ECE35 WIN10 Quiz 3 solution

ECE35 WIN10 Quiz 3 solution - v 1(5/8 =-4 or v 1 =-32/5 so...

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For the familiar circuit at right, (1) Write KCL for a supernode including both voltage supplies, supplies using v 1 , v 2 and v 3 , as variables. (2) Write the same equation using only v 1 as a variable, then solve for v 1 . (3) For the circuit below, where the current supply has been replaced with a 4V voltage supply, write KCL for a supernode including all three voltage supplies, using v 1 as variable. QUIZ 3 ECE35 Win’10 NAME _______________________ SID ______________________ Solution v 1 /4 + v 2 /8 + v 3 /4 – 2 + 2 = 0 v 1 /4 + (v 1 +8)/8 + (v 1 +12)/4 = 0, or, collecting terms, v 1 (1/4+1/8+1/4) = -4, so v 1 = 0V
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Unformatted text preview: v 1 (5/8) = -4, or v 1 = -32/5 , so v 1 = - 6.4V 8V +-12V 4V 4 Ω 8 Ω 4 Ω v 3 v 1 v 2 The voltages at the three lines leading out of the supernode haven’t changed relative to v 1 . The only difference is the current bridging the supernode above (which cancelled) is gone. So: v 1 /4 + (v 1 +8)/8 + (v 1 +12)/4 = 0 (no change) 8V 12V 2A 4 Ω 8 Ω 4 Ω v 3 v 1 v 2 One comment: 4V, pointed the way it is, was the only voltage setting that would have satisfied KVL around the outer loop – it was the only voltage setting that would have been physically possible....
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