chap02

# chap02 - Chapter 2 Solution 1 v = iR i = v/R =(16/5 mA =...

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Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Solution 2 p = v 2 /R R = v 2 /p = 14400/60 = 240 ohms Chapter 2, Solution 3 R = v/i = 120/(2.5x10 -3 ) = 48k ohms Chapter 2, Solution 4 (a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6 n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7 7 elements or 7 branches and 4 nodes, as indicated. 30 V 1 20 2 3 +++- 2A 30 60 40 10 4 + -

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Chapter 2, Solution 8 d c b a 9 A i 3 i 2 12 A 12 A i 1 8 A 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 Chapter 2, Solution 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 Chapter 2, Solution 10 At node 1, 4 + 3 = i 1 i 1 = 7A At node 3, 3 + i 2 = -2 i 2 = -5A 3 2 -2A 3A 1 4A i 2 i 1
Chapter 2, Solution 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 Chapter 2, Solution 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 1 +v 2 +v 3 = 0 v 3 = 30v + 15v - loop 3 loop 2 loop 1 + 20v - + 10v - 25v + + v 2 - + v 1 - + v 3 - Chapter 2, Solution 13 2A I 2 7A I 4 1 2 3 4 4A I 1 3A I 3

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At node 2, 3 7 0 10 2 2 + + =  → = − I I A 12 2A 2 5A At node 1, I I I I A 1 2 1 2 2 2 + =  → = = At node 4, 2 4 2 4 4 4 = +  → = = − I I At node 3, 7 7 4 3 3 + =  → = = I I I Hence, I A I A I A I 1 2 3 4 12 10 5 2 = = − = = , , , A V 11 8 Chapter 2, Solution 14 + + - 3V V 1 I 4 V 2 - I 3 - + 2V - + - + V 3 - + + 4V I 2 - I 1 + - 5V For mesh 1, + + =  → = V V 4 4 2 5 0 7 For mesh 2, + + + =  → = − = − 4 0 4 7 3 4 3 V V V V For mesh 3, + =  → = + = − 3 0 3 1 3 1 3 V V V V V For mesh 4, =  → = − = V V V V V 1 2 2 1 2 0 2 6 Thus, V V V V V V V 1 2 3 4 8 6 11 = − = = − V 7 = , , ,
Chapter 2, Solution 15 + + + 12V 1 v 2 - - 8V + - v 1 - 3 + 2 - v 3 10V - + For loop 1, 8 12 0 4 2 2 + =  → = v v V V V For loop 2, =  → = − v v 3 3 8 10 0 18 For loop 3, + + =  → = − v v v 1 3 1 12 0 6 Thus, v V v V v 1 2 3 6 4 = − = = − , , V 18 Chapter 2, Solution 16 + v 1 - + - + - 6V - + loop 1 loop 2 12V 10V + v 1 - + v 2 - 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

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Chapter 2, Solution 17 + v 1 - + - + - + 10V 12V 24V loop 2 + v 3 - v 2 - loop 1 - + 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 Chapter 2, Solution 18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A -V ab + 5I + 8 = 0 V ab = 28V
Chapter 2, Solution 19 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 Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i 0 + 5i 0 = 0 i 0 = 4A Chapter 2, Solution 21 10 + - 45V - + + v 0 - Apply KVL to obtain -45 + 10i - 3V 0 + 5i = 0 But v 0 = 10i, 3v 0 -45 + 15i - 30i = 0 i = -3A P 3 = i 2 R = 9 x 5 = 45W 5

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Chapter 2, Solution 22 4 + v 0 - 10A 2v 0 6 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 Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below.
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