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Quiz1soln2 - MAE140 Linear Circuits Quiz # 1 - October 17,...

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MAE140 Linear Circuits Quiz # 1 --- October 17, 2002 Solutions Question 1 --- Circuit Concepts (a) Begin by computing the current and voltage of the load resistors. R 1 = 40 W : P = 40 W , i = 1 A , v = 40 V R 1 = 10 W : P = 40 W , i = 2 A , v = 20 V Now use the circuit properties. Do this first with the Norton-type circuit on the left. i eq = ( G 1 + G eq ) v = ( 1 40 + G eq )40 = 40 G eq + 1, case R 1 = 40 W i eq = ( 1 10 + G eq ) v = ( 1 10 + G eq )20 = 20 G eq + 2, case R 1 = 10 W Subtract second equation from first: G eq = 1 20 S , R eq = 20 W Solve for i eq = 3 A . Using the Thévenin variant on the right yields a different set of equations. 40 = 40 R eq + 40 v eq or 40 R eq + 1600 = 40 v eq or R eq + 40 = v eq , case R 1 = 40 W 20 = 10 R eq + 10 v eq or 20 R eq + 200 = 10 v eq or 2 R eq + 20 = v eq , case R 1 = 10 W Subtract first equation from second: R eq = 20 W . Then solve for v eq = 60 V . Of course one solution is transformable into the other using a source transformation. (b) 1. Transform the left (20V-100 W ) pair into their Norton (200mA-100 W ) equivalent. Now rearrange the parallel elements. 100 W 200mA 100 40 40 200mA 100 200mA 100 40 40 200mA
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By inspection or calculation (use nodal analysis to show that the voltage at the top node is 0V), we see that the current in all of the resistive elements is 0A. The current
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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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Quiz1soln2 - MAE140 Linear Circuits Quiz # 1 - October 17,...

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