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Unformatted text preview: malsam (wgm329) hw13 turner (56705) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Consider the circuit a b A 11 V 10 V I 1 4 . 7 4 I 3 5 I 2 4 . 1 5 . 1 Find the current through the Amp meter, I 3 . Correct answer: 0 . 083756 A. Explanation: a b A E 1 E 2 I 1 r 1 r 2 I 3 r 3 I 2 r 4 r 5 Let : E 1 = 11 V , E 2 = 10 V , r 1 = 4 . 7 , r 2 = 4 , r 3 = 5 , r 4 = 4 . 1 , and r 5 = 5 . 1 . We consider R 1 = r 1 + r 2 = 4 . 7 + 4 = 8 . 7 , and R 2 = r 4 + r 5 = 4 . 1 + 5 . 1 = 9 . 2 . From the junction rule, I 1 = I 2 + I 3 . Applying Kirchhoffs loop rule, we obtain two equations. E 1 I 1 R 1 I 3 r 3 = 0 E 1 = I 1 R 1 + I 3 r 3 (1) E 2 I 3 r 3 + I 2 R 2 = 0 E 2 = I 2 R 2 I 3 r 3 = ( I 1 I 3 ) R 2 I 3 r 3 = I 1 R 2 I 3 ( R 2 + r 3 ) , (2) Multiplying Eq. (1) by R 2 and Eq. (2) by R 1 and adding, E 1 R 2 = I 1 R 1 R 2 + I 3 r 3 R 2 (3) E 2 R 1 = I 1 R 1 R 2 + I 3 R 1 ( R 2 + r 3 ) (4) E 1 R 2 E 2 R 1 = I 3 [ r 3 R 2 + R 1 ( R 2 + r 3 )] . Since r 3 R 2 + R 1 ( R 2 + r 3 ) = (9 . 2 ) (5 ) +(8 . 7 ) (5 + 9 . 2 ) = 169 . 54 , then I 3 = E 1 R 2 E 2 R 1 r 3 R 2 + R 1 ( R 2 + r 3 ) (5) = (11 V) (9 . 2 ) (10 V) (8 . 7 ) 169 . 54 = . 083756 A . 002 (part 2 of 3) 10.0 points Find the current I 1 . Correct answer: 1 . 21623 A. Explanation: From Eq. 1 and I 3 from Eq. 5, we have I 1 = E 1 I 3 r 3 R 1 = 11 V (0 . 083756 A) (5 ) 8 . 7 = 1 . 21623 A . 003 (part 3 of 3) 10.0 points Find the current I 2 . malsam (wgm329) hw13 turner (56705) 2 Correct answer: 1 . 13248 A. Explanation: From Kirchhoffs junction rule, I 2 = I 1 I 3 = 1 . 21623 A . 083756 A = 1 . 13248 A . 004 (part 1 of 4) 10.0 points An energy plant produces an output potential of 1600 kV and serves a city 88 km away. A highvoltage transmission line carries 1520 A to the city. The effective resistance of a trans mission line [wire(s)] is 1 . 74 / km times the distance from the plant to the city. What is the potential provided to the city, i.e. , at the end of the transmission line? Correct answer: 1367 . 26 kV. Explanation: Let : V plant = 1600 kV , = 88 km , I = 1520 A , and = 1 . 74 / km . The potential drop on the wire is V = I R = I = (1520 A) (1 . 74 / km) (88 km) = 232 . 742 kV , so the potential delivered to the city is V city = V plant V wire = 1600 kV 232 . 742 kV = 1367 . 26 kV . 005 (part 2 of 4) 10.0 points How much power is dissipated due to resistive losses in the transmission line?...
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This note was uploaded on 11/22/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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