ch_02solu - Problem 2.1 v = iR Problem 2.2 p = v2/R i = v/R...

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Problem 2.1 v = iR i = v/R = (16/5) mA = 3.2 mA Problem 2.2 p = v 2 /R R = v 2 /p = 14400/60 = 240 ohms Problem 2.3 R = v/i = 120/(2.5x10 -3 ) = 48k ohms Problem 2.4 (a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA Problem 2.5 n = 9; l = 7; b = n + l – 1 = 15 Problem 2.6 n = 12; l = 8; b = n + l –1 = 19 Problem 2.7 There are 7 elements which implies that there are 7 branches . There are 5 nodes , as indicated below. Problem 2.8 At node a, 8 = 12 + i 1 i 1 = - 4A At node c, 9 = 8 + i 2 i 2 = 1A At node d, 9 = 12 + i 3 i 3 = -3A - + 3 5 2 3 6 4 2 1 5i 10 V 5 i i 1 12 A 12 A i 2 i 3 8 A 9 A a b c d
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Problem 2.9 Applying KCL, i 1 + 1 = 10 + 2 i 1 = 11A 1 + i 2 = 2 + 3 i 2 = 4A i 2 = i 3 + 3 i 3 = 1A Problem 2.10 At node 1, 4 + 3 = i 1 i 1 = 7A At node 3, 3 + i 2 = -2 i 2 = -5A Problem 2.11 Applying KVL to each loop gives -8 + v 1 + 12 = 0 v 1 = 4v -12 - v 2 + 6 = 0 v 2 = -6v 10 - 6 - v 3 = 0 v 3 = 4v -v 4 + 8 - 10 = 0 v 4 = -2v Problem 2.12 For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v For loop 2, -10 +15 -v 2 = 0 v 2 = 5v For loop 3, +v 2 +v 3 = 0 v 3 = 30v i 1 i 2 4A 1 3A -2A 2 3 + - + - - + loop 1 + v 2 - + v 1 - + v 1 - loop 2 6V 12V 10V + v 1 - + v 2 - + v 3 - 25v + + 10v - + 15v - + 20v - loop 1 loop 2 loop 3
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Problem 2.13 Applying KVL around loop 1, –6 + v 1 + v 1 – 10 – 12 = 0 v 1 = 14V Applying KVL around loop 2, 12 + 10 – v 2 = 0 v 2 = 22V Problem 2.14 It is evident that v 3 = 10V Applying KVL to loop 2, v 2 + v 3 + 12 = 0 v 2 = -22V Applying KVL to loop 1, -24 + v 1 - v 2 = 0 v 1 = 2V Thus, v 1 = 2V , v 2 = -22V , v 3 = 10V - + loop 1 v 2 + v 3 - loop 2 24V 12V 10V - + + v 1 - - +
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Problem 2.15 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A -V ab + 5I + 8 = 0 V ab = 28V Problem 2.16 Applying KVL around the loop, we obtain -12 + 10 - (-8) + 3i = 0 i = -2A Power dissipated by the resistor: p 3 = i 2 R = 4(3) = 12W Power supplied by the sources: p 12V = 12 (- -2) = 24W p 10V = 10 (-2) = -20W p 8V = (- -2) = -16W Problem 2.17 Applying KVL around the loop, -36 + 4i 0 + 5i 0 = 0 i 0 = 4A Problem 2.18 Apply KVL to obtain -45 + 10i - 3V 0 + 5i = 0 But v 0 = 10i, -45 + 15i - 30i = 0 i = -3A P 3 = i 2 R = 9 x 5 = 45W Problem 2.19 4 + v 0 - - + 3v 0 45V 10 5 i
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At the node, KCL requires that 0 0 v 2 10 4 v + + = 0 v 0 = –4.444V The current through the controlled source is i = 2V 0 = -8.888A and the voltage across it is v = (6 + 4) i 0 = 10 111 . 11 4 v 0 - = Hence, p 2 v i = (-8.888)(-11.111) = 98.75 W Problem 2.20 (a) I 0 = 2 1 s R R V + α - = 1 V I 0 ( 29 4 3 R R = 4 3 4 3 2 1 s R R R R R R V + + α - ( 29 ( 29 4 3 2 1 4 3 1 R R R R R R Vs V + + α - = (b) If R 1 = R 2 = R 3 = R 4 = R, 10 4 2 R R 2 V V S 0 = α = α = α = 40 Problem 2.21 V 0 = 5 x 10 -3 x 10 x 10 3 = 50V Using current division, 6 2v 0 + v 0 - 10A
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I 20 = + = ) 50 01 . 0 ( 20 5 5 x 0.1 A V 20 = 20 x 0.1 kV = 2 kV p 20 = I 20 V 20 = 0.2 kW Problem 2.22 Using current division, i 1 = = + ) 20 ( 6 4 4 8 A i 2 = = + ) 20 ( 6 4 6 12 A Problem 2.23 Using voltage division, v 1 = = + ) 24 ( 9 3 3 6 V v 2 = = + ) 24 ( 9 3 9 18 V Problem 2.24 We first combine the two resistors in parallel = 10 15 6 We now apply voltage division, v 1 = = + ) 40 ( 6 14 14 20 V v 2 = v 3 = = + ) 40 ( 6 14 6 12 V Hence, v 1 = 28 V , v 2 = 12 V, v s = 12 V Problem 2.25 The series combination of 6 and 3 resistors is shorted. Hence i 2 = 0 = v 2
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v 1 = 12, i 1 = = 4 12 3 A Hence v 1 = 12 V , i 1 = 3 A , i 2 = 0 = v 2
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Problem 2.26 By current division, = + = ) 9 ( 12 6 12 i 6 A = = = - = 1 1 i 4 v , A 3 6 9 i 4 x 3 = 12 V p 6 = 1 2 R = 36 x 6 = 216 W Problem 2.27 The 5 resistor is in series with the combination of = + 5 ) 6 4 ( 10 .
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This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.

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ch_02solu - Problem 2.1 v = iR Problem 2.2 p = v2/R i = v/R...

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