HW5%20solution

HW5%20solution - R 1 = 1200 Ω. Using (1), R s = 600 Ω....

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ECE 201 Spring 2010 Homework 5 Solutions Problem 26 Since V 2 = 60 V , the current through 60 Ω branch is 1 A. The resistors 90 Ω and 180 Ω are in parallel. Their equivalent resistance is R eq = 90 × 180 90 + 180 = 60 Now 60 Ω and 60 Ω are in series. Thus the voltage drop across the parallel combination of 40 Ω and 120 Ω is 120 V. Thus current through 40 Ω resistor is 120/40=3 A. Hence, I s = 3 + 1 =4 A Applying KVL around the loop containing the source and 40 Ω resistor, V s - 180 × 4 - 120 = 0 V s = 840 V Power delivered by the source is given by, P s = V s × I s = 840 × 4 = 3360 W Problem 32(a) Using voltage division, the following equations can be written, R 1 + R 2 + R s = 2400 (1) R 1 + R 2 R 1 + R 2 + R s =0 . 75 (2) R 2 R 1 + R 2 + R s =0 . 25 (3) 1
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Using (1) and (3), R 2 = 600 Ω. Using (2),
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Unformatted text preview: R 1 = 1200 Ω. Using (1), R s = 600 Ω. Problem 40 The resistors 9 k Ω and 18 k Ω are in parallel. Thus their equivalent resistance of 6 k Ω is in series with 6 k Ω. Now the combination 4 k Ω and 12 k Ω are in parallel. The current I in divides into I 1 and I 2 through the parallel resistors in inverse ratio of their resistances. Similarly, the current I 2 divides in inverse ratio of resistances 9 k Ω and 18 k Ω in parallel. Thus we get, I 1 = 120 × 12 12 + 4 = 90 mA I 2 = 120-90 = 30 mA I 3 = 30 × 9 9 + 18 = 10 mA V in = 0 . 09 × 4 × 10 3 = 360 V P source = V in I in = 43 . 2 W 2...
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This note was uploaded on 01/22/2012 for the course IE 370 taught by Professor Chunghorng,r during the Spring '08 term at Purdue.

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HW5%20solution - R 1 = 1200 Ω. Using (1), R s = 600 Ω....

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