# hw8a - Chapter 4, Problem 40. Find the Thevenin equivalent...

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Chapter 4, Problem 40. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.107. Figure 4.107 For Prob. 4.40. Solution. To obtain V Th , we apply KVL to the loop. 70 (10 20) 4 0 o kI V - + + + = But 10 o V kI = 70 70 1 kI I mA = → = 70 10 0 60 V Th kI V V - + + = = To find R Th , one approach is to remove the 70-V source and apply a 1-V source at terminals a-b, as shown in the circuit below. 1 V We notice that V o = -1 V. 1 1 1 20 4 0 0.25 mA o kI V I - + + = = 2 1 1 0.35 mA 10 V I I k = + = 2 1 1 2.857 k 0.35 V R k I = = Ω = 1 + _ 70 V + V o 4 V o + 10 k 20 k b a b 20 4 V o 10 k + _ + V o I 2 a I 1 i i +

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Chapter 4, Problem 43. Find the Thevenin equivalent looking into terminals a - b of the circuit in Fig. 4.110 and solve for i x . The 6 ohm resistor is a load resistance. Figure 4.110 Solution. To find R Th , consider the circuit in Fig. (a). R Th = 10||10 + 5 = 10 ohms To find V Th , consider the circuit in Fig. (b). v b = 2x5 = 10 V, v a = 20/2 = 10 V -v a + V Th + v b = 0, or V Th = v a – v b = 0 volts and i x = 0 A 2 (a) 10 R Th a b 10 5 (b) 10 20V + - 2 A 10 V Th + + v a + v b - 5 b a
Chapter 4, Problem 48.

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## This note was uploaded on 02/13/2010 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.

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hw8a - Chapter 4, Problem 40. Find the Thevenin equivalent...

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