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Unformatted text preview: We first combine the two resistors in parallel = 10 15 6 We now apply voltage division, v 1 = = + ) 40 ( 6 14 14 28 V v 2 = v 3 = = + ) 40 ( 6 14 6 12 V Hence, v 1 = 28 V , v 2 = 12 V, v 3 = 12 V Problem 4 Find V in the circuit below and the power dissipated by the controlled source Chapter 2, Solution 22 At the node, KCL requires that v 2 10 4 v + + = 0 v = 4.444V The current through the controlled source is i = 2V = 8.888A and the voltage across it is v = (6 + 4) i (where i = v /4) = 10 111 . 11 4 v= Hence, p 2 v i = (8.888)(11.111) = 98.75 W...
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This note was uploaded on 02/20/2010 for the course EE Electrical taught by Professor Unknown during the Spring '10 term at City.
 Spring '10
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