hw14-solutions

hw14-solutions - marout(ssm979 – hw13 – turner...

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Unformatted text preview: marout (ssm979) – hw13 – turner – (56725) 1 This print-out should have 17 questions. Multiple-choice 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 8 V 13 V I 1 3 . 4 Ω 4 . 6 Ω I 3 4 . 9 Ω I 2 2 . 7 Ω 4 . 3 Ω Find the current through the Amp meter, I 3 . Correct answer: − . 370656 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 = 8 V , E 2 = 13 V , r 1 = 3 . 4 Ω , r 2 = 4 . 6 Ω , r 3 = 4 . 9 Ω , r 4 = 2 . 7 Ω , and r 5 = 4 . 3 Ω . We consider R 1 = r 1 + r 2 = 3 . 4 Ω + 4 . 6 Ω = 8 Ω , and R 2 = r 4 + r 5 = 2 . 7 Ω + 4 . 3 Ω = 7 Ω . From the junction rule, I 1 = I 2 + I 3 . Applying Kirchhoff’s 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 ) = (7 Ω) (4 . 9 Ω) +(8 Ω) (4 . 9 Ω + 7 Ω) = 129 . 5 Ω , then I 3 = E 1 R 2 −E 2 R 1 r 3 R 2 + R 1 ( R 2 + r 3 ) (5) = (8 V) (7 Ω) − (13 V) (8 Ω) 129 . 5 Ω = − . 370656 A . 002 (part 2 of 3) 10.0 points Find the current I 1 . Correct answer: 1 . 22703 A. Explanation: From Eq. 1 and I 3 from Eq. 5, we have I 1 = E 1 − I 3 r 3 R 1 = 8 V − ( − . 370656 A) (4 . 9 Ω) 8 Ω = 1 . 22703 A . 003 (part 3 of 3) 10.0 points Find the current I 2 . marout (ssm979) – hw13 – turner – (56725) 2 Correct answer: 1 . 59768 A. Explanation: From Kirchhoff’s junction rule, I 2 = I 1 − I 3 = 1 . 22703 A − − . 370656 A = 1 . 59768 A . 004 (part 1 of 4) 10.0 points An energy plant produces an output potential of 1900 kV and serves a city 158 km away. A high-voltage transmission line carries 900 A to the city. The effective resistance of a trans- mission line [wire(s)] is 1 . 88 Ω / 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: 1632 . 66 kV. Explanation: Let : V plant = 1900 kV , ℓ = 158 km , I = 900 A , and ρ = 1 . 88 Ω / km . The potential drop on the wire is V = I R = I ρ ℓ = (900 A) (1 . 88 Ω / km) (158 km) = 267 . 336 kV , so the potential delivered to the city is V city = V plant − V wire = 1900 kV − 267 . 336 kV = 1632 . 66 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 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas.

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hw14-solutions - marout(ssm979 – hw13 – turner...

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