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HW3 Solution

# HW3 Solution - EE302 Homework#3 Chapter 2 Problem 26 For...

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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. 2 Ω 4 Ω 8 Ω i o 16 Ω i x 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 = = = =0.960 kW

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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 Ω = 40 10 = 50 40 x 10 8 Ω 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 3.20 A = = ) 8 ( 50 30 i 2 4.80 A = = ) 12 ( 50 10 i 3 2.40A = = ) 12 ( 50 40 i 4 9.60 A
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 Ω 10 Ω 20 Ω Chapter 2, Solution 34. 40//(10 + 20 + 10)= 20 Ω , 40//(8+12 + 20) = 20 Ω 20 20 40 eq R = + = Ω 2 12 12/ 40, 3.6 W 40 eq V I P VI R = = = = = =3.60W + _ 40 Ω 40 Ω 12 V 20 Ω 12 Ω 10 Ω

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Chapter 2, Problem 38. Find R eq and i o in the circuit of Fig. 2.102. 5 Ω 6 Ω + _ 15 Ω 20 Ω 40 V 60 Ω 12 Ω 80 Ω R eq i o Chapter 2, Solution 38. 20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4 The circuit is reduced to that shown below.

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