2 T0 2 ( 2 - t ) cos ( n t ) dt + T 2 0 0 2 - cos ( n 0 t ) dt 0 sin ( n0 t ) n 0 0 3 t sin ( n 0 t ) cos ( n 0 t ) sin ( n 0 t ) + + + 2 2n0 2n 0 4 2 ( n 0 ) 2 n3 sin - sin ( n 2 ) sin ( n ) cos ( n ) 2 - - + 2 n 0 2n 0 2 ( n0 ) n sin 2 2 an = 2 2 (1 - cos ( n ) ) - n n 2 2 2 0 bn = ( 2 - t ) sin ( n0 t ) dt - sin ( n0 t ) dt 0 T0 T0 -1 bn = bn = bn = cos ( n 0 t ) n 0 n 0 0 0 t cos ( n0 t ) sin ( n 0 t ) cos ( n 0 t ) + - + 2 2n 0 2n 0 -1 2 ( n 0 ) 0 2 2 1 - cos ( n ) + cos ( n ) n 0 - sin ( n ) 2 2 ( n0 ) + 1 - cos ( -n 0 ) 2n 0 1 1 3 - cos ( n 0 ) = n 2n 0 1 4 a0 = n 3 - cos 2 n sin 2 2 an = 2 2 (1 - cos ( n ) ) - n n bn = 1 n n 3 - cos 2 Problem 14.15 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution T0 = 2 sec an = 2 T0 0 = r / s -t cos ( n 0 t ) dt + 2 T0 a0 = 2 1 1 ( -1)(1) 2 = (1)(1) = 4 T0 1 0 0 2 1 cos ( n 0 t ) dt 2 cos ( n t ) t sin ( n t ) sin ( n t ) an = + + ( n )2 n n 1 1 an = an = 1 ( n ) 1 2 2 - cos ( n ) ( n ) 2 - sin ( n ) n + sin ( n 2 ) - sin ( n ) n (1 - cos ( n ) ) n 2 2 1 2 bn = -t sin ( n 0 t ) dt + T0 0 T0 2 1 sin ( n 0 t ) dt 1 t cos ( n 0 t ) sin ( n 0 t ) cos ( n 0 t ) bn = - + 2 n 0 n 0 ( n0 ) 0 2 1 bn = a0 = cos ( n ) n 1 4 - sin ( n ) cos ( n ) cos ( 2 n ) 2 1 _ - = cos ( n ) - 2 n n n n ( n ) 1 n 2 2 an = (1 - cos ( n ) ) bn = 1 2cos ( n ) - 1 n Problem 14.16 Find the trigonometric Fourier series coefficients for the waveform shown. Suggested Solution T0 = 2 sec an = 0 = r / s a0 = 1 1 1 2 (1)( -1) + (1)(1) = 4 T0 2 2 1 ( t - 1) cos ( n0t ) dt + 1 cos ( n0t ) dt T0 0 1 1 2 cos ( n t ) t sin ( n t ) sin ( n t ) sin ( n t ) an = + + - ( n )2 n n n 0 1 0 so, an = an = cos ( n ) ( n ) 1 2 2 + sin ( n ) n - 1 ( n ) 2 - sin ( n ) n + sin ( 2 n ) n - sin ( n ) n cos ( n ) - 1 n 2 2 1 2 bn = ( t - 1) sin ( n 0 t ) dt + T0 0 T0 1 2 1 sin ( n 0 t )dt 1 1 sin ( n 0 t ) t cos ( n 0 t ) cos ( n 0 t ) cos ( n 0 t ) bn = - + + ( n )2 n 0 n 0 n0 0 2 0 0 bn = bn = sin ( n ) ( n ) 2 - cos ( n ) n + cos ( n ) n an = - cos ( 2 n ) n + cos ( n ) n - 1 n 1 4 1 cos ( n ) - 2 n cos ( n ) - 1 n 2 2 1 a0 = Problem 14.17 Derive the trigonometric Fourier series for the function shown. Suggested Solution Half wave symmetry an = 4 T0 a0 = 0 an = bn = 0 for n even T0 2 0 2vt 8v co ( nw0 t ) dt = 2 T0 T0 T0 2 0 T0 t cos ( n0 t ) dt 2 8v t sin ( n0 t ) co ( n 0 t ) an = 2 + 2 T0 n 0 ( n0 ) 0 an = 8v 2 T0 T 0 2 sin n cos ( n ) - 1 + 2 n 0 ( nw0 ) 8v an = 2 T0 -2T0 2 -4v 2 2= 2 2 4n n T0 For n odd , 4 To T0 2 0 bn = bn = 2 2v 8v sin ( n 0 t ) t cos ( n 0 t ) t sin ( n0 t ) dt = 2 - 2 T0 n0 T0 ( n0 ) 0 8v sin ( n ) T0 cos ( n ) - = 0 2 2 2 n 0 T0 ( n0 ) n =1 n odd for n odd , bn = 2v n v (t ) = n si ( n t ) - n 0 2 2v 4v 2 cos ( n 0 t )V Problem 14.18 Derive the trigonometric Fourier series for the function v(t) = A|sin t| as shown. Suggested Solution To = even function, 1 a0 = T0 an = 4 T0 0 = 2 bn = 0 T0 v ( t ) = A sin | t | 0 0 T0 2 0 A 2A A sin ( t ) dt = cos t = T0 T0 T 0 A sin ( t ) cos ( n 0 t ) dt sin ( + ) 2 sin ( + ) 2 Use trig. identity , sin cos = Now, an = 4 A 20 sin ( 2n + 1) t + sin (1 - 2n ) t dt 2T0 0 2 A cos (1 + 2n ) t cos (1 + 2n ) t + 1 = 2n 1 - 2n T0 2 0 T + an = 1 - cos (1 = 2n ) 1 - cos (1 - 2n ) 2 2 = + 1 + 2n 1 - 2n Since an = cos (1 + 2n ) = cos (1 - 2n ) = 0 for all n, 2 2 2 A 1 - 2n + 1 + 2n 4A = (1 - 2n )(1 + 2n ) (1 - 4n 2 ) cos ( 2nt ) 2A 1 + 2 2 n =1 1 - 4n 2A 4A a0 = an = (1 - 4n 2 ) f (t ) = bn = 0 Problem 14.19 Derive the trigonometric Fourier series for the waveform shown. Suggested Solution T0 = 2 an = 2 T0 0 = a0 = 1 T0 1 0 A sin ( t ) dt = 0 A A cos ( t ) 1 = 2 1 0 A sin ( t ) cos ( n t ) dt sin ( - ) + sin ( + ) 2 Trig. identity sin cos = an = 1 A ( sin (1 - n ) t + sin (1 + n ) t ) dt 2 0 0 A cos (1 - n ) t cos (1 + n ) t + an = 2 (1 - n ) (1 + n ) so, an = 1 A 1 - cos (1 - n ) 1 - cos (1 + n ) + 2 (1 - n ) (1 + n ) an = A 2 if n is odd an = 0 If n is even, 2 T0 1 2 2A 2 1 - n + 1 + n = 1 - n 2 ( ) for n = 1 bn = 0, forn 1 A sin ( t ) sin ( n t ) dt = 1 2 0 A0 sin ( t ) dt 1 (1 - cos 2 ) , 2 1 Since sin 2 = A 1 A sin ( 2 t ) A bn = (1 - cos ( 2 t ) ) dt = t - = [1] 2 0 2 2 0 2 so, f (t ) = A + A 2A sin ( t ) + cos ( n 0 t ) 2 2 n = 2 (1 - n ) n even Problem 14.20 Use PSpice to determine the Fourier series of the waveform shown in the form. VS ( t ) = a0 + bn sin ( n 0t + n ) n =1 Suggested Solution Transient specifics were Final Time=10ms, Step Ceiling=10us, Center Frequency=100 Hz and Number of Harmonics=10 FOURIER COMPONENTS OF TRANSIENT RESPONSE V(Vs) DC COMPONENT= 3.750000E+00 HARMONIC FREQUENCY FOURIER NO (HZ) PHASE(DEG) 1 1.000E+02 0.000E+00 2 2.000E+02 1.971E+02 3 3.000E+02 4.185E+02 4 4.000E+02 6.395E+02 5 5.000E+02 8.020E+02 6 6.000E+02 9.645E+02 7 7.000E+02 8.256E+02 8 8.000E+02 1.047E+03 9 9.000E+02 1.244E+03 10 1.000E+03 1.303E+03 NORMALIZED COMPONENT 3.954E+00 2.016E+00 1.247E+00 6.417E-01 2.027E-01 2.852E-01 2.291E-01 1.260E-01 4.883E-02 1.993E-08 PHASE COMPONENT 1.000E+00 5.099E-01 3.154E-01 1.623E-01 5.126E-02 7.213E-02 5.793E-02 3.188E-02 1.235E-02 5.040E-09 NORMALIZED (DEG) -1.424E+02 -8.774E+01 -8.735E+00 6.991E+01 9.000E+01 1.101E+02 -1.713E+02 -9.226E+01 -3.759E+01 -1.206E+02 TOTAL HARMONIC DISTORTION = 6.310255E+01 PERCENT Problem 14.21 Use PSpice to determine the Fourier series of the waveform shown in the form vs ( t ) = a0 + bn sin ( n0t = n ) n =1 Suggested Solution Transient specifics were Final Time=10ms, Step Ceiling=10us, Center Frequency=100 Hz and Number of Harmonics=10 FOURIER COMPONENTS OF TRANSIENT RESPONSE V(Vs) DC COMPONENT= 1.997500E+00 HARMONIC NO FREQUENCY (HZ) FOURIER COMPONENT NORMALIZED COMPONENT PHASE (DEG) NORMALIZED PHASE(DEG) 1 2 3 4 5 6 7 8 9 10 2.000E+01 4.000E+01 6.000E+01 8.000E+01 1.000E+02 1.200E+02 1.400E+02 1.600E+02 1.800E+02 2.000E+02 4.401E+00 1.669E+00 8.810E-01 4.518E-01 3.183E-01 3.001E-01 2.046E-01 1.829E-01 1.913E-01 1.592E-01 1.000E+00 3.792E-01 2.002E-01 1.027E-01 7.233E-02 6.819E-02 4.648E-02 4.157E-02 4.347E-02 3.617E-02 1.049E+02 -3.975E+01 -1.480E+02 3.521E+01 1.791E+02 2.212E+01 -1.565E+02 -2.297E+01 1.625E+02 -1.800E+00 0.000E+00 -2.495E+02 -4.627E+02 -3.844E+02 -3.454E+02 -6.073E+02 -8.907E+02 -8.621E+02 -7.816E+02 -1.051E+03 TOTAL HARMONIC DISTORTION= 4.597325E+O1 PERCENT Problem 14.22 The discrete line spectrum for a periodic function f(t) is shown. Determine the expression for f(t). Suggested Solution f 0 = 10 HZ Dn = n = an - jbn all n = 90 f ( t ) = -4sin ( 20 t ) - 5sin ( 40 t ) - 3sin ( 60 t ) - 2sin ( 80 t ) - sin (100 t ) an = 0 and bn = - Dn 0 = 20 r / s Dn = 2 | Cn | Problem 14.23 The amplitude and phase spectra for a periodic function v(t) that has only a small number of terms is shown. Determine the expression for v(t) if T0 = 0.1 s. Suggested Solution T0 = 0.1 sec n 1 2 3 4 an 8 6 4 2 0 = 20 r / s n 80 80 0 -45 v ( t ) = 8cos ( 20 t + 80 ) + 6cos ( 40 t + 80 ) + 4cos ( 60 t ) + 2cos ( 80 t - 45 ) Problem 14.24 Plot the first four terms of the amplitude and phase spectra for the signal f (t ) = n =1 n odd n sin -2 n 2 6 sin ( n0t ) cos ( n 0t ) + n Suggested Solution f (t ) = an = n =1 n odd n sin -2 6 n cos n0t + sin ( n0t ) 2 n -2 n sin n 2 6 bn = n Dn = an + bn n 1 2 3 4 5 6 7 Dn 2 2 2 n odd n = tan -1 Dn 2.01 0 0.67 0 0.40 0 0.29 n () -108 0 -72 0 -108 0 -72 -bn an n 0 -20 -40 1 -60 -80 -100 n 1 3 5 7 -110 1 3 5 7 n Problem 14.25 Determine the steady-state response of the current io(t) in the circuit shown if the input voltage is described by the waveform shown in Problem 14.8. 1 -+ Vs(t) 2H 2 I0 Suggested Solution Z = jzn ||1 = jzn 1 + jzn From 14.8, T0 = 2 , 0 = 1 r / s n +1 20 cos ( nt - 90 )V n n =1 I0 Z jzn jn = = = Vs Z + 2 jzn + z + j 4n 1 + j 3n vs ( t ) = ( -1) let G ( n ) = jn and n = G ( n ) 1 + j 3n n +1 I 0( n ) = Vs ( n ) G ( n ) i0 ( t ) ( -1) = 20 n | G ( n ) | cos ( nt - 90 + n ) Problem 14.26 If the input voltage in Problem 14.25 is vs ( t ) = 1 - n sin ( 0.2 nt )V n =1 2 1 find the expression for the steady-state current io(t). Suggested Solution prob. 14.25 circuit w / vs ( t ) = 1 - n sin n =1 2 1 nt V 5 Now, 0 = s , jn and n = G ( n ) 5 + j 3n From the circuit , i0 dc component is zero, 50, Now, G ( n ) = i0 ( t ) = n =1 jn I0 j 2n0 s = jn = = Vs 2 + j 6n0 1 + j 3n 5 + j 3n s -2 nt | G ( n ) | cos - 90 + n A n s Problem 14.27 Determine the first three terms of the steady-state voltage vo (t) if the input voltage is a periodic signal of the form vs ( t ) = 1 1 + ( cos ( n - 1) ) sin ( nt )V 2 n =1 n 1H 1 V(t) -+ + 1F 1 Vo (t) _ Suggested Solution VTH = V 1 + jn Z TH = V0 1 = = VTH 1 + jn + Z TH (1 + jn )2 + 1 V0 = V (n) 2 - n + j 2n 2 1 1 + jn 1 + jn 1 and 2 - n + j 2n 2 let +2 H (n) = 90 n = H ( n ) 1 V0 = 4 V0 ( 2 ) = 0 V0 = 1 4 V0 (1) = 1 + j2 = 0.28 26.6 2 -49 V0 ( 3) = 3 = 0.023 -49 -7 + j 6 V0 (1) = 0.28 26.6 V0 ( 2 ) = 0 V0 ( 3) = 0.023 -49.4 Problem 14.28 The current Is (t) shown is applied to the circuit shown. Determine the expression for the steady-state current io (t) using the first four harmonics. Suggested Solution T0 = 1 0 = 2 is ( t ) = ( -1) n =1 n +1 From Table 15.1, 10 sin ( n 0 t ) \ n 100 n n I0 = I s = Let H ( n ) = n - j100 100k 100 - j n - j100 n n 1 2 3 | H (n) | 10-3 2 10-3 3 10-3 89.9 89.9 89.8 H (n) 10 bn ( for is lt ) -5 10 3 -5 4 4 10-3 89.8 2 i0 = 3.18sin ( 2 t + 89.9 ) - 3.18 sin ( 4 t + 89.9 ) + 3.18sin ( 6 t + 89.8 ) -3.18sin ( 8 t + 89.8 ) mA Problem 14.29 Find the average power absorbed by the network shown if V(t) = 50 + 25 cos(377t + 45) + 24 cos(754t - 60) V. 24 30mH -+ V(t) 12 50F Suggested Solution v ( t ) = 60 + 36cos ( 0 t + 45 ) + 24 ( 2 0 t - 60 )V 0 = 377 r / s I 1 1 53.1 + jn12 = = = V 24 + jn11.31 + 12 || 53.1 24 - jn11.31 + 636.6 1911 - n 2 135.7 + jn600 jn 53.1 + jn12 Vdc = 60V V (1) = 36V V ( 2 ) = 24V P = 60 (1.67 ) + P = 127.8W 53.1 I dc 60 = 1.67 A 1911 v1 = 45 I (1) = 1.05 39.1 v = -60 2 I ( 2 ) = 0.77 -76.9 24 ( 0.77 ) 2 cos (16.9 ) 36 (1.05) 2 cos ( 5.9 ) + Problem 14.30 Find the average power absorbed by the 12- resistor in the network of Problem 14.29 if V (t) = 50 = 25 cos(377t - 45) + 12.5 cos(754t + 45) V. Suggested Solution V ( t ) = 50 + 25cos (0 - 45 ) + 12.5cos ( 2 0t + 45 ) 0 = 377 rad s 53.1 637.2 Z = 12 || = jn 53.1 + j12n V0 Z 637.2 = = V Z + 24 + jn11.31 637.2 + 1274.4 - n 2137.7 + j888.6n 1 3 - n ( 0.22 ) + j1.39n 2 V0 ( n ) = V ( n ) = VDC = 50 V (1) = 25 V ( 2 ) = 12.5 I0 (n ) = V0 ( n ) 12 V0DC = 50 3 1 = -45 50 = 1.39 36 V0 (1) = 8.04 -71.6 I 0DC = V0 ( 2 ) = 3.58 -7.67 I 0 (1) = 0.67 -71.6 I 0 ( 2 ) = 0.30 -7.67 2 = 45 (8.04 ) 0.67 + ( 3.58) 0.30 = 26.40 W 50 P = 1.39 + 2 2 3 Problem 14.31 Determine the Fourier transform of the waveform shown. Suggested Solution F ( ) = - 0 f (t )e - j t dt T 0 T A - j t 0 e + e - j t -T 0 j V ( ) = - Ae - j t dt + Ae - j t dt = -T { } V ( ) = A A {1 - e- jt + 1 - e- jt } = 2j {1 - cos ( t )} j Problem 14.32 Derive the Fourier transform for the following functions: (a) f (t) = e-2t cos 4tu (t). (b) f (t) = e-2t sin 4tu (t). Suggested Solution PART A f ( t ) = e -2 t cos ( 4t ) u ( t ) 1 -( 2- j 4+ j )t - 2 + j 4 + j )t dt +e ( 0 e 0 2 0 0 - 2 - j 4 + j )t - 2 + j 4 + j )t 1 e ( 2 + j e( = F ( ) = + 2 2 - j 4 + j 2 + j 4 + j ( 2 + j - j 4 )( 2 + jw + j 4 ) 2 + j F ( ) = 2 ( 2 + j ) + 16 F ( ) = e -2 t cos ( 4t ) e - j t dt = ( ) PART B f ( t ) = e -2 t sin ( 4t ) u ( t ) = F ( ) = 1 - ( 2 - j 4 )t - 2+ j 4 t - e ( ) u (t ) e j2 1 -( 2- j 4+ j )t - 2 + j 4 + j )t e +e ( dt j 2 0 - 2 - j 4 + j t 0 - 2 + j 4 + j )t 1 e( j8 = + 2 + j 4 + j 0 j 2 ( 2 + j - j 4 )( 2 + jw + j 4 ) ( ) ) 1 e( F ( ) = j 2 2 - j 4 + j 4 F ( ) = 2 ( 2 + j ) + 16 Problem 14.33 Show that F f1 ( t ) f 2 ( t ) = 1 2 - F1 ( x )F2 ( - x ) dx Suggested Solution Let G = 1 2 - F1 ( x ) F2 ( - x ) dx x =- x =- x =- F -1 [ G ] = F -1 [ G ] = F -1 [G ] = ( 2 ) 2 1 1 1 F1 ( x ) F1 ( x ) =- F2 ( - x ) e j t d dx let u = - x and du = d F2 ( u ) e jut e jxt dudx ( 2 ) 2 u =- ( 2 ) 2 F1 ( x ) e jxt dx 1 2 u =- F2 ( u ) e jut du = f1 ( t ) f 2 ( t ) F f1 ( t ) f 2 ( t ) = G = - F1 ( x ) F2 ( - x ) dx Problem 14.34 Find the Fourier transform of the function f (t) = e-a|t|. Suggested Solution f (t ) = e - a|t| e( ) F ( ) = a - j a - j t F ( ) = 0 - f ( t ) e - j t dt = e( - 0 0 a - j )t dt + e 0 - ( a - j )t dt e( ) + a + j - - a - j t = 1 1 2a + = 2 a - j a + j a + 2 Problem 14.35 Find the Fourier transform of the function f (t0 = 12e-2|t| cos 4t. Suggested Solution f ( t ) = 12e -2|t| cos 4t From table 14.3, AND, F ( ) = Let g ( t ) = 12e -2|t| F g ( t ) cos 0t = G ( ) = 48 4 +2 1 G ( - 0 ) + G ( + 0 ) 2 1 48 48 24 24 + = + 2 2 2 2 4 + ( - 4 ) 4 + ( + 4 ) 4 + ( - 4 ) 4 + ( + 4 ) 2 Problem 14.36 Determine the output signal vo(t) of a network with input signal vi(t) = 3e-tu(t) and network impulse response h(t) = e-2t u(t). Assume that all initial conditions are zero. Suggested Solution 3 1 + j 1 h ( t ) = e -2 t u ( t ) H ( ) = 2 + j 3 3 3 V0 ( ) = Vi ( ) H ( ) = = - (1 + j )( 2 + j ) 1 + jw 2 + j vi ( t ) = 3e - t u ( t ) Vi ( ) = v0 ( t ) = 3 ( e - t - e -2 t ) u ( t ) V Problem 14.37 The input signal to a network is vi(t) = e-3t u(t). The transfer function of the network is H(j) = 1/(j + 4). Find the output of the network vo(t) if the initial conditions are zero. Suggested Solution vi ( t ) = e -3t u ( t ) Vi ( ) = H ( j ) = 1 4 + j 1 3 + j V0 ( ) = Vi ( ) H ( ) = ( 3 + j )( 4 + j ) 3 = 1 1 - 3 + jw 4 + j v0 ( t ) = ( e -3t - e -4 t ) u ( t ) V Problem 14.38 The input signal for the network shown is vi(t) = 10e-5t (t) V. Determine the total 1- energy content of the output vo (t). 1 + Vi(t) _ 1F + Vo (t) _ Suggested Solution vi ( t ) = 10e -5t u ( t ) Vi ( ) = 1 j H ( ) = 1 1+ j 10 5 + j V0 = 10 (1 + j )( 5 + j ) 100 100 100 = 24 2 - 24 2 | V0 ( ) |2 = 2 2 (1 + )( 25 + ) 1 + w 25 + d 25 d - - 2 - - 25 + w 2 12 1 + w 5 25 -1 1 -1 25 W= tan ( ) - - tan = - 5 = 3 J 12 5 5 - 12 W= 1 2 | V0 ( ) |2 d = Problem 14.39 Use the Fourier transform to find i (t) in the network shown if vi (t) = 2e-t u (t). 1 i(t) V1 (t) -+ 1H 1 Suggested Solution Z = j ||1 = Vi = I= 2 1 + j j 1 + j I= V = Vi j Z = Vi Z +1 1 + 2 j V j = 1 (1 + j )(1/ 2 + j ) 2 1 - 1 1 + j + j 2 -t - t 2 i ( t ) = 2e - e u ( t ) A Problem 14.40 Determine the voltage vo(t) in the circuit shown, using the Fourier transform if vi (t) = 2e-4t u(t). 1 + ii(t) 1 1F Vo (t) _ Suggested Solution V0 1 1 ||1 = = = H ( ) Ii j j + 1 2 2 V0 = 3 - 3 j + 1 j + 4 2 v0 ( t ) = e - t - e -4 t u ( t ) V 3 V0 = ( j + 1)( j + 4 ) 2 Problem 14.41 Compute the 1- energy content of the signal vo (t) in Problem 14.38 in the frequency range from = 2 to = 4 rad/s. Suggested Solution FROM 14.38 4 -1 4 1 -1 tan ( ) 2 - tan = 0.106 J 5 5 2 25 W= 12 Problem 14.42 Determine the 1- energy content of the signal vo(t) in Problem 14.38 in the frequency range from 0 to 1 rad/s. Suggested Solution FROM 14.38 1 -1 1 1 -1 tan ( ) 0 - tan = 0.50 J 5 5 0 25 W= 12 Problem 14FE-1 Given the waveform shown, determine which of the trigonometric Fourier coefficients have zero value, which have nonzero value and why. A T0/4 -T0/4 -T0/2 T0/2 t -A Suggested Solution a0=0 Average value is zero an=0 for all n since this is an odd function bn=0 n-even because of halfwave symmetry bn=finite number for n-odd Problem 14FE-2 Given the waveform shown, describe the type of symmetry and its impact on the trigonometric coefficients in the Fourier series, i.e. a0, an, bn. f(t) -T0/2 -T0/4 T0/4 3T0/4 t Suggested Solution a0=0 Average value is zero an=0 for all n since this is an odd function bn=0 n-even because of halfwave symmetry bn=finite number for n-odd Problem 15.1 Given the two networks shown, find the Y parameters for the circuit in (a) and Z parameters for the circuit in (b). I1 + V1 I2 ZL + V2 - (a) I2 + ZL V2 - I1 + V1 - (b) Suggested Solution Y11 = I1 V1 I1 V2 = V 2=0 (a) Y12 = 1 ZL 1 ZL Y21 = Y22 = I2 V1 I2 V =- V 2=0 1 ZL =0=- V1 = V 1= 0 1 ZL Z11 = (b) Z12 = V1 = ZL I1 I 2 = 0 V1 I2 = ZL I 1= 0 Z 21 = Z 22 = V2 = ZL I1 I 2 = 0 V2 I2 = ZL I 1= 0 Problem 15.2 Find the Y parameters for the two-port network shown . 12 12 12 Suggested Solution Y11 = Y12 = I1 V2 I1 V2 = V 2=0 S 6 S 12 Y21 = Y12 = I2 V1 I2 V2 =- V 2=0 S 12 =- V 1= 0 = V 1= 0 S 6 Problem 15.3 Find the Y parameters for the two-port network shown. 12 6 3 Suggested Solution Y11 = Y12 = I1 V2 I1 V2 = V 2=0 1 S = 12 + 3 6 14 V 1-V 4 = ,V = 3 6 12 7 = I1 , when, V2 = 1, V 1= 0 So, I1 = Y12 = C= -4 / 7 -1 A = 12 21 -1 S 21 = I 2 , When,V 1 = 1, V 2=0 I2 V1 V 1-V 1 = ,V = 12 3 6 7 -V / 7 - A -S = , Y21 = 3 21 21 I S 1 = = Y22 = 2 V2 V 1= 0 3 + 6 12 7 So, I 2 = Problem 15.4 If a 12-A source is connected at the input port of the network shown, find the current in a 4-ohm load resistor. 12 12 12 Suggested Solution 12 12A 12 12 4 I 1/ 6 -1/12 V1 The equations for the two-port are: 1 = ,But the terminal conditions are I 2 -1/12 1/ 6 V2 1 1 12 = V1 - V2 V2 = 16V , I 2 = -4 A -V 6 12 I1 = 12 A, I 2 = 2 , :. 1 5 4 0 = - V1 + V2 12 12 V2 = 16V , I 2 = -4 A Problem 15.5 Find the Y parameters for the two-port network shown. I1 Z2 V1 Z1 I2 V1 V2 Suggested Solution Y11 = Y12 = I1 V1 I1 V2 = V 2=0 1 21 Y22 = =0 V 1= 0 I2 V2 = V 1= 0 1 Z2 Y21 = I2 V1 = V 2=0 8 Z2 Problem 15.6 Determine the Y parameters for the network shown. 3 + V1 6 6 V2 Suggested Solution Y11 = Y12 = I1 V1 I1 V2 = V 2=0 1 1 = S 3+ 6 6 6 V2 - V 1 V = ,V = 6 6 3 4 = I1 , WHEN , V2 = 1, V 1= 0 THEN, I1 = Y21 = I2 V1 I2 V2 -V / 4 -1 -1 S = A and Y12 = 3 12 12 = I 2 , WHEN , V1 = 1, V 1-V 1 -V / 2 - A -S = ,V = = , Y21 = THEN, I 2 = 3 6 6 2 6 21 12 V 2=0 Y22 = = V 1= 0 S 1 = 6+3 6 8 Problem 15.7 Determine the admittance parameters for the network shown. C2 I1 + V1 + R C1 gV1 C3 V2 - Suggested Solution Y11 = jw(C1 + C2 ), Y21 = g - jwC2 Y12 = - jwC2 , Y22 = 1 + jw(C2 + C3 ) R Problem 15.8 Find the Z parameters of the two port network shown. I1 + V1 21 I2 + V2 10.5 - 42 Suggested Solution I1 + 42 21 + V2 10.5 Z11 = Z 21 = V2 = V1 I1 V2 I1 = I 2 =0 (42)(31.5) = 18 42 + 31.5 I 2 =0 10.5V1 10.5 = (18 I1 ) 21 + 10.5 31.5 V2 = 6 I1 Z 21 = 6 Z 22 = Z12 = V2 I2 V1 I2 = I1 = 0 (10.5)(63) = 9 42 + 31.5 (42 / 63)(V2 ) 2 = Z 22 = 6 3 I2 = I1 = 0 Problem 15.9 Find Z parameters of the network in Problem 15.5. Suggested Solution Z11 = Z 21 = V1 I1 V2 I1 = Z1 , Z12 = I 2 =0 V1 I2 =0 I1 = 0 = - Z1 , Z 22 = I2 =0 V2 I2 = Z2 I1 = 0 Problem 15.10 Determine the Z parameters for the two port network shown. R2 R3 V1 R1 V1 V2 Suggested Solution R2 V1 R1 V1 V2 I2 = 0 I1 = V1 + V1 R1 1 + )V1 R1 I1 = ( 1 + R1 R1 )V1 Z11 = I1 = ( 1 + R1 R1 R1 ) I1 V2 = V1 - V1 R2 V2 = V1 (1 - R2 ) = (1 - R2 )( 1 + R1 Z 21 = R1 (1 - R2 ) 1 + R1 Z 21 = R1 (1 - R2 ) 1 + R1 R2 R3 V1 R1 V1 V2 I1 = 0 I2 = V1 + V1 R1 1 1 + R1 R1 I2 + )V1 = ( )V1 V1 = R1 R1 1 + R1 R1 1 + R1 V1 R ) R2 + V1 = I 2 R3 + V1 ( 2 + 1) R1 R1 I2 = ( Z12 = V2 = I 2 R3 + ( Problem 15.11 Find the Z parameters for the two port shown. Determine the voltage gain of the entire circuit with a 4kohm load attached. 60I1 500 + +- V1 I1 50 20K 4K V2 - Suggested Solution Z11 = 550, Z 21 V2 = 50 I - 60(20 K ) I1 Z 21 = 1200 K Z12 = 50, Z 22 = 20 K + 50 = 20 K -V 0 , V2 = V 0 4K V1 = 550 I1 + 50 I 2 I2 = V2 = -1200 KI1 + 20 KI 2 V1 = 550 - 50 V0 4K V0 = -1200 KI1 - 20 K VC 4K 550 I1 = V1 + 0.0125V0 1200 KI1 = -V0 - 5VC 1200 KI1 = 2181.82V1 + 2727V0 1200 KI1 = -6V0 -6V0 = 2181.82V1 + 27.27V0 -33.27V0 = 2181.82V1 V0 = -65.6 V1 V0 = -65.6 V1 Problem 15.12 Find the Z parameters for the two port network shown. I1 + 1K V1 (3)10-4 V2 40I2 20S V2 + I2 Suggested Solution I1 I2 V1 1K 40I1 V2 20V 3E-4V2 h11 = 1K , h21 = 40, h12 = 3E - v, h22 = 20 H = h11h22 - h12 h21 = 8E - 3 Z11 = Z12 = Z 21 = Z 22 = h 8 E - 3 = = 400 h22 2 E - 5 h12 3E - 4 = = 15 h22 2 E - 5 -h21 -40 = = -2 E - 6 h22 20 1 = 50 K h22 Problem 15.13 Find the voltage gain of the two port network in Problem 15.12 with a 12k-ohm load. Suggested Solution V1 = 400 I1 + 15 I 2 V2 = -2 X 10-6 I1 + 50 X 103 I 2 V1 = 400 I1 - 15V2 12 K V2 = -2 X 106 I1 - 50 X 103 V2 12 X 103 6 5000V1 = 2 X 10 - 6.25V2 5000V1 + V2 = -10.42V2 -11.42 = 5000V1 V2 = -438 V1 V2 = -438 V1 V2 = -2 X 106 I1 - 4.17V2 Problem 15.14 Find the input impendence of the network in Problem 15.12 Suggested Solution From problem 15.13 solution: 5000V1 = 2 X 106 I1 - 6.25(-4.38V1 ) (5000 - 2737.5)V1 = 2 X 106 I1 Z in = V1 2 X 106 = = 884 I1 2262.5 Z in = 884 Problem 15.15 Find the Z parameters of the two port network shown. I1 + V1 - I2 + jwL1 jwL2 V2 - Suggested Solution I1 + V1 I2 + V2 - V1 = I1 j L1 + I1nj V2 = I1nj + I 2 j L2 Z11 = j L1 , Z12 = j n, Z 21 = j n, Z 22 = j L2 Z11 = j L1 , Z12 = j n, Z 21 = j n, Z 22 = j L2 Problem 15.16 Determine the Z parameters of the two port network shown. R2 1:n R1 Suggested Solution R2 I1 + V1 + R1 1:n I2 + V2 - V1 = nI 2 R1 Z12 = nR1 Z11 = R1 V2 = nR1 I1 Z 21 = nR1 Z 22 = n 2 ( R1 + R2 ) Problem 15.17 Compute the hybrid parameters for the network shown. I1 + V1 I2 ZL + V2 - (a) I1 + V1 ZL I2 + V2 - (b) Suggested Solution 21 42 10.5 h11 = h12 = h21 = V1 I1 V1 V2 I1 I2 = 21 || 42 = 14 V 2=0 = I 1= 0 42 2 = 42 + 21 3 -42 -2 = 42 + 21 3 1 1 = S 10.5 || (21 + 42) 9 = V 2=0 I h22 = 2 V2 = I 1= 0 Problem 15.18 Consider the network shown. The two port network is a hybrid model of a basic transistor . Determine the voltage gain of entire network, V2/Vs, if a source Vs with internal resistance R1 is applied at the input to the two port and a load RL is connected. R1 I1 h11 I2 + h11V2 VS V1 h21I 1/h22 + V11 = h11 I1 + h12V2 I 2 = h21 I1 + h22V2 Also V2 = - RL I 2 orI 2 = -V2 andV1 = VS - R1 I1 RL VS - R1 I1 = h11 I1 + h12V2 -V2 = h21 I1 + h22V2 RL VS = ( R1 + h11 ) I1 + h12V2 h21 I1 = -(h22 + 1 )V2 RL I1 = -(h22 + VS = [h12 - VS = [h12 - VS = [ V2 h 21RL = VS h12 h21 RL - (1 + h22 RL )( R1 + h11 ) +- V2 RL Suggested Solution 1 1 )V2 RL h21 1 1 (h22 + )( R1 + h11 )]V2 h21 RL (1 + h22 RL )( R1 + h11 ) ]V2 h21 RL h12 h21 RL - (1 + h22 RL )( R1 + h11 ) ]V2 h21 RL Problem 15.19 Find the hybrid parameters for the network shown. 12 12 12 Suggested Solution h11 = h12 = h21 = h22 = V1 I1 V1 V2 I1 I2 I2 V2 = 12 ||12 = 6 V 2=0 = I 1= 0 12 = 0.5 12 + 12 -12 = -0.5 12 + 12 1 1 = S 12 || (12 + 12) 8 = V 2=0 = I 1= 0 Problem 15.20 Determine the hybrid parameters for the network shown. I1 I1 + V1 - R1 1/jwC I2 + R2 V2 - Suggested Solution 1/jwC + V1 R2 I2 + V2 - I1 = 0 V1 = R2V2 1 R2 + j C j CR2 h12 = 1 + j CR2 I2 = V2 = j CV2 j C h22 = 1 + j CR2 1 + j CR2 1 R2 + j V2 = 0 R2 ( I1 + I 2 ) + 1 ( I1 + I 2 ) = 0 j C 1 ( R2 + ) I1 + ( R2 + )I2 = 0 j C j C ( + j R2 C ) I1 = -(1 + j R2 C ) I 2 I2 = [ -( + j R2 C ) ]I1 1 + j R2 C h21 = -( + j R2 C ) 1 + j R2 C R2 ( + j R2 C ) ]I1 1 + j R2 C 2 V1 = ( R1 + R2 ) I1 + R2 I 2 = [ R1 + R2 - 2 h11 = R1 j R1 R2 C + R2 + j R2 C - R2 - j R2 C 1 + j R2 C h11 = R1 + R2 (1 - ) + j R1 R2 C 1 + j R2 C Problem 15.21 Find the ABCD parameters for the networks shown. I1 ZL + V1 + V2 I2 + I1 V1 ZL I2 + V2 - Suggested Solution V1 V2 A= B= C= D= =1 I 2=0 I1 -I2 I1 V2 = Z2 V 2=0 =0 I 2=0 I1 -I2 V1 V2 =1 V 2=0 A= =1 I 2=0 B=0 1 C= Z2 D =1 Problem 15.22 Find the transmission parameters for the network shown. -j1 1 Suggested Solution A =1 B = -1 j C = 1s 1( I1 ) = -I2 D = 1 - j 1- j Problem 15.23 Find the transmission parameters for the network shown. I1 + V1 6 12 3 I2 + V2 - Suggested Solution A= C= B= D= V1 I1 I1 V2 =3 I 2=0 = I 2=0 I1 1 = S 6 I1 6 -V1 I2 - I1 I2 V 2=0 = V 2=0 - I1 3 = 6 I1 2 9 I1 = 1.5I 2 V1 = 12 I1 + 6( I1 + I 2 ) = 18I1 + 6 I 2 = -21I 2 B= -21I 2 = 21 I2 Problem 15.24 Find ABCD parameters for the network shown. I1 V1 Z2 I2 V2 Z1 V1 Suggested Solution V1 V2 V1 -1 = V1 = V 2=0 A= B= = I 2=0 V1 -I2 I1 V2 -Z2 V1 = -(V2 + V1 ) Z2 C= D= = I 2=0 I1 -1 = - V2 Z1 = -Z2 I1 = - V1 Z1 Z2 I1 -I2 V 2=0 Problem 15.25 Determine the transmission parameters for the network shown. I1 + V1 R2 R1 V1 + V2 - Suggested Solution I1 + V1 R2 R1 R3 I1 I2 + V2 - I2 = 0 V2 = ( + R2 ) I1 C= 1 + R2 R1 + R1 + R2 V1 = ( R1 + R2 ) I1 A = V2 = 0 I1 + ( R3 + R2 ) I 2 + R2 I1 = 0 ( + R2 ) I1 = -( R3 + R2 ) I 2 I1 = -( R3 + R2 R + R2 )I2 D = 3 + R2 + R2 R3 + R2 )( R1 + R2 ) I 2 + R2 I 2 + R2 V1 = ( R1 + R2 ) I1 + R2 I 2 = -( V 1 = -[ B = ( R3 + R2 )( R1 + R2 ) - R2 ]I 2 + R2 2 2 R3 R1 + R3 R2 + R2 R1 + R2 - R2 - R2 R R + R3 R2 + R2 R1 - R2 = 3 1 + R2 + R2 R3 R1 + R3 R2 + R2 R1 - R2 + R2 Problem 15.26 Find the transmission parameters for the circuit shown. jwL1 jwL2 Suggested Solution ForI 2 = 0, I1 = AND V2 = j MI1 V2 = Since V2 = 0, V1 = j L1 I1 + j MI 2 0 = j MI1 + j L2 I 2 L2 (- I 2 ) M L V1 = - j L1 ( 1 )(- I 2 ) - j M (- I 2 ) M V1 LL -M2 B = = j[ 1 2 ] M -I2 I1 = D= I1 L = 2 -I2 M L1 M 1 j M j L1 L2 - M 2 M L2 M j MV1 V L ,A= 1 = 1 j L1 V2 M V1 j L1 A B = C D Problem 15.27 Find the transmission parameters for two port network and then find I0 using the terminal conditions. -j2 V1200 4 6300V 1 2 0 1 2 - j1 j (8 - 1) 4 1 4 - 2 j 1 0 THEN 1 2 2 0 1 - j1 V1 = 1200 V V2 = 6300 V -I2 = I0 V1 = AV2 + B (- I 2 ) 1200 = (2 - 2 j )6300 + (12 - 5 j ) I 0 Solving I 0 = -0.48 - 22.350 A j (8 - 1) 1 4 - 2 j 2 - 2 j 12 - 5 j = 4 0 1 -1 j 2-4j +- +- j2 j4 Suggested Solution Problem 15.28 Following are the hybrid parameters for a network. h11 h 21 11 h12 5 = h22 -2 5 2 5 1 5 Determine the Y parameters of the network Suggested Solution Using Table 15.1 y11 = y12 = y21 = y22 = 1 5s = h11 11 - h12 -2 s = h11 11 h21 -2 s = h11 11 H 3s = h11 11 Problem 15.29 If the Y parameters for a network are known to be y11 y 21 5 y12 11 = y22 -2 11 -2 11 3 11 Find the Z parameters. Suggested Solution 5 Y = 11 -2 11 Y = Then y22 3 /11 = = 3 Y 1/11 -y 2 /11 = 2 Z12 = 12 = Y 1/11 -y Z12 = 21 = 2 Y y Z 22 = 11 = 5 Y Z11 = -2 11 3 11 3x5 (-2)(-2) 1 - = y11 y22 - y21 y12 = |2| |2| 11 Problem 15.30 Find the Z parameters in terms of the ABCD parameters Suggested Solution The ABCD Parameters equations are: V1 = AV1 - BI 2 I1 = CV2 - DI 2 V2 = 1 [ I1 + DI 2 ] C I DI AI AD V1 = A[ 1 + 2 ] - BI 2 = 1 + ( - B) I 2 C C C C A C AD - BC Z12 = C 1 Z 21 = C D Z 22 = C Z11 = Problem 15.31 Find the Hybrid parameters in terms of the Impendance parameters Suggested Solution V1 = Z11 I1 + Z12 I 2 V2 = Z 21 I1 + Z 22 I 2 I2 = - Z 21 V I1 + 2 Z 22 Z 22 Substitute this value into the first equation yields V1 = Z11 Z 22 - Z12 Z 21 Z I1 + 12 V2 Z 22 Z 22 h11 = h12 = h21 = h22 = Z Z 22 Z12 Z 22 - Z 21 Z 22 1 Z 22 Where Z = Z11 Z 22 - Z12 Z 21 Problem 15.32 Find the Y parameters of the network shown, by considering the network to be a parallel connection of two port networks. 21 42 Nb 10.5 Na Suggested Solution ForNA 1 YA = 21 -1 21 NB 1 0 42 YB = 1 0 10.5 1 14 Y1 = YA + YB = -1 21 -1 21 1 7 -1 21 1 21 Problem 15.33 Find the Y parameters of the two port shown. Find the input admittance of the network when the capacitor is connected to the output port j2 2 Yin 2 2 -j2 Suggested Solution 2 2 2 j2 1 j2 Y1 = 1 - j 2 1 Y2 = -1 2 - 1 j2 1 j2 -1 2 1 -1 1 - 2 j2 1 1+ j2 1 1+ j2 YT = -1 1 2 - j2 -1 1 1 1 ( - )(- - ) Y12Y21 2 2j 2 2j YIN = Y11 - = 1 - 0.5 J - [ ] Y22YL 1 - 0.5 j + 0.5 j = 1s Problem 15.34 Find the Y parameters for the network shown. Z1 Z2 Use a parallel interconnection Z1 Z2 Z1 Suggested Solution Z2 Z2 Z2 y11 = 1 2 Z1 1 2 Z1 1 2Z 1 y12 = - y21 = - y22 = 1 2Z1 1 2Z 2 1 2Z 2 1 2Z 2 1 2Z 2 1 1 2Z + 2Z 2 = 1 -1 1 2Z + 2Z 1 2 1 -1 + 2 Z1 2Z 2 1 1 + 2 Z1 2Z 2 y11 = y12 = y21 = y22 = y11 y 21 y12 y22 TOTAL Problem 15.35 Determine the Y parameters for the network shown. Z2 Z1 Z4 Z5 Z6 Z3 Suggested Solution For one T network Y11 = Z + Z3 1 = 2 Z1 + Z 2 || Z 3 Z123 WHERE Z123 = Z1 Z 2 + Z 2 Z 3 + Z1 Z 3 Y12 = Y21 = Y22 = -Z3 Z123 Z1 + Z 3 Z123 Then for the total network Z 2 + Z3 Z5 + Z6 + Z Z 456 123 YT = -Z3 Z - 6 Z123 Z 456 Z 2 + Z3 Z5 + Z6 + Z123 Z 456 -Z3 Z - 6 Z123 Z 456 Problem 15.36 Find the Z parameters of the network shown by considering the circuit to be a series connection of two port networks. 12 Na 3 Nb 6 Suggested Solution ForNA 12 0 ZA = 0 3 ForNB 6 6 ZB = 6 6 18 6 ZT = Z A + Z B = 6 9 Problem 15.37 Find the transmission parameters of the network shown by considering the circuit to be a cascade connection of three two port networks. 12 3 6 Na Nb Nc Suggested Solution ForN a 1 12 0 1 ForN b 1 0 1 1 6 ForN c 1 3 0 1 1 0 3 1 12 1 3 = 1 ABCDT = 1 1 0 1 0 1 6 6 21 3 2 Problem 15.38 Find the ABCD parameters for the circuit shown. 1 1H 1F Suggested Solution For the Resistor A B 1 1 C D = 0 1 For the Inductor A B 1 1 C D = j 1 For the Capacitor A B 1 C D = 0 j 1 AB 1 + j 1 + j - 2 CD = 1- 2 j Problem 15.39 Find the transmission parameters for the two port network shown. -j1 1 -j 4 -j3 Suggested Solution A =1 C =1 B=-j D=-j A = -1 C =-j D=3 B = j11 A B 1 - j 4 C D = 1 1 - j - j j8 j11 3 = 3 3 - j1 3 + j8 Problem 15.40 Find the transmission parameters of the two port and then use the terminal conditions to find I. 1 -j1 1:2 2 - j1 +Suggested Solution +- 600V (4-j4)V A B 1 1 - j 0.5 0 1 2 - j1 0.5 3 - J 2.5 = C D = 0 1 0 2 0 1 0 2 V1 = 0.5V2 + (3 - j 2.5)(- I 2 ) I1 = 2(- I 2 ) V1 = 60 V2 - 4 - j 4 I0 = -I2 60 = 0.5(4 - j 4) + (3 - j 2.5)( I 0 ) I0 = 4 + j2 = 1.1466.380 A 3 - j 2.5 Problem 15FE-1 A two port network is know to have the following parameters Y11 = 1/14s Y12 = Y21=-1/21s Y22 = 1/7s 2A 2A 2-PORF 7 +Suggested Solution 12V If 2 A current source is connected to the input terminals find the voltage across this source. I1 = Y11V1 + Y12V2 I 2 = Y21V1 + Y22V2 Z= 1 1 V1 - V2 14 21 -1 1 V1 + V2 21 7 0= 1 1 V1 1 7 21 2 V = 1 0 2 Y 1 21 14 WHERE 1 1 1 1 3.5 Y = ( )( ) - ( )( ) = 14 7 21 21 441 CHECK Y22 1/ 3 = = 18 Y 3.5 / 441 V1 = (2)( Z11 ) = 36V Z11 = V1 = 36V Problem 15FE-2 Find the Thevenin equivalent circuit at the output terminals of the network. of Problem 15FE1. Suggested Solution See previous solution. V2 = V0C = 12V ZTA = 1 = 7 Y22 The Thevenin equivalent circuit is Problem 16.1 A diode has Is = 10(-15) A and n= 1. Using the diode equation find Id for Vd = 0V, 0.25V, 0.5V, 0.75V. Suggested Solution I D = I S [eVD / nVT - 1] = 10-15 [e39VD - 1] Vd(V) 0 0.25 0.50 0.75 Id(A) 0 1.7x10-11 2.9x10-7 5.0x10-3 Problem 16.2 Plot an Id-Vd curve for the diode in Problem 16.1 over the Vd range 0 t +0.65V. On the same graph, add curves for constant voltage and piece wise linear models for current between 10 to 100 uA. What are your estimates for Von Vf and Rd. Suggested Solution At this level of study, selecting values for Von, Vf and Rd is essentially a matter of preference. From the plot, the following values produce reasonable fits to the diode curve. Constant voltage model = Von = 0.625V Piecewise linear model Vf = 0.585V and Rd = 740 ohm. Recall Rd is the reciprocal of slope of piecewise linear model I-V curve. Problem 16.3 Repeat 16.1 using the constant voltage model with Von = 0.6V Suggested Solution a. Diode is Forward biased V1= 10 0.6 = 9.4V 10V 40 + V1 0.6V I b. Diode is Reverse biased V1= (0.1)(100) = 10V + 0.1A 0A 100 V1 - c. Diode is Forward biased 4=0.6+V1 V1=3.4V d. Diode is Reverse biased I = (4/3k+1k) = 1mA V1=-I(1k)=-1V C. 0.6V I 4V I1 3K + 1K I2 V1 - 0A D. I 4V 1K + V1 3K Problem 16.4 For the circuit shown find V0 using a)the ideal model of diode, b) the constant voltage model with Von = 0.7 V, and c) the piecewise linear model with Vf=0.6V and Rd=10ohm. Give your self an attaboy if you can use the mathematical diode equation with Is=1e-15A, n=1 and R = 1Kohm. R + 10V 2K VO 6V Suggested Solution a. Assume Diode is Reverse biased R + VO + 10V 2K VO 6V - 2R ] = 6.67V 2R + R VD = V0 - 6 = 0.67V V0 = 10[ Since VD > 0, is our assumption was wrong. Try again with forward bias b. Assume reverse bias from a) VD = 0.67V, since VD < 0.7V our assumption is good. c. Assume reverse bias from a) VD = 0.67V. Since VD > 0.6V, our assumption is bad. Try forward biased. By superposition V0 = 10[ 2000 ||10 1000 || 2000 ] + 6.6[ ] = 6.6V (2000 ||10) + 1000 (1000 || 2000) + 10 R + 10V 2K VO - 10 0.6V 6V To use the diode equation, start by applying KCL at Vo R + 10V 2K VO 6V 10 - V 0 V0 = + I0 2R R aLSO V0 = VD + 6 Thus 10 - VD - 6 VD + 6 1 - 1.5VD - = ID = 2R R R From the diode equation VD = nVT ln[ and 1 - (3 / 2)nVT ln[ ID = R ID + 1] IS ID + 1] IS can not be solved in closed form. We must iterate. Procedure: 1. Guess a value for Id 2. Substitute in right side of equation 3. Get answer for Id 4. Put that value back in right side 5. When the difference between successive calculation is small, we have an answer. This procedure is easy to automate in Excel. The excel values are shown Id (mA) Guess 1.00E-01 2.58 E-02 7..79 E-02 3.54 E-02 6.57 E-02 4.20 E-02 5.92 E-02 4.60 E-02 5.57 E-02 4.83 E-02 5.38 E-02 4.97 E-02 5.27 E-02 5.04 E-02 5.22 E-02 5.09 E-02 5.18 E-02 5.11 E-02 5.16 E-02 5.12 E-02 5.15 E-02 5.13 E-02 5.15 E-02 5.14 E-02 5.15 E-02 5.14 E-02 Id (mA) Result 2.58E-02 7.79 E-02 3.54 E-02 6.57 E-02 4.20 E-02 5.92 E-02 4.60 E-02 5.57 E-02 4.83 E-02 5.38 E-02 4.97 E-02 5.27 E-02 5.04 E-02 5.22 E-02 5.09 E-02 5.18 E-02 5.11 E-02 5.16 E-02 5.12 E-02 5.15 E-02 5.13 E-02 5.15 E-02 5.14 E-02 5.15 E-02 5.14 E-02 5.14 E-02 Vd(V) 0.438 0.466 0.446 0.462 0.450 0.459 0.452 0.457 0.454 0.456 0.454 0.456 0.455 0.456 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 These are the final results. Problem 16.5 The circuit in fig, is called half wave rectifier. Plot Vin and Vout verses time for 3 cycles of input voltage using a)the ideal diode model, b)constant voltage model with Von = 1.0V + VO Vin(t)5Sin(wt) - Suggested Solution Using the ideal model procedures the circuits and b for forward and reverse bias conditions respectively. When Vin(t) is positive, the diode is forward biased and V0(t) = 5sin(wt). Alternatively, when the diode is reversed biased Vin(t) is negative and V0(t)= 0. The circuits, c and d model forward and reverse bias conditions respectively for the constant voltage model. When Vin(t) is greater than Von, the diode is forward biased and V0 (t ) = 5sin( t ) - 1 Else V0(t) = 0, Plot V0 for both models. + VO Vin(t)5Sin(wt) - + Vin(t) VO - + Vin(t) VO - + Vin(t) VO - Problem 16.6 The MOSFET in fig, has Gm = 1mS, and Rds = 100kohm. Draw the small signal circuit and find out the voltage gain Vo/Vin. RD 10K + VIN VGS VO VDD - Suggested Solution VGS = VIN V0 = - g M vgs [rds || RD ] VO = - g m [rds || RD ] = -(1m)[100 K ||10 K ] = -9.1 VGS G + VIN Vgs S gmVgs rds RD VO D + V0 = -9.1 VIN Problem 16.7 For the amplifier in problem 16.6 plot the gain V0/Vin verses Rd for Rd between 1Kohm and 1Mohm. Why doesn't the gain increase linearly with Rd. Suggested Solution The small-signal model resistance Rds is in parallel with Rd. For small values of Rd, it will dominate the parallel combination and the gain will be nearly linear with Rd. As Rd increases, the affect of Rds becomes stronger until it dominates the combination and thus the gain. Problem 16.8 The amplifier in fig is a comman drain configuration. Find Vo/Vin and the output resistance, Ro in terms of Gm,Rds,Rs. If Gm = 10mS and Rds=1Kohm. What is the value of Ro. What possible utility could this configuration have? G + VIN Vgs S RS VO RO gmVgs rds D Suggested Solution VGS = VIN - VO GM VGS = RS || RDS GM (VIN - VO ) = V0 RDS || RS V GM 1 ] 0 = 1 RDS || RS VI N G + M RDS || RS V0 V + 0 RDS RS GM VIN = V0 [GM + + Vgs gmVgs - S ix rds Rs VX A= V0 GM ( RS || RDS ) = VI N GM ( RS || RDS ) + 1 Gain is positive and less than 1. Rout : Eliminate Vin and apply a test source Vx at point of interest. RO = VX IX VX = VGS = -VGS ByKCL I X + GM VGS = I X = VX [GM + VX RS || RDS 1 ] RS || RDS R0 = REG VX = I X GM REG + 1 Where REG = RS || RDS + Vgs gmVgs - S ix rds Rs VX If GmReg >>1, then Ro= 1/Gm, which is generally small. For Gm=10mS, Rds = 1k-ohm and Rs = 1 K-ohm R0 = 83.3 ohm The comman drain configuration has a voltage near 1 and a low output resistance. It can serve as a voltage buffer much like the unity gain buffer op-amplifier. Problem 16.9 For MOSFET in fig, Gm= 2mS and assume Rds is infinity. Using superposition find Vo/(V2-V1) RD 20K V1 Vgs V2 VDD Suggested Solution V1 VGS gmVgs Rd G Vgs gmVgs RD VO2 V2 ForV 1 VGS = V1 V01 = -GM RD V01 = -V1GM RD = -40V1 ForV 2 VGS = -V2 GM VGS + V02 =0 RD V02 = V2GM RD = 40V2 V0 = V01 + V02 = 40(V2 - V1 ) V0 = 40 V2 - V1 V0 = 40 V2 - V1 Problem 16.10 The amplifier in Fig is called as a differential amplifier. MOSFETs M1 and M2 are identical devices with Gm values and V1 and V2 small signal inputs. Find Gain V0/V2-V1 in terms of Gm and Rd. Assume Rds is infinite. VDD RD 50K VO V2 V1 VSS TAIL Suggested Solution Use Superposition V2 = VGS 2 - VGS 1 GM 1 = GM 2 = GM GM VGS 2 + GM VGS 1 = 0 VGS 2 = -VGS 1 V2 = -VGS 1 2 G R G R V02 = -GM 1VGS 1 RD = M D V2 V02 = V2 ( M D ) 2 2 V2 = 2VGS 2 VGS 2 = G2 D2 + Vgs2 V2 gm2Vgs2 - RD + VO2 gm1Vgs1 - Vgs1 S2 V1 = VGS 1 - VGS 2 GM 1 = GM 2 = GM GM VGS 2 + GM VGS 1 = 0 VGS 2 = -VGS 1 V01 = -GM VGS 1 RD = -V 1( GM RD ) 2 GM RD (V2 - V1 ) = V0 2 V0 = V01 + V02 = GM RD (V2 - V1 ) = V0 2 Problem 16.11 Find an expression for the complex gain A(jw) = Vo/Vin for the amplifier in terms of Gm, Rds, Rd and C. Find expression for the gain at dc and pole zero location. For Gm=500uS and Rds=400kohm, graph bode plot over frequency range 1Hz-100KHz. RD 100K VDD Vin VGS VO Suggested Solution 1/jwc G + Vin S Vgs gmVgs R VO R = RDS || RD = 80k VGS = VIN KCL : (VIN - V0 ) j C = GM VGS + V0 1 VIN ( j C - GM ) = V0 ( j C + ) R R j C - 1) V0 GM A( j ) = = (GM R ) VIN ( j CR + 1) ( Problem 16.12 The MOSFET in fig, acts as a switch to turn red LED on or off. On-resistance of MOSFET is given as Ron=2/(Vgs-Vth) where Vth=1V. A LED is nothing more than a diode which glows when forward biased. The desired operating point for the LED is 2.0V at 20mA. Find Ron and the required value of R1. D + 2V ID R1 5V 5V 20mA Suggested Solution VON 2V R1 5V 20mA RON Equivalent Circuit KCL: 5 = 2 + 0.02( R1 + RON ) RON 2 2 = = = 0.5 VGS - VT 5 - 1 RON = 0.5 R1 = 149.5 Problem 16.13 The MOSFET in fig are used to turn on a green ar red LED. Vgs is switched to either 0 or 3 V. Which voltage level turns on which LED? Use the constant voltage model for diodes with Von = 2.1 V model the ON resistance of the MOSFET as Ron=17/(Vgs-0.75). Find the values of R1 such that diode current never exceeds 10mA. Suggested Solution OFF 3V R1 2.1V RON 0.9V RON 0.9V OFF 0.01A 2.1V VGS = 3V , VGS 1 = 3V , VGS 2 = 0V , M 1 = ON , M 2 = OFF RON = KVL : 3 = 2.1 + 0.01( R1 + RON ) R1 82.4 17 = 7.55 3 - 0.75 VGS = 0V , VGS 1 = 0V , VGS 2 = -3V , M 1 = OFF , M 2 = ON RON = 7.55 KVL : 3 = 0.01( R1 + RON ) + 2.1 R1 82.4 RON = 7.55 R1 82.4 Problem 16.14 The MOSFETin fig, is used as a switch to turn on motor. It is desired that circuit operate at an efficiency at least 95% where efficiency is defined as Motor Power (W) n = --------------------------Total Power (W) Find the maximum allowable value of Ron. Suggested Solution 12V 0.5A RON 1 Total Power = 12( ) = 6W 2 Power in FET = PFET = I 2 R ON = R ON 4 RON 1.2 Total Power = PMOTOR + PFET 6W = R 6 - PFET > 0.95 PFET < 0.3W = ON 6 4 Problem 16.15 Fig shows an electric car motor can be controlled by a MOSFET. For the conditions shown find the efficiency of the circuit and MOSFET power when Ron = 5mohm. Efficiency is defined as Motor Power (W) n = --------------------------Total Power (W) Find the values of Ron required for 99% efficiency. As mentioned in the text, MOSFETs, with Ron values as low as 2mohm are commercially available (circa 2000). How can you produce the required Ron value. Suggested Solution 12V 0.5A RON Total Power = PMOTOR + PFET = 50(600) + I 2 RON = 30000 + 6002 (5M ) = 31800W = for 30000 = 94.3% 31800 = 99% = 30k PFET = 303W 30k + PFET RON = PFET PFET = = 0.84m I2 6002 = 30000 = 94.3% 31800 RON = Less than currently available 2m units. Solution: PFET PFET = = 0.84m I2 6002 Use several identical devices in parallel. If 5 are used with Ron = 4 m, then the combined RON = 4 m = 0.8m 5 Problem 16FE-1 For the circuit sketch voltage Vo verses Vin, over the range 10V to 10V. What are the bias conditions for the diodes at Vin=-10, 0 and +10 V. Assume diodes are ideal. 200 VIN 300 D1 D2 VO 2V 6V Suggested Solution When Vin is greater than 6 V. D1 is forward biased and D2 is reversed biased. The circuit reduces to that where VIN = 500 I + 6 V0 = 300 I + 6 V0 = 6 + 0.6(VIN - 6) When Vin is less than 2 V, D2 is forward biased and D1 is reverse biased. Under these conditions V0= 2V For Vin between 2 V and 6 V, both diodes are reverse biased, no flows anywhere. And Vo= Vin. A plot is shown Problem 16FE-2 The MOSFET in the circuit acts as a switch with an resistance of 1ohm and diode is ideal. a. Assume the MOSFET is on the network ahs reached steady state. What is the value of Inductor current. b. With the current operating as described in a) the MOSFET is turned off at time 0. Find expression for iL(t) and VL(t), for t>=0. Suggested Solution A. The figure shown models the circuit when MOSFET is ON. In steady state, the inductor voltage is 0, yielding, 20=5IL IL = 4 A B. The circuit models transients in the circuit when the MOSFET turns OFF. The initial condition of the inductor current is found a) to be 4A. When MOSFET turns off, the inductor current cannot go to 0 instantaneously. It flows through diode and discharges inductor. The inductor current and voltage can be given as iL (t ) = K1 + K 2 e -t VL (t ) = K 3 + K 4 e -t where iL (0) = 4 A and iL() = 0 A. This K1 = 0 and K2 = 4A. Similarly, Vl(0) = -Il(0)Rl = -16Vand VL()=0V. Thus K3 = 0 and K4 = -16V. The time constant =25micro sec. The resulting expressin for current and voltage are iL (t ) = 4e -40000t VL (t ) = -16e -40000t iL(t) L 100H RL 4 + L 100H + VL(t) RL 4 VL(t) RON 1 ...
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This note was uploaded on 05/01/2010 for the course ECE 300 taught by Professor Watson during the Spring '09 term at Carson-Newman.

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