Chapter 4, Solution 1. i i o 5 Ω 8 Ω 1 Ω + − 1 V 3 Ω Ω=+ 4)35(8 , 5 1 41 1 i = + = === 10 1 i 2 1 i o 0.1A Chapter 4, Solution 2. ,3)24(6 Ω=+ A 2 1 i 21 ==i , 4 1 i 2 1 i 1o == == oo i2v 0.5V 5 Ω 4 Ω i 2 8 Ω i 1 i o 6 Ω 1 A 2 Ω If i s = 1µA, then v o = 0.5µV Chapter 4, Solution 3. R + v o − 3R i o 3R 3R R + − 3R + − 1 V 1.5R V s (b) (a) (a) We transform the Y sub-circuit to the equivalent ∆ . ,R 4 3 R 4 R3 R3R 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 = 10Ω, v o = 5V, i o = 10/(10) = 500mA Chapter 4, Solution 4. If I o = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω resistor is 2A. + v 1 − 3A I s 2 Ω 4 Ω i 1 3A 1A I s 2A 6 Ω 4 Ω 3 Ω 2 Ω 2 Ω (a) (b) Ω= 263 , v o = 3(4) = 12V, .A3 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 Chapter 4, Solution 5. If v o = 1V, V21 3 1 V 1 =+       = 3 10 v 3 2 2V 1s =+       = If v s = 3 10 v o = 1 Then v s = 15 v o = =15x 10 3 4.5V v o 3 Ω 2 Ω + − 6 Ω 6 Ω 6 Ω v 1 V s Chapter 4, Solution 6 Let s T T oT V RR R V RR RR RRR 132 32 32 then ,// + = + == 133221 32 1 32 32 32 32 1 RRRRRR RR R RR RR RR RR RR R V V k T T s o ++ = + + + = + == Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find V Th , consider the circuit below. 3V x 5Ω 5 Ω + + 4V 15 Ω V Th - 6 Ω - + V x - From the figure, V3)4( 515 15 ,0 = + == Thx VV To find R Th, consider the circuit below: 3V x 5 Ω 5 Ω V 1 V 2 + 4V 15 Ω 1A - 6 Ω + V x - At node 1, 12 2111 73258616, 515 3 5 4 VVxV VVV V V xx −=→== − ++= − (1) At node 2, 950 5 31 21 21 −=→= − ++ VV VV V x (2) Solving (1) and (2) leads to V 2 = 101.75 V mW 11.22 75.1014 9 4 ,75.101 1 2 max 2 ===Ω== xR 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 Ω i 1 + − 20V i 2 5 A 4 Ω 6 Ω 4 Ω (a) (b) i 1 = ,A3)5( 46 6 = + A2 46 20 2 = + =i Thus i = i 1 + i 2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i 2 x 1 xx ii += where i is due to 15V source and i is due to 4A source, 1 x 2 x 12 Ω -4A 40Ω 10 Ω i x2 40 Ω i 10 Ω i x1 12 Ω + − 15V (a) (b) 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 Ω + − 3v ab2 2 A 10 Ω + − + 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 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. 12V 4A 2Ω 2 Ω i x2 6Ω 4A 3 Ω 2Ω i 2 3 Ω i o (a) 2 Ω i 1 6 Ω + − (b) For i 1 , consider Fig. (a). 2||3 = 2x3/5 = 6/5, i o = 12/(6 + 6/5) = 10/6 i 1 = [3/(2 + 3)]i o = (3/5)x(10/6) = 1 A For i 2 , consider Fig. (b), 6||3 = 2 ohm, i 2 = 4/2 = 2 A i = 1 + 2 = 3 A Chapter 4, Solution 12. Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 are due to the 2-A, 12-V, and 19-V sources respectively. For v o1 , consider the circuit below. 5 Ω 5 Ω + v o1 − i o 2A 2A 3Ω 4 Ω 6Ω 12 Ω 5 Ω + v o1 − 6||3 = 2 ohms, 4||12 = 3 ohms. Hence, i o = 2/2 = 1, v o1 = 5io = 5 V For v o2 , consider the circuit below. 6 Ω 5 Ω 4 Ω 6 Ω 5 Ω + v o2 − 3 Ω 3 Ω + v 1 − + − 12V + v o2 − 12 Ω + − 3 Ω 12V 3||8 = 24/11, v 1 = [(24/11)/(6 + 24/11)]12 = 16/5 v o2 = (5/8)v 1 = (5/8)(16/5) = 2 V For v o3 , consider the circuit shown below. 4 Ω 5 Ω 5 Ω 4 Ω 2 Ω + v o3 − 12 Ω + v 2 − + − 19V + − 19V 6Ω + v o3 − 12 Ω 3 Ω 7||12 = (84/19) ohms, v 2 = [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 v o = 5 + 2 – 7.125 = -125 mV , Chapter 4, Solution 13 Let iiii 321o + + = where i 1 , i 2 , and i 3 are the contributions to i o due to 30-V, 15-V, and 6-mA sources respectively. For i 1 , consider the circuit below. 1 kΩ 2 k Ω 3 k Ω + i 1 30V - 4 k Ω 5 k Ω 3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i 1 30 mA 4 k Ω 0.7949 k Ω Using current division, mA 4.973mA)30( 7949.4 7949.0 1 ==i For i 2 , consider the circuit below. 1 k Ω 2 k Ω 3 k Ω i 2 - 15V 4 k Ω 5 k Ω + After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i 2 4 k Ω 0.7949 k Ω Using current division, mA 4012.0mA)42.2( 7949.4 7949.0 2 −=−=i For i 3 , consider the circuit below. 6mA 1 k Ω 2 k Ω 3 k Ω i 3 4 k Ω 5 k Ω After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i 3 4 k Ω 0.7949 k Ω mA 5134.0mA)097.3( 7949.4 7949.0 3 −=−=i Thus, mA 058.4 321 =++= iiii o Chapter 4, Solution 14. Let v o = v o1 + v o2 + v o3 , where v o1 , v o2 , and v o3 , are due to the 20-V, 1-A, and 2-A sources respectively. For v o1 , consider the circuit below. 6 Ω 20V + − + v o1 − 4 Ω 2 Ω 3 Ω 6||(4 + 2) = 3 ohms, v o1 = (½)20 = 10 V