Circuits-solutions

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

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R + R i = 116.1 ?) (5.77 V) 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 - 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: 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 . Let : E = 10.1 V , I = 1.57 A , and R i = 1.1 Ω . Thus the voltage across the resistance R is V = I R = R E 6.1 Ω + 0.388 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 = 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 ΔV 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 = 0.2 A . 005 (part 2 of 3) 10.0 points Find the potential difference across the first resistor.
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Correct answer: 5 V. Explanation: gilvin (jg47854) - 13. Circuits - meyers - (21235) 2 Before the wire is connected, I A = I B = V 2 R , so that ΔV 1 = I 1 R 1 = (0.2 A)(25 Ω) = 5 V . ( P A = ) 2 V 2 R · R = V 2 4 R . 006 (part 3 of 3) 10.0 points Find the potential difference across the second resistor. Correct answer: 1 V. After the wire is connected, I A = R and I B = 0 , so ( ) 2 Explanation: ΔV 2 = I 2 R 2 P ′ A = V R · R = V 2 R = 4 P A . = (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. A B 008 (part 2 of 2) 10.0 points and bulb B 1. will burn twice as brightly as before. 2. will burn half as brightly as before. 3. will burn nearly four times as brightly as before. 4. will burn as brightly as before. E Bulb A 1. will burn half as brightly as before. 2. will burn twice as brightly as before. 3.
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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.

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

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