ECE35 WIN10 Quiz 3 solution

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

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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 8 4 v 3 v 1 v 2 The voltages at the three lines leading out of the supernode havent 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 8 4 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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