EE1002_CG1108_Tutorial_solutions___dtd_5th_sep_2011

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Unformatted text preview: Tutorial 1 1. Use Kirchoff’s current law to determine the unknown currents in the circuit of Figure 1Figure 1. Assume that I0=‐2A, I1=‐4A, Is=8A and Vs=12V. R4 I1 I0 R0 I2 R1 R2 VS IS I3 R3 Figure 1 Analysis: KCL can be used to find one unknown current if all the other currents into and out of a node are known. Solution: Applying KCL at node (a), sum of currents leaving node (a)=0 I 0 I1 I 2 0 I 2 ( I 0 I1 ) (2 4) 6 A Applying KCL at super node (b), sum of currents leaving = 0 I 0 I s I1 I 3 0 I 3 I 0 I s I1 2 4 8 2 A 2. Apply KVL to find the voltages v1 and v2 in Figure 2. EE1002/CG1108 Tutorials 1 to 5 Solutions 1 3V 10V v2 v1 Figure 2 Analysis: According to KVL, the algebraic sum of voltage rises (or voltage drops) around a closed loop is zero. Solution: Applying sum of voltage rises around the loop ABCDA, we get 5 3 v2 0 v2 2V Applying sum of voltage falls (just to show as an alternate way) around the loop ABCEDA: 5 3 10 v1 0 v1 12V 3. For the circuit given in Figure 3, a. Determine which components are absorbing power and which are delivering power. b. Is conservation of power satisfied? 3V 2A B A 5V C D 3A E 10V Figure 3 EE1002/CG1108 Tutorials 1 to 5 Solutions 2 Analysis: We can do circuit analysis to find the voltage and current associated with all the elements in the circuit. Then we can find the power delivered each element. We can label the unknown voltages and currents for the circuit following passive sign convention (current entering into the positive reference terminal for voltage). Solution: If we consider the loop containing ABEC, and apply KVL (sum of voltage rises): VA 3 10 5 0 VA 12V VD VE 10V Applying KCL at the junction (node) of B, D and E; sum of outgoing currents =0: I B 2 3 0 I B 5 A As elements A, B and C are in series, I A I B 5 A IC I A 5 A With voltages and currents known for all the elements, following the passive sign convention. We can calculate the power as product of voltage and current: PA VA I A 12*(5) 60W i.e. element A is delivering power. PB VB I B 3*(5) 15W i.e. element B is delivering power. PC VC I C 5*5 25W i.e. element C is absorbing power. EE1002/CG1108 Tutorials 1 to 5 Solutions 3 PD VD I D 10*3 30W i.e. element D is absorbing power. PE VE I E 10* 2 20W i.e. element E is absorbing power. Total power supplied = 60+15=75W Total power absorbed = 25+30+20=75W Conservation of power is proved. 4. In the circuit given in Figure 4, the power absorbed by the 15‐Ohm resistor is 15W. Find R. R 4 6 15 4 24 4 Figure 4 Analysis: We can apply series/parallel equivalent for resistors to find the equivalent resistance parallel to the We can then apply current divider principle to find the current in the branch parallel to the 15Ohm branch. We can also find the total current through the unknown resistor. As the voltage drop across the 15 Ohm resistor is known, from KVL, we can find the voltage drop across the unknown resistor R. Then we can determine the unknown resistor R. Solution: EE1002/CG1108 Tutorials 1 to 5 Solutions 4 R 15, P 15W P I 2R I 15 P 1A 15 R We can use series/parallel rules for resistors to find the new equivalent circuit: 4 R 6 15 25V 4 4 6 4 24 4 6 8 2 4 4 24 8 8 || 24 6 10 vR R 25V v15 IR 15 I15 10 I12 Applying current division principle: EE1002/CG1108 Tutorials 1 to 5 Solutions 5 10 10 IR 10 15 25 15 15 I10 I R IR 10 15 25 I10 15 15 I10 I15 1.5 A I15 10 10 I15 I R I R I15 25 2.5 A 10 Applying KVL around the loop containing the power supply, R and 15 Ohm resistor: v15 15V 25 vR v15 0 vR 25 15 10V We can now calculate the resistance R: R vR 10 4 I R 2.5 EE1002/CG1108 Tutorials 1 to 5 Solutions 6 EE1002/CG1108 Tutorial 2 5. For the circuit shown in the Figure 5, find: a. The currents i1 and i2 . b. The power delivered by the 3A current source and by the 12V voltage source. c. The total power dissipated by the circuit. R1=25 ohm, R2= 10ohm, R3=5 ohm, R4=7 ohm. v1 R1 1 R3 2 i2 i1 R2 R4 Figure 5 Solution: We can find the currents if we know the voltage at node1. Let the voltage at node 1 as v1 . i1 v1 R2 12 v1 i2 R3 EE1002/CG1108 Tutorials 1 to 5 Solutions 7 a) Applying the KCL at node 1, algebraic sum of currents entering =0. 3 0 v1 12 v1 0 R2 R3 3R2 R3 v1 R3 12 R2 v1 R2 (4 R3 ) R2 0 v1 3 R2 R3 R2 R3 Putting the values of the resistances, v1 18V . Therefore, 18 1.8 A 10 12 v1 12 18 i2 1.2 A R3 5 i1 b) To find the power delivered by the current source, we need to know the votlage across it. We have labeled the voltages across the elements in the mesh shown in the figure. We can apply KVL around the loop to find this voltage. v1 R1 R3 1 vR1 vI 1 i2 vR2 R2 i 2 R4 Starting from node1 going clock‐wise along the mesh, the algebraic sum of voltage rises around the mesh: EE1002/CG1108 Tutorials 1 to 5 Solutions 8 v R 2 v I v R1 0 i1 R2 v I 3R1 0 v I 1.8 10 3 25 18 75 93V p vi 93 3 279W We have labeled the voltage and current reference directions according to the passive sign convention. Thus negative power implies that power is delivered by the current source. To find the power associated with the voltage source, we need to know the current through the voltage source. For this, we assign the current in resistance R4 and the source as shown. v1 R1 R3 1 vR1 vI 1 i i2 vR2 R2 iR4 2 is R4 Applying KCL at node2: Algebraic sum of currents entering the node =0. is i2 i R 4 0 i s i2 i R 4 1.2 12 514.3mA R4 Power associated with the voltage source = 12 (0.5143) 6.17W i.e. power is delivered by the source. c) Total power dissipated by the circuit can be obtained from the conservation of energy principle, by equating with the total power delivered by the two sources. Thus, total power dissipated = 279+6.17=285.7W. EE1002/CG1108 Tutorials 1 to 5 Solutions 9 6. Given the circuit of Figure 6: a) Determine the power delivered by the dependent current source. Ans. 108W b) Determine the power delivered by the voltage source. Ans. 0W v15 15 0.5i 2 vI 7 i 5 24V Figure 6 Analysis: To solve the voltage and current values for the dependent source. We can find the current i from the circuit directly. Then we can find the value of the dependent current source. Then, we can apply KVL in the left mesh and find the voltage across the dependent current source. Solution: i 24 2A 75 Value of the current source = 0.5 4 2 A Current through the 15 Ohm resistor = 2A Applying KVL around the left mesh: We have labeled the voltages according to passive sign convention. Starting with the negative terminal of the voltage source, summing the voltage rises: EE1002/CG1108 Tutorials 1 to 5 Solutions 10 v I v15 24 0 v I 2 15 24 0 v I 54V Power associated with the dependent current source = v I 2 54 2 108W i.e power is delivered by the dependent current source. Please note that there is current through the 24V source and hence no power delivered by the voltage source. 7. Consider NiMH hobbyist batteries shown in the circuit of Figure 7Error! Reference source not found.: a. If V1=12.0V, R1=0.15 ohm, RL=2.55 ohm, find the load current IL and the power dissipated by the load. b. If we connect a second battery in parallel with battery 1 that has voltage V2=12V and R2=0.28 ohm, will the load current IL increase or decrease? By how much? Use mesh current analysis method. IL R1 RL V1 Battery 1 Load I1 I2 IL R2 R1 RL V2 V1 Battery 2 Battery 1 Load Figure 7 EE1002/CG1108 Tutorials 1 to 5 Solutions 11 Analysis: We can find the current with one battery easily. When the second battery is connected, we can apply mesh current analysis to find the load current. Solution: a) With the first battery, the load current: i L V1 12 12 4.44 A . R1 R L 0.15 2.55 2.70 2 Power in the load = i L R 4.44 4.44 2.55 50.4W b) We can consider two meshes as shown in Figure: I1 I2 ia R2 R1 V2 IL ib RL V1 Then, we can find the branch currents and voltages in terms of the mesh currents: I1 R2 V2 ia ia R2 I2 (ib ia)R 1 (ib ia) ib RL IL ib RL V1 EE1002/CG1108 Tutorials 1 to 5 Solutions 12 Then, we apply the KVL around mesh a and mesh b to obtain the two independent equations. V2 ia R2 (ib ia ) R1 V1 0 V1 (ib ia ) R1 ib RL 0 Rearranging the two equations: i a ( R1 R 2 ) ib R1 V1 V 2 ia R1 ib ( R1 RL ) V1 (1) (2) Putting the values of the resistors and voltages we get: 0.15 ib 0.43 eqn(2) : 0.15ia 2.70ib 12 ib 4.53 A, ia 1.58 A eqn(1) : 0.43ia 0.15ib 0 ia The new load current = 4.53 A The load current increases by 4.53‐4.44=0.09A 8. Using node voltage analysis in the circuit of the figure, find the current i through the voltage source. 0.5 i 0.5 0.25 0.33 Figure 8 EE1002/CG1108 Tutorials 1 to 5 Solutions 13 Solution: 0.5 i 0.5 v1 0.25 0.33 v2 The negative terminal of the voltage source is marked as the reference. Two nodes are marked as node voltage v1 , v 2 . The KCL at the two nodes will be: v1 3 v1 2 0 (1) 0.5 0.5 v2 v2 3 2 0 (2) 0.25 0.33 Eqn(1)and (2) give: v1 2V v2 2 * 0.25 * 0.33 3 * 0.25 1.01V 0.25 0.33 The current through the voltage source, by apying KCL: i 3 v1 3 v 2 2 6.03 8.03 A 0 .5 0.33 EE1002/CG1108 Tutorials 1 to 5 Solutions 14 9. Using KCL, perform node analysis n the circuit shown in Figure 9 and determine voltage across R4. Note that one source is a controlled voltage source! Let Vs=5V, Av=70, R1=2.2kohm, R2=1.8kohm, R3=6.8kohm, R4=220ohm. vR1 R1 R2 R3 Vs AV VR1 R4 Figure 9 Solution: There are two unknown node voltages as shwon in the figure: v2 vR1 R1 R2 R3 5V v1 AV VR1 R4 The dependent source voltage can be written in terms of the node voltages: AvV R1 Av (5 v 2 ) EE1002/CG1108 Tutorials 1 to 5 Solutions 15 Writing the KCL at the two nodes: At the super node: v1 v1 v 2 v1 Av (5 v 2 ) v 2 0 R4 R3 R2 1 1 Av 1 5 Av 1 1 v2 v1 R R R R3 R2 R2 3 2 4 At node2: v 2 5 v 2 v1 v 2 (v1 Av (5 v 2 )) 0 R1 R3 R2 1 1 v1 R R 3 2 1 1 1 Av v2 R R R 3 2 1 5 Av 5 R R 2 1 Putting the values (factoring our 10‐3), we get: 5.2481v1 39.592v 2 194.44 0.7026v1 40.046v 2 196.717 Solving these equaiton swe get: v1 8.7572mV v 2 4.9124V EE1002/CG1108 Tutorials 1 to 5 Solutions 16 EE1002/CG1108 Tutorial 3 10. Determine, using superposition, the voltage v across R in the circuit of Figure 10. I B 3 A, RB 1,VG 15V , RG 1, R 2 RG RB IB R VG v Figure 10 Analysis: To evaluate the voltage with only one source at a time. Finally add up the two voltages. Kill the voltage source by replacing it with a short. Kill the current source by replacing it with an open circuit. Solution: Kill the voltage source by shorting it as shown here. Obtain the voltage due to the current source alone. 2 3A 1 1 v1 3A The three resistors in parallel can be reduced to their equivalent resistance: EE1002/CG1108 Tutorials 1 to 5 Solutions Req v1 17 1 111 2 2.5 Req Req 1 2 1 5 26 v1 3 1.2V 55 Then, kill the current source by opening it as shown here. Obtain the voltage due to the current source alone. 1 1 v2 15V 1 2 v2 15V Rx Then, the two resistors (1,2) in parallel can be reduced to its equivalent resistance Rx. 1 113 2 Rx Rx 1 2 2 3 Then, voltage divider principle can be used to find the voltage across Rx. v2 15 RRx1 15 x 23 6V 2 3 1 Finally, the original voltage in question can be obtained as v v1 v2 1.2 6 7.2V . 11. Find the Thevenin equivalent circuit that the load ( R L ) sees for the circuit of Error! Reference source not found.. 1k 10V 1 1k 3 1mA RL Figure 11 EE1002/CG1108 Tutorials 1 to 5 Solutions 18 Analysis: We need to find the open circuit voltage between points a and b, which will be the Thevenin’s voltage. Then we need to find the Thevenin resistance. Solution 1k v2 v1 1 3 10V 1k voc 1mA Let us apply node voltage analysis method. Note the reference node and the two unknown node voltages, v1 , v2 . Writing KCL at v1 : v1 v2 0.001 0 v1 v2 0.001 1 Writing KCL at node v 2 : v2 10 v v v 2 2 1 0 1000 1000 1 (1) (2) Putting the eqn(1) in eqn(2), and multiplying both sides by 1000, we get: v2 10 v2 1 0 v2 9 4.5V 2 Putting value of v2 in eqn(1), we get v1 v2 0.001 4.499V . To find the value of the Thevenin resistance: As the circuit contains independent sources only, we can kill the sources and calculate the equivalent resistance between the points a and b . EE1002/CG1108 Tutorials 1 to 5 Solutions 19 1k v2 v1 1 3 R eq 1k The equivalent resistance can be calculated as: Req 3 1 1000 1000 504 . 1000 1000 The Thevenin equivalent circuit is: 504 4 . 499 V RL 12. For the circuit given in Figure 12: (i) Obtain the Thevenin’s equivalent for the circuit which contains a dependent voltage source. (ii) What should be the optimum value of a load resistor RL to be connected between a and b so that the power delivered to it by the network is maximum? (iii) What is the maximum power? EE1002/CG1108 Tutorials 1 to 5 Solutions 20 4A Figure 12 To find : Thevenin's equivalent and then obtain the load at which power transfer is maximum. Analysis: We can apply node voltage analysis to find the Thevenin's voltage. As there is a dependent source, we can use test source method to obtain the Thevenin's resistance. Solution: (i) Find the open circuit voltage. Apply node voltage analysis. v2 v1 voc 4A Note the choice of the reference node and the unknown node voltages v1 , v2 . We can write the current i x as i x v1 . 40 Then the dependent voltage source becomes 160i x 160 v1 4v1 . 40 Applying KCL at the super node surrounding v1 : v1 v1 v1 160i x v2 0 40 80 20 Replacing i x we get v1 160i x v1 160 Rewriting equation (1), we get: EE1002/CG1108 Tutorials 1 to 5 Solutions (1) v1 5v1 40 21 1 5 v 23 1 v1 2 0 v1 v2 0 4 40 80 20 20 Applying KCL at node v2 : v v1 160i x v 4 2 2 0 60 20 Replacing (v1 160i x ) by 5v1 : v1 v2 (1) vv 5 1 1 v1 v2 4 1 2 4 20 4 15 60 20 (1) (2) 23 : 23 23 15 v2 v2 . 4 23 v2 15 2 Putting the value of v2 in eqn (1), we get: (2) 4 30V 23 Thus the open circuit voltage is voc v1 30V . Find the Thevenin resistance using 'Test source method'. We shall kill the independent current source of 4A and connect a current source of 1A to the terminals and then calculate the voltage across them. v2 20Ω 60Ω 160iX +- v1 a 80Ω 40Ω iX v? 1A b Applying node voltage analysis method. Note the choice of the reference node and the unknown node voltages. We can write the current i x as i x v1 . 40 Then the dependent voltage source becomes 160i x 160 EE1002/CG1108 Tutorials 1 to 5 Solutions v1 4v1 . 40 22 Applying KCL at the super node surrounding v1 : 1 v1 v1 v1 160i x v2 1 5 v 23 1 0 v1 2 1 v1 v2 20 40 80 20 4 40 80 20 20 (1) Applying KCL at v2 : v2 v2 (v1 160i x ) 0 60 20 Replacing (v1 160i x ) by 5v1 : vv 5 1 1 v1 v2 0 1 2 0 20 4 15 60 20 (2) (1)+(2)X23=> 23 20 15 20 v2 15 8 Putting the value of v2 in eqn(2), we get: v2 v2 v1 20 15 0 v1 10V 4 15 8 v Thevenin resistance is 10 . 1 Thevenin equivalent is: 10 30 V (ii). According to maximum power transfer theorem, the load resistance for maximum power transfer would be same as the soruce resistance (Thevenin resistance). RL 10 (iii) Maximum power delivered to the load at RL 10 : 2 PL (max) 9 30 10 10 22.5W 4 10 10 EE1002/CG1108 Tutorials 1 to 5 Solutions 23 EE1002/CG1008 Tutorial 4 13. If the switch in the circuit of Figure 13 is closed at t=0, i) Determine the current flow through the resistors and the capacitor when t=0+. ii) What will be the current flow under steady state condition? iii) Determine the voltage across the capacitor under steady state condition. iv) Find an expression for the capacitor voltage as a function of time t>0. Assume that the capacitor is initially uncharged. t0 20 i1 i2 20 vc 2 F i3 Figure 13 Solution: As capacitor voltage cannot change instantaneously, vc (0) vc (0) , i.e. the voltage immediately before the switch is closed will be same as the voltage immediately after the switch is closed. vc (0) can be obtained by the DC analysis before switch is closed. As can be seen, the capacitor will be discharged before switch is closed i.e. vc (0) 0 . Thus, vc (0 ) vc (0) 0 . i) Current through the resistors immediately after switch is closed: EE1002/CG1108 Tutorials 1 to 5 Solutions 24 100 0 5A 20 0 i 2 (0 ) 0A 20 i3 (0) 5 0 5 A i1 (0) ii) At steady‐state, the capacitor will be open circuited. 20 i1 vc i2 20 2 F i3 i3 () 0 i1 i2 100 2.5 A 20 20 iii) At steady‐state , the capacitor voltage can be obtained using voltage divider principle: vc () 100 20 50V 20 20 iv) To find the capacitor voltage as a function of time t i.e. vc (t ) at t 0 , we shall put the circuit in a standard form i.e. the Thevenin’s equivalent of rest of the circuit seen from the capacitor. R th Vth vc (t) 2 F Rth 10,Vth 50V Time constant of the circuit Rth C 10 2 10 6 20 S EE1002/CG1108 Tutorials 1 to 5 Solutions 25 vc (0) vc (0) 0V vc () Vth 50V vc (t ) vc (0 )e t t vc ()1 e t 501 e 20106 V 14. For the circuit shown in Figure 14, assume that switch S1 was closed and switch S2 was opened for a long time. Then, at time t=0, switch S1 is opened and switch S2 is closed. a. Find the capacitor voltage vc(t) at t=0+. b. Find the time constant for t>=0. c. Find an expression for vc(t), and sketch the function. d. Find vc(t) for each of the following values of t zero, the time constant, twice the time constant, five times the time constant and ten times the time constant. 4 S1 5 4F vc 4F S2 6 6 Figure 14 Analysis: First, we need to find the capacitor voltage vc (0) before time t=0, which will be same as vc (0) . Then, we need to find the Thevenin equivalent of the circuit after time t=0, so that we can use the standard formula for capacitor voltage transient. Solution: With only S1 closed, and S2 open, the circuit is reduced to: EE1002/CG1108 Tutorials 1 to 5 Solutions 26 4 4 5 8F v c 6 5 v c 6 The two capacitors in parallel are put as their equivalent. As the capacitor behaves like an open circuit in steady‐state, we can use voltage divider principle: vc (0) 30 10 20V vc (0 ) 15 Next when switch S1 is opened and S2 is closed, the circuit becomes: 4 8F vc 4 6 6 8F vc 3 We can use source transformation, to replace the 4A current source with 3 parallel resistor by a voltage source of 12V and series resistance of 3 . 4 7 3 8F 8F Time constant of the circuit after t 0 , RC 7 8S 56 S . vc () 12V EE1002/CG1108 Tutorials 1 to 5 Solutions 27 vc (t ) vc (0)e t t vc ()1 e 12 (20 12)e t vc () vc (0 ) vc () e t 5610 6 12 8e t 5610 6 15. For the circuit given in Figure 15, switch S2 was closed for a long time before t=0. At t=0, the switch S1 is closed and S2 is opened. a. Find the inductor current i(t) at t=0+. b. Find the time constant for t>=0. c. Find an expression for i(t). d. Find i(t) for each of the following values, the time constant, twice the time constant, five times the time constant and ten times the time constant. Sketch the function 20 S1 S2 20 20 L 30 H i(t) 2A 50 Figure 15 Analysis: When switch S2 is closed for a long time, the circuit should be in steady‐state. We can consider the inductor as a short‐circuit and then determine the current in it. Also, inductor current cannot change instantaneously and hence, the current immediately before t=0 will be same as immediately after t=0. Solution: a) When S1 is open and S2 is closed, the circuit would be like: i(t) L 30 H 2A 50 EE1002/CG1108 Tutorials 1 to 5 Solutions 28 The inductor would act as a short circuit, and hence the current through the inductor will be 2A. i (0 ) i ( 0 ) 2 A At t=0, when S1 is closed and S2 is opened, the circuit is reduced which is then converted to its Thevenin equivalent form: Rth 20 20 20 i(t) Vth i(t) b) The Thevein voltage and Thevenin resistance for the circuit on the left can be obtained by inspection alone. 20 15V 20 20 20 Rth 20 30 2 Vth 30 The time constant for t>0 is L 30H 1S R 30 c) i (t ) i (0) e t t i () i (0) i () e A t 1 e i ( ) As the inductor will behave like short circuit in steady‐state, i () Vth 15 0.5 A Rth 30 So the current will be decaying exponentially from 2A to 0.5A. d) i (t ) t 0.5 1.5 e 1 1.0518 A i (t ) t 2 0.5 1.5 e 2 0.7030 A i (t ) t 3 0.5 1.5 e 3 0.5498 A i (t ) t 5 0.5 1.5 e 5 0.5101A EE1002/CG1108 Tutorials 1 to 5 Solutions 29 EE1002/CG1108 Tutorial 5 16. For the circuit in Figure 16, a) Find the expression for v R (t ) . b) If the sinusoidal has a frequency of 10 kHz, and the inductor is 1 mH , what is the value of R for phase difference between v s (t ) and v R (t ) to be 45 deg? c) Draw the phasor diagram showing the v s (t ) and v R (t ) for part (b). vL L vs (t) Vm sin( t) R vR Figure 16 Analysis: We need to replace the voltage source by its phasor and the circuit components by their impdances. Then we can use the DC circuit laws to solve the AC circuit. Solution. Given v s (t ) Vm sin t we can replace it by its phasor i.e. Vm 0 0 . Inductor L can be replaced by its impedance i.e. Z L jL L90 0 . Z R R R0 0 vL j L Vm00 R vR EE1002/CG1108 Tutorials 1 to 5 Solutions 30 By applying voltage divider rule, RVm R R Vm Vm 2 2 22 R jL R L R 2 L2 L tan 1 R RVm v R (t ) sin(t ) 2 R 2 L2 VR b) Phase difference between v s (t ) and v R (t ) is . f 10000 2f 2 10000 62832 450 tan 1 L R R L 62832 *110 3 62.832 c) Vm00 VR 450 17. Determine the current i(t) in the circuit shown in Figure 17. vs (t ) 636 cos 3000t 12 R1 2.3k , R2 1.1k L 190mH , C 55nF i(t ) R1 R2 v s (t ) L C Figure 17 EE1002/CG1108 Tutorials 1 to 5 Solutions 31 Solution: Find the phasor for the voltage source and the impedances for the circuit components. v s (t ) 636 3000 12 Z L jL j 3000 190 10 3 j 570 1 1 ZC j j j 6060 C 3000 55 10 9 Z R1 2300 Z R 2 1100 I 636 2300 1100 12 j 570 j 6060 We can use the common techniques like series/parallel reduction for solving the circuit. 636 I 12 2300 j 570 1100 j6060 636 I 12 Z Z (2300 j570) (1100 j 6060) 2369.613.920 6159 79.7 0 2300 j570 1100 j 6060 3400 j5490 Z 2260 7.56 0 Vs 636150 0.281422.56 0 Z 2260 7.56 0 0 i (t ) 0.2814 cos(3000t 22.56 ) I EE1002/CG1108 Tutorials 1 to 5 Solutions 32 18. Find the Thevenin equivalent of the circuit as seen from terminals a‐b for the circuit shown in Figure 18. j 8 a j8 5 300 8 b Figure 18 Analysis: The circuit given ha the ac sources in phasor form and the circuit elements in their impedances. We can use the same techniques as DC circuit analysis, except that we have to deal with complex numbers. Solution: Open circuit voltage can be obtained using the voltage divider principle: Vth Voc 5 30 0 8 j8 8 j8 5 30 0 5 30 0 (1 j1) 8 j8 j8 8 5 30 0 2450 5 2150 Thevenin impedance can be obtained by killing the voltage source and finding the equivalent impedance between a and b. j 8 a j 8 Z eq 8 b j8 8 j8 j8 8 j8 Z eq j (8 j8) 8 j8 8 j8 j8 8 8 j8 The Thevenin equivalent of the circuit is: a 5 2 0 15 b EE1002/CG1108 Tutorials 1 to 5 Solutions 33 EE1002/CG1108 Tutorials 1 to 5 Solutions 34 ...
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