homework11 - Rodriguez, Brandon Homework 11 Due: Sep 28...

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Unformatted text preview: Rodriguez, Brandon Homework 11 Due: Sep 28 2006, 11:00 am Inst: Andrew J Rader 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 8 points 4 . 4 3 . 4 15 . 6 26 . 8 V 13 . 4 V Find the current through the 15 . 6 lower- right resistor. Correct answer: 1 . 09836 A. Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I Let : E 1 = 26 . 8 V , E 2 = 13 . 4 V , r 1 = 4 . 4 , r 2 = 3 . 4 , and R = 15 . 6 . From the junction rule, I = i 1 + i 2 . Applying Kirchhoffs loop rule, we obtain two equations: E 1 = i 1 r 1 + I R (1) E 2 = i 2 r 2 + I R = ( I- i 1 ) r 2 + I R =- i 1 r 2 + I ( R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = i 1 r 1 r 2 + r 2 I R E 2 r 1 =- i 1 r 1 r 2 + I r 1 ( R + r 2 ) Adding, E 1 r 2 + E 2 r 1 = I [ r 2 R + r 1 ( R + r 2 )] I = E 1 r 2 + E 2 r 1 r 2 R + r 1 ( R + r 2 ) = (26 . 8 V) (3 . 4 ) + (13 . 4 V) (4 . 4 ) (3...
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.

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homework11 - Rodriguez, Brandon Homework 11 Due: Sep 28...

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