Lemuel physics 2 9

Lemuel physics 2 9 - Wong Lemuel – Homework 9 – Due Apr...

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Unformatted text preview: Wong, Lemuel – Homework 9 – Due: Apr 2 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE: HW is due 11pm central time, which is midnight our time. 001 (part 1 of 3) 10 points 5 . 9 V 2 . 3 V 4 . 1 V I 1 . 2 Ω 2 . 1 Ω I 2 6 Ω I 3 8 . 2 Ω Find the current I 1 in the 0 . 2 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 1 . 01173 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation of Charge) I 1 + I 2- I 3 = 0 . (1) Kirchhoff’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoff’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 0 . 2 Ω , R B = 2 . 1 Ω , R C = 6 Ω , R D = 8 . 2 Ω , E 1 = 5 . 9 V , E 2 = 2 . 3 V , and E 3 = 4 . 1 V . Using determinants, I 1 = fl fl fl fl fl fl 1- 1 E 1 + E 2 R D E 3 R C R D fl fl fl fl fl fl fl fl fl fl fl fl 1 1- 1 R A + R B R D R C R D fl fl fl fl fl fl Expanding along the first row, the numera- tor is D 1 = fl fl fl fl fl fl 1- 1 E 1 + E 2 R D E 3 R C R D fl fl fl fl fl fl = 0- 1 fl fl fl fl E 1 + E 2 R D E 3 R D fl fl fl fl + (- 1) fl fl fl fl E 1 + E 2 E 3 R C fl fl fl fl =- [( E 1 + E 2 ) R D-E 3 R D ]- [ R C ( E 1 + E 2 )- 0] = R D ( E 3-E 1-E 2 )- R C ( E 1 + E 2 ) = (8 . 2 Ω)(4 . 1 V- 5 . 9 V- 2 . 3 V)- (6 Ω)(5 . 9 V + 2 . 3 V) =- 82 . 82 V Ω . Expanding along the first column, the de- nominator is D = fl fl fl fl fl fl 1 1- 1 R A + R B R D R C R D fl fl fl fl fl fl = 1 fl fl fl fl R D R C R D fl fl fl fl- ( R A + R B ) fl fl fl fl 1- 1 R C R D fl fl fl fl + 0 = 0- R C R D- ( R A + R B )( R D + R C ) = (6 Ω)(8 . 2 Ω)- (0 . 2 Ω + 2 . 1 Ω)(8 . 2 Ω + 6 Ω) =- 81 . 86 Ω 2 , and Wong, Lemuel – Homework 9 – Due: Apr 2 2007, 11:00 pm – Inst: Matt Mackie 2 I 1 = D 1 D =- 82 . 82 V Ω- 81 . 86 Ω 2 = 1 . 01173 A . 002 (part 2 of 3) 10 points Determine I 2 . Correct answer:- . 295505 A. Explanation: Using determinants, I 2 = fl fl fl fl fl fl 1- 1 R A + R B E 1 + E 2 R D E 3 R D fl fl fl fl fl fl D Expanding the numerator along the first row, D 2 = fl fl fl fl fl fl 1- 1 R A + R B E 1 + E 2 R D E 3 R D fl fl fl fl fl fl = 1 fl fl fl fl E 1 + E 2 R D E 3 R D fl fl fl fl- + (- 1) fl fl fl fl R A + R B E 1 + E 2 E 3 fl fl fl fl = R D ( E 1 + E 2 )- R D E 3- [ E 3 ( R A + R B )- 0] = (8 . 2 Ω)(5 . 9 V + 2 . 3 V)- (8 . 2 Ω)(4 . 1 V)- (4 . 1 V)(0 . 2 Ω + 2 . 1 Ω) = 24 . 19 V Ω , so I 2 = D 2 D = 24 . 19 V Ω- 81 . 86 Ω 2 =- . 295505 A . 003 (part 3 of 3) 10 points Determine I 3 . Correct answer: 0 . 716223 A....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Lemuel physics 2 9 - Wong Lemuel – Homework 9 – Due Apr...

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