13. Circuits-solutions

13. Circuits-solutions - gilvin(jg47854 13 Circuits...

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gilvin (jg47854) – 13. Circuits – meyers – (21235) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. This homework is due ±riday, march 25, at midnight Tucson time. 001 (part 1 of 2) 10.0 points A battery with an emF oF 10 . 1 V and internal resistance oF 1 . 1 Ω is connected across a load resistor R . IF the current in the circuit is 1 . 57 A, what is the value oF R ? Correct answer: 5 . 33312 Ω. Explanation: Let : E = 10 . 1 V , I = 1 . 57 A , and R i = 1 . 1 Ω . The electromotive Force E is given by E = I ( R + R i ) R = E I R i = 10 . 1 V 1 . 57 A 1 . 1 Ω = 5 . 33312 Ω . 002 (part 2 of 2) 10.0 points What power is dissipated in the internal re- sistance oF the battery? Correct answer: 2 . 71139 W. Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (1 . 57 A) 2 (1 . 1 Ω) = 2 . 71139 W . 003 10.0 points What potential di²erence is measured across a 16 . 1 Ω load resistor when it is connected across a battery oF emf 5 . 77 V and internal resistance 0 . 388 Ω? Correct answer: 5 . 63422 V. Explanation: Let : E = 5 . 77 V , R = 16 . 1 Ω , and R i = 0 . 388 Ω . The total resistance is R T = R + R i = E I , so I = E R + R i . Thus the voltage across the resistance R is V = I R = R E R + R i = (16 . 1 Ω) (5 . 77 V) 16 . 1 Ω + 0 . 388 Ω = 5 . 63422 V . 004 (part 1 of 3) 10.0 points A 25 Ω resistor and a 5 Ω resistor are con- nected in series to a 6 V battery. ±ind the current in each resistor. Correct answer: 0 . 2 A. Explanation: Let : R 1 = 25 Ω , R 2 = 5 Ω , and Δ V 2 = 6 V . ±or the series resistors R eq = R 1 + R 2 = 25 Ω + 5 Ω = 30 Ω . I 1 = I 2 = I = 6 V 30 Ω = 0 . 2 A . 005 (part 2 of 3) 10.0 points ±ind the potential di²erence across the frst resistor.
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gilvin (jg47854) – 13. Circuits – meyers – (21235) 2 Correct answer: 5 V. Explanation: Δ V 1 = I 1 R 1 = (0 . 2 A)(25 Ω) = 5 V . 006 (part 3 of 3) 10.0 points Find the potential diference across the second resistor. Correct answer: 1 V. Explanation: Δ V 2 = I 2 R 2 = (0 . 2 A)(5 Ω) = 1 V . 007 (part 1 of 2) 10.0 points Two identical light bulbs A and B are con- nected in series to a constant voltage source. Suppose a wire is connected across bulb B as shown. E A B Bulb A 1. will burn hal± as brightly as be±ore. 2. will burn twice as brightly as be±ore. 3. will go out. 4. will burn as brightly as be±ore. 5. will burn nearly ±our times as brightly as be±ore. correct Explanation: The electric power is given by P = I 2 R . Be±ore the wire is connected, I A = I B = V 2 R , so that P A = p V 2 R P 2 · R = V 2 4 R . A±ter the wire is connected, I A = V R and I B = 0 , so P A = p V R P 2 · R = V 2 R = 4 P A . 008 (part 2 of 2) 10.0 points and bulb B 1. will burn twice as brightly as be±ore. 2. will burn hal± as brightly as be±ore. 3. will burn nearly ±our times as brightly as be±ore. 4. will burn as brightly as be±ore. 5. will go out. correct Explanation: Since there is no potential diference be- tween the two ends o± bulb B, it goes out.
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13. Circuits-solutions - gilvin(jg47854 13 Circuits...

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