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Unformatted text preview: total equivalent power, and we get that P = 1 2 1 2 = 1 4 ! 2. problem 2 (a) The total current that enters the VS, 20Ω and 5Ω resistors is, by KCL, the current induces by the current source. The current that goes through the 20Ω and 5Ω resistors, is induced by the voltage drop upon them, hence by the voltage source. So, by KCL, the current through the voltage source is: i vs = i csV s 5ΩV s 20Ω =12 52 20 =1 . 5[ A ]. The power is: P = 2 · 1 . 5 = 3[ W ] (b) To ﬁnd the voltage drop on the current source, we use KCL on the loop VS1ΩCS. We get, V s + i cs · 1ΩV cs = 0 ⇒ V cs = 2 + (1) · 1 = 1[ V ]. So P = 1[ A ] 1...
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This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.
 Fall '08
 PingLi

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