hw14-solutions

# Hw14-solutions - howard(cah3459 hw13 turner(56705 This print-out should have 17 questions Multiple-choice questions may continue on the next column

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howard (cah3459) – hw13 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 10.0 points Consider the circuit a b A 26 V 9V I 1 3 . 6 Ω 2 . 5 Ω I 3 3 . 8 Ω I 2 3 . 7 Ω ±ind the current through the Amp meter, I 3 . Correct answer: 1 . 65506 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 = 26 V , E 2 = 9 V , r 1 =3 . , r 2 =2 . , r 3 . , r 4 . , and r 5 = 6 Ω . We consider R 1 = r 1 + r 2 . 6 Ω + 2 . 5 Ω = 6 . 1 Ω , and R 2 = r 4 + r 5 . 7 Ω + 6 Ω = 9 . . ±rom the junction rule, I 1 = I 2 + I 3 . Applying Kirchho²’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 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 . 7 Ω) (3 . 8 Ω) +(6 . 1 Ω) (3 . 8 Ω + 9 . 7 Ω) = 119 . 21 Ω , then I 3 = E 1 R 2 2 R 1 r 3 R 2 + R 1 ( R 2 + r 3 ) (5) = (26 V) (9 . 7 Ω) - (9 V) (6 . 1 Ω) 119 . 21 Ω = 1 . 65506 A . 002 (part 2 of 3) 10.0 points ±ind the current I 1 . Correct answer: 3 . 23127 A. Explanation: ±rom Eq. 1 and I 3 From Eq. 5, we have I 1 = E 1 - I 3 r 3 R 1 = 26 V - (1 . 65506 A) (3 . 8 Ω) 6 . = 3 . 23127 A . 003 (part 3 of 3) 10.0 points ±ind the current I 2 .

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howard (cah3459) – hw13 – turner – (56705) 2 Correct answer: 1 . 57621 A. Explanation: From Kirchhof’s junction rule, I 2 = I 1 - I 3 =3 . 23127 A - 1 . 65506 A = 1 . 57621 A . 004 (part 1 of 4) 10.0 points An energy plant produces an output potential o± 17000 kV and serves a city 222 km away. A high-voltage transmission line carries 1180 A to the city. The efective resistance o± a trans- mission line [wire(s)] is 4 . 09 Ω / km times the distance ±rom the plant to the city. What is the potential provided to the city, i.e. , at the end o± the transmission line? Correct answer: 15928 . 6 kV. Explanation: Let : V plant = 17000 kV , ± = 222 km , I = 1180 A , and ρ =4 . 09 Ω / km . The potential drop on the wire is V = IR = I ρ ± = (1180 A) (4 . 09 Ω / km) (222 km) = 1071 . 42 kV , so the potential delivered to the city is V city = V plant - V wire = 17000 kV - 1071 . 42 kV = 15928 . 6 kV . 005 (part 2 of 4) 10.0 points How much power is dissipated due to resistive losses in the transmission line? Correct answer: 1 . 26427 × 10 9 W. Explanation: P = I 2 R = I 2 ρ± = (1180 A) 2 (4 . 09 Ω / km) (222 km) = 1 . 26427 × 10 9 W . 006 (part 3 of 4) 10.0 points Assume the plant charges \$ 0 . 117 / kW · hr ±or electric energy. At this rate, how much does it cost to trans- mit energy to the city (by the transmission line heating the atmosphere) each hour? Correct answer: 1 . 4792 × 10 5 dollars / hr. Explanation: Let : R = \$ 0 . 117 / kW · hr . The cost o± energy is Cost = P [( R ) (time)] = (1 . 26427 × 10 9 W) × (0 . 117 / kW · hr) (1 hr) ± 1 kW 1000 W ² = 1 . 4792 × 10 5 dollars / hr . 007 (part 4 of 4) 10.0 points Consider the money lost by the transmission line heating the atmosphere each hour. As- sume the energy plant produces the same amount o± power; however, the output electric potential o± the energy plant is 20% greater.
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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Hw14-solutions - howard(cah3459 hw13 turner(56705 This print-out should have 17 questions Multiple-choice questions may continue on the next column

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