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# solution_pdf 13 - malsam(wgm329 hw13 turner(56705 This...

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malsam (wgm329) – hw13 – turner – (56705) 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 11 V 10 V I 1 4 . 7 Ω 4 Ω I 3 5 Ω I 2 4 . 1 Ω 5 . 1 Ω 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 Ω , r 2 = 4 Ω , r 3 = 5 Ω , r 4 = 4 . 1 Ω , and r 5 = 5 . 1 Ω . We consider R 1 = r 1 + r 2 = 4 . 7 Ω + 4 Ω = 8 . 7 Ω , and R 2 = r 4 + r 5 = 4 . 1 Ω + 5 . 1 Ω = 9 . 2 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 ) = (9 . 2 Ω) (5 Ω) +(8 . 7 Ω) (5 Ω + 9 . 2 Ω) = 169 . 54 Ω , 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 Ω) (10 V) (8 . 7 Ω) 169 . 54 Ω = 0 . 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 Ω) 8 . 7 Ω = 1 . 21623 A . 003 (part 3 of 3) 10.0 points Find the current I 2 .

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malsam (wgm329) – hw13 – turner – (56705) 2 Correct answer: 1 . 13248 A. Explanation: From Kirchhoff’s junction rule, I 2 = I 1 I 3 = 1 . 21623 A 0 . 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 high-voltage transmission line carries 1520 A to the city. The effective resistance of a trans- mission line [wire(s)] is 1 . 74 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: 1367 . 26 kV. Explanation: Let : V plant = 1600 kV , = 88 km , I = 1520 A , and ρ = 1 . 74 Ω / km . The potential drop on the wire is V = I R = I ρ ℓ = (1520 A) (1 . 74 Ω / 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? Correct answer: 3 . 53768 × 10 8 W. Explanation: P = I 2 R = I 2 ρ ℓ = (1520 A) 2 (1 . 74 Ω / km) (88 km) = 3 . 53768 × 10 8 W . 006 (part 3 of 4) 10.0 points Assume the plant charges \$ 0 . 087 / kW · hr for 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: 30777 . 9 dollars / hr. Explanation: Let : R = \$ 0 . 087 / kW · hr . The cost of energy is Cost = P [( R ) (time)] = (3 . 53768 × 10 8 W) × (0 . 087 / kW · hr) (1 hr) parenleftbigg 1 kW 1000 W parenrightbigg = 30777 . 9 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 of power; however, the output electric potential of the energy plant is 20% greater.
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solution_pdf 13 - malsam(wgm329 hw13 turner(56705 This...

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