EE302_spring_2008_hw8a

# EE302_spring_2008_hw8a - EE 302 HW 8 solutions Chapter 4,...

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EE 302 HW 8 solutions 1 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 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 Th V R k I = = Ω = Ω + _ 70 V + V o 4 V o + 10 k Ω 20 k Ω m m b a b 20 Ω 4 V o 10 k Ω + _ + V o I 2 a I 1 t T + I

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EE 302 HW 8 solutions 2 Chapter 4, Problem 42. For the circuit in Fig. 4.109, find Thevenin equivalent between terminals a and b . Figure 4.109
EE 302 HW 8 solutions 3

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EE 302 HW 8 solutions 4 Chapter 4, Problem 44. For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals
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## This note was uploaded on 04/18/2008 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.

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EE302_spring_2008_hw8a - EE 302 HW 8 solutions Chapter 4,...

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