chap04

chap04 - Chapter 4 Solution 1 1 i 5 io 1V 8 3 8(5 3 = 4 i =...

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Chapter 4, Solution 1. i i o 5 8 1 + 1 V 3 = + 4 ) 3 5 ( 8 , 5 1 4 1 1 i = + = = = = 10 1 i 2 1 i o 0.1A Chapter 4, Solution 2. , 3 ) 2 4 ( 6 = + A 2 1 i 2 1 = = i , 4 1 i 2 1 i 1 o = = = = o o i 2 v 0.5V 5 4 i 2 8 i 1 i o 6 1 A 2 If i s = 1 µ A, then v o = 0.5 µ V Chapter 4, Solution 3. R + v o 3R i o 3R 3R R + 3R + 1 V 1.5R V s (b) (a)
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(a) We transform the Y sub-circuit to the equivalent . , R 4 3 R 4 R 3 R 3 R 2 = = R 2 3 R 4 3 R 4 3 = + 2 v v s o = independent of R i o = v o /(R) When v s = 1V, v o = 0.5V , i o = 0.5A (b) When v s = 10V, v o = 5V , i o = 5A (c) When v s = 10V and R = 10 , v o = 5V , i o = 10/(10) = 500mA Chapter 4, Solution 4. If I o = 1, the voltage across the 6 resistor is 6V so that the current through the 3 resistor is 2A. + v 1 3A I s 2 4 i 1 3A 1A I s 2A 6 4 3 2 2 (a) (b) = 2 6 3 , v o = 3(4) = 12V, . A 3 4 v o 1 = = i Hence I s = 3 + 3 = 6A If I s = 6A I o = 1 I s = 9A I o = 6/(9) = 0.6667A
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Chapter 4, Solution 5. If v o = 1V, V 2 1 3 1 V 1 = + = 3 10 v 3 2 2 V 1 s = + = If v s = 3 10 v o = 1 Then v s = 15 v o = = 15 x 10 3 4.5V v o 3 2 + 6 6 6 v 1 V s Chapter 4, Solution 6 Let s T T o T V R R R V R R R R R R R 1 3 2 3 2 3 2 then , // + = + = = 1 3 3 2 2 1 3 2 1 3 2 3 2 3 2 3 2 1 R R R R R R R R R R R R R R R R R R R R V V k T T s o + + = + + + = + = =
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Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find V Th , consider the circuit below. 3V x 5 5 + + 4V 15 V Th - 6 - + V x - From the figure, V 3 ) 4 ( 5 15 15 , 0 = + = = Th x V V To find R Th, consider the circuit below: 3V x 5 5 V 1 V 2 + 4V 15 1A - 6 + V x - At node 1, 1 2 2 1 1 1 7 3 258 6 1 6 , 5 15 3 5 4 V V x V V V V V V x x = → = = + + = (1) At node 2,
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95 0 5 3 1 2 1 2 1 = → = + + V V V V V x (2) Solving (1) and (2) leads to V 2 = 101.75 V mW 11 . 22 75 . 101 4 9 4 , 75 . 101 1 2 max 2 = = = = = x R V p V R Th Th Th Chapter 4, Solution 8. Let i = i 1 + i 2 , where i 1 and i L are due to current and voltage sources respectively. 6 i 1 + 20V i 2 5 A 4 6 4 (a) (b) i 1 = , A 3 ) 5 ( 4 6 6 = + A 2 4 6 20 2 = + = i Thus i = i 1 + i 2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i 2 x 1 x x i i + = where i is due to 15V source and i is due to 4A source, 1 x 2 x 12 -4A 40 10 i x2 40 i 10 i x1 12 + 15V (a) (b)
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For i x1 , consider Fig. (a). 10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75 i x1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For i x2 , consider Fig. (b). 12||40 = 480/52 = 120/13 i x2 = [(120/13)/((120/13) + 10)](-4) = -1.92 i x = 0.6 – 1.92 = -1.32 A p = vi x = i x 2 R = (-1.32) 2 10 = 17.43 watts Chapter 4, Solution 10. Let v ab = v ab1 + v ab2 where v ab1 and v ab2 are due to the 4-V and the 2-A sources respectively. + v ab2 10 + 3v ab2 2 A 10 + + v ab1 + 3v ab1 4V (a) (b) For v ab1 , consider Fig. (a). Applying KVL gives, - v ab1 – 3 v ab1 + 10x0 + 4 = 0, which leads to v ab1 = 1 V For v ab2 , consider Fig. (b). Applying KVL gives, - v ab2 – 3v ab2 + 10x2 = 0, which leads to v ab2 = 5 v ab = 1 + 5 = 6 V
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Chapter 4, Solution 11. Let i = i 1 + i 2 , where i 1 is due to the 12-V source and i 2 is due to the 4-A source.
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