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hw1_ext_sol

hw1_ext_sol - total equivalent power and we get that P = 1...

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EE 101 Handout # Prof. A. El Gamal Jan 24, 2003 Extra problem solution of Home-work #1 1. problem 1 (a) The first step is to find the equivalent resistance of the circuit. we get that R eq = 1Ω+2Ω || R L Ω = 1+2 || 2 = 1+1 = 2Ω The current i in is then, i in = 1 2 and is positive because of reverse associated reference direction. (b) The total power given by the source is P = V I = 1 · 0 . 5 = 0 . 5[ W ]. the junction of R L and the 2Ω resistor is a current divider, so the current that flows through R L is i R L = i in 2 = 0 . 25 and the power is, P = I 2 R = 1 16 2 = 1 8 . So the fraction of the power dissipated by the resistor is 25%. Another Way to look at it would be using symmetry. The 1Ω resistor is cascaded by an equivalent of 1Ω = 2 || R L resistor, so each one takes half the power. Because the parallel resistors are the same, each one takes half of their
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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 cs-V s 5Ω-V s 20Ω =-1-2 5-2 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 VS-1Ω-CS. We get, V s + i cs · 1Ω-V cs = 0 ⇒ V cs = 2 + (-1) · 1 = 1[ V ]. So P = 1[ A ] 1...
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