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Unformatted text preview: Reilly (mar3978) hw13 turner (56725) 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 21 V 15 V I 1 3 . 6 3 . 2 I 3 3 I 2 4 . 9 5 . 3 Find the current through the Amp meter, I 3 . Correct answer: 0 . 932203 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 = 21 V , E 2 = 15 V , r 1 = 3 . 6 , r 2 = 3 . 2 , r 3 = 3 , r 4 = 4 . 9 , and r 5 = 5 . 3 . We consider R 1 = r 1 + r 2 = 3 . 6 + 3 . 2 = 6 . 8 , and R 2 = r 4 + r 5 = 4 . 9 + 5 . 3 = 10 . 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 ) = (10 . 2 ) (3 ) +(6 . 8 ) (3 + 10 . 2 ) = 120 . 36 , then I 3 = E 1 R 2 E 2 R 1 r 3 R 2 + R 1 ( R 2 + r 3 ) (5) = (21 V) (10 . 2 ) (15 V) (6 . 8 ) 120 . 36 = . 932203 A . 002 (part 2 of 3) 10.0 points Find the current I 1 . Correct answer: 2 . 67697 A. Explanation: From Eq. 1 and I 3 from Eq. 5, we have I 1 = E 1 I 3 r 3 R 1 = 21 V (0 . 932203 A) (3 ) 6 . 8 = 2 . 67697 A . 003 (part 3 of 3) 10.0 points Find the current I 2 . Reilly (mar3978) hw13 turner (56725) 2 Correct answer: 1 . 74477 A. Explanation: From Kirchhoffs junction rule, I 2 = I 1 I 3 = 2 . 67697 A . 932203 A = 1 . 74477 A . 004 (part 1 of 4) 10.0 points An energy plant produces an output potential of 2800 kV and serves a city 105 km away. A highvoltage transmission line carries 1300 A to the city. The effective resistance of a trans mission line [wire(s)] is 2 . 85 / 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: 2410 . 97 kV. Explanation: Let : V plant = 2800 kV , = 105 km , I = 1300 A , and = 2 . 85 / km . The potential drop on the wire is V = I R = I = (1300 A) (2 . 85 / km) (105 km) = 389 . 025 kV , so the potential delivered to the city is V city = V plant V wire = 2800 kV 389 . 025 kV = 2410 . 97 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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 Spring '10
 TURNER,J

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