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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 = 12090 = 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.
 Spring '08
 Chunghorng,R

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