EE 302 - HW Solution - 3

EE 302 - HW Solution - 3 - EE302 Homework #3 Chapter 2,...

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EE302 Homework #3 Chapter 2, Problem 26. For the circuit in Fig. 2.90, i o =2 A. Calculate i x and the total power dissipated by the circuit. Chapter 2, Solution 26. If i 16 = i o = 2A, then v = 16x2 = 32 V 8 4 A 8 v i = = , 4 2 8 A, i 16 4 2 v v i = = = = 2 4 8 16 16 8 4 2 30 A x i i i i i = + + + = + + + = 2 2 2 2 2 16 2 8 4 4 8 2 16 960 W P i R x x x x = = + + + = or 30 32 960 W x P i v x = = = 2 4 8 i o 16 i x
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Chapter 2, Problem 32. Find i 1 through i 4 in the circuit in Fig. 2.96. Chapter 2, Solution 32. We first combine resistors in parallel. = 30 20 = 50 30 x 20 12 = 40 10 = 50 40 x 10 8 Using current division principle, A 12 ) 20 ( 20 12 i i , A 8 ) 20 ( 12 8 8 i i 4 3 2 1 = = + = + = + = = ) 8 ( 50 20 i 1 3.2 A = = ) 8 ( 50 30 i 2 4.8 A = = ) 12 ( 50 10 i 3 2.4A = = ) 12 ( 50 40 i 4 9.6 A
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Chapter 2, Problem 34. Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit of Fig. 2.98. Find the overall dissipated power. 8
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This note was uploaded on 03/22/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

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EE 302 - HW Solution - 3 - EE302 Homework #3 Chapter 2,...

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