EE302_spring_2008_hw7a

# EE302_spring_2008_hw7a - EE302 HW 7 Solutions Chapter 4,...

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EE302 HW 7 Solutions 1 Chapter 4, Problem 25. Obtain v o in the circuit of Fig. 4.93 using source transformation. Figure 4.93 Chapter 4, Solution 25. Transforming only the current source gives the circuit below. Applying KVL to the loop gives, –(4 + 9 + 5 + 2)i + 12 – 18 – 30 – 30 = 0 20i = –66 which leads to i = –3.3 v o = 2i = –6.6 V 12V + + + + i 2 Ω 4 Ω + v o 18 V 9 Ω 5 Ω 30 V 30 V

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EE302 HW 7 Solutions 2 Chapter 4, Problem 30. Use source transformation on the circuit shown in Fig 4.98 to find i x . Figure 4.98 Chapter 4, Solution 30. Transform the dependent current source as shown below. i x 24 Ω 60 Ω 10 Ω + + 12V 30 Ω 7i x - - Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below. i x 24 Ω + 12V 30 Ω 70 Ω 0.1i x -
Solutions 3 Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below. i

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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_hw7a - EE302 HW 7 Solutions Chapter 4,...

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