Q2soln - MAE140 Linear Circuits Quiz 2 Answers Question...

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# 2 --- February 21, 2002 Answers Question 1 (Thomas & Rosa 4-7) Nodal analysis yields: Node C: v D =v S Node D: grounded v D =0 Node A: v A =v X 0 10 10 2 1 10 10 1 10 10 1 10 4 1 10 2 1 3 3 3 3 3 = + × × × + × + × A S B A v v v v Node B: 0 10 10 20 1 10 10 1 10 10 1 3 3 3 = × + × + × A B A v v v This may be jointly written as S B A S B A v v v v v v = × = × × × × 0 10 3 001 , 200 2 017 , 200 0 10 20 10 10 20 3 10 20 002 , 200 10 20 2 10 20 017 , 200 3 3 3 3 3 Now note that v B =v O . So S S S B O v v v v v 10 9978 . 9 3 002 , 200 2 017 , 200 0 002 , 200 10 017 , 200 = = = Question 2 (Thomas & Rosa 4-34) (a) Node B: () . 0 , 0 1 2 1 2 = + + = C B A n v G v G G
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This note was uploaded on 01/17/2009 for the course MAE MAE 140 taught by Professor Mauricio during the Fall '04 term at UCSD.

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Q2soln - MAE140 Linear Circuits Quiz 2 Answers Question...

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