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Unformatted text preview: gilvin (jg47854) – 13. Circuits – meyers – (21235) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This homework is due Friday, 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 difference 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. Find 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 . For the series resistors R eq = R 1 + R 2 = 25 Ω + 5 Ω = 30 Ω . I 1 = I 2 = I = 6 V 30 Ω = . 2 A . 005 (part 2 of 3) 10.0 points Find the potential difference across the first resistor. 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 difference 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 half as brightly as before. 2. will burn twice as brightly as before. 3. will go out. 4. will burn as brightly as before. 5. will burn nearly four times as brightly as before. correct Explanation: The electric power is given by P = I 2 R . Before the wire is connected, I A = I B = V 2 R , so that P A = parenleftbigg V 2 R parenrightbigg 2 · R = V 2 4 R ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.
 Spring '11
 doc
 Magnetism, Work

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