EE302_spring_2008_hw9a

# EE302_spring_2008_hw9a - HW 9 solutions Chapter 4, Problem...

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HW 9 solutions 1 Chapter 4, Problem 50. Obtain the Norton equivalent of the circuit in Fig. 4.116 to the left of terminals a - b . Use the result to find current i Figure 4.116 Chapter 4, Solution 50. From Fig. (a), R N = 6 + 4 = 10 ohms From Fig. (b), 2 + (12 – v)/6 = v/4, or v = 9.6 V -I N = (12 – v)/6 = 0.4, which leads to I N = -0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c). i = [10/(10 + 5)] (4 – 0.4) = 2.4 A 6 Ω 4 Ω (a) 12V + I sc = I N 6 Ω 4 Ω (b) 2A i 10 Ω (c) 0.4A 5 Ω 4A

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HW 9 solutions 2 Chapter 4, Problem 54. Find the Thèvenin equivalent between terminals a - b of the circuit in Fig. 4.120. Figure 4.120 Chapter 4, Solution 54. To find V Th =V x , consider the left loop. x o x o V i V i 2 1000 3 0 2 1000 3 + = → = + + (1) For the right loop, o o x i i x V 2000 40 50 = = (2) Combining (1) and (2), mA 1 3000 4000 1000 3 = → = = o o o o i i i i 2 2 2000 = → = = Th o x V i V To find R Th , insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below. 1 k
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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_hw9a - HW 9 solutions Chapter 4, Problem...

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