hw14-solutions

# hw14-solutions - Reilly(mar3978 – hw13 – turner...

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Unformatted text preview: Reilly (mar3978) – 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 21 V 15 V I 1 3 . 6 Ω 3 . 2 Ω I 3 3 Ω I 2 4 . 9 Ω 5 . 3 Ω 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 Ω , r 2 = 3 . 2 Ω , r 3 = 3 Ω , r 4 = 4 . 9 Ω , and r 5 = 5 . 3 Ω . We consider R 1 = r 1 + r 2 = 3 . 6 Ω + 3 . 2 Ω = 6 . 8 Ω , and R 2 = r 4 + r 5 = 4 . 9 Ω + 5 . 3 Ω = 10 . 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 ) = (10 . 2 Ω) (3 Ω) +(6 . 8 Ω) (3 Ω + 10 . 2 Ω) = 120 . 36 Ω , 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 Ω) − (15 V) (6 . 8 Ω) 120 . 36 Ω = . 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 Ω) 6 . 8 Ω = 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 Kirchhoff’s 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 high-voltage transmission line carries 1300 A to the city. The effective resistance of a trans- mission line [wire(s)] is 2 . 85 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: 2410 . 97 kV. Explanation: Let : V plant = 2800 kV , ℓ = 105 km , I = 1300 A , and ρ = 2 . 85 Ω / km . The potential drop on the wire is V = I R = I ρ ℓ = (1300 A) (2 . 85 Ω / 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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hw14-solutions - Reilly(mar3978 – hw13 – turner...

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