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Unformatted text preview: EE 101 Handout # 15 Professor Abbas El Gamal Feb. 14, 2003 Homework #5 Solutions 1. This problem was designed to be very hard to solve without using superposition, and quite reasonable with superposition. (a) The power supplied by the 2A current source is just 2 times the the voltage v appearing across it (as shown in the schematic below). v s v i s 2A 2Ω 2Ω 1Ω 1Ω So we need to find v , which we’ll do by superposition. First turn off i s and v s and leave on the 2A source, which yields: v 2A 2Ω 2Ω 1Ω 1Ω The two 2Ω resistors are in parallel and so form a 1Ω resistor, which is in series with the righthand 1Ω resistor, so the total resistance seen by the current source is 2Ω. Hence v = 4 in this case. Now turn off i s and the 2A current source, and leave on v s , which yields: 1 v s v 2Ω 2Ω 1Ω 1Ω No current flows through the righthand (or bottom) 1Ω resistors, so there is no voltage drop across the righthand resistor. By a voltage divider argument, it follows that v = v s / 2 in this case. Finally, we turn off v s and the 2A current source, and leave on i s , which yields: v i s 2Ω 2Ω 1Ω 1Ω In this case no current flows through the two 2Ω resistors (by a current divider argument). The voltage v is just the voltage across a 1Ω resistor with current i s flowing in it, so we have v = i s in this case. Putting it all together we find that v = 4 v s / 2 + i s with all three sources on. Hence we have: p = 8 + v s 2 i s . (b) Since we can adjust v s and i s independently, and their effect of p is linear, we simply choose each to maximize the terms it appears in. To maximize the term v s appearing p we set it to its maximum allowed value, i.e. , v s = 5V. To maximize the term 2 i s appearing p we set it to its minimum allowed value, i.e. , i s = 0A. 2. By linearity ( i.e. , superposition), the voltage across the 1Ω resistor can be expressed as αi 1 + βi 2 , where α and β are some constants with dimensions of resistance. Therefore, the power dissipated in the resistor ( p = v 2 R ) due to i 1 and i 2 is ( αi 1 + βi 2 ) 2 . From the result of the first experiment, we get the equation 0 . 25 = α 2 , which implies α = ± . 5 2 (we don’t know whether α = 0 . 5 or α = . 5). Similarly, the second experiment yields . 25 = β 2 , or β = ± . 5. For the third experiment, we obtain the equation 0 = ( α β ) 2 , that is α = β . From the latter, we have p = (3 α + β ) 2 = 16 α 2 , and since  α  = 0 . 5, we conclude that p = 4W....
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 Fall '08
 PingLi

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