16 - fryer(vdf96 – Circuits 2 – graves –(6 1 This...

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Unformatted text preview: fryer (vdf96) – Circuits 2 – graves – (6) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points In the circuit shown below, A , B , C , and D are identical light bulbs. Assume that the battery maintains a con- stant potential difference between its termi- nals ( i.e. , the internal resistance of the battery is assumed to be negligible) and the resistance of each lightbulb remains constant. E A C B D Which of the following is the correct rela- tionships of the brightnesses ( i.e. , the power consumed by) of the light bulbs? 1. P B = P C > P A > P D 2. P A > P B = P C > P D 3. P A > P B > P C > P D 4. P D > P B = P C > P A 5. P A > P D > P B = P C correct 6. P D > P A > P B = P C Explanation: The circuit diagram is as follows. A D B C E Observing the currents I A = I D + I BC , so I A > I D and P A = I 2 A R > I 2 D R = P D . Observing potential difference relationships, V D = V B + V C = 2 V B = 2 V C . Therefore V D > V B and P D = V 2 D R > V 2 B R = P B = P C . Thus P A > P D > P B = P C . 002 (part 2 of 4) 10.0 points If the emf of the battery is 16 V, and each resistance is 2 Ω, what is the power consumed by bulb B ? Correct answer: 5 . 12 W. Explanation: Let : E = 16 V and R = 2 Ω . R B and R C are in series, so R BC = R B + R C = 2 R. R BC and R D are in parallel, so R BCD = R BC R D R BC + R D = 2 R 3 . R A and R BCD are in series, so R ABCD = R A + R BCD = 5 R 3 . The current in the battery (and through R A ) is I A = E R ABCD = 3 E 5 R . R BC = 2 R D so I D = 2 I BC . fryer (vdf96) – Circuits 2 – graves – (6) 2 I BC = I A 3 = E 5 R , so P B = I 2 BC R B = parenleftbigg E 5 R parenrightbigg 2 R = E 2 25 R = (16 V) 2 25 (2 Ω) = 5 . 12 W . 003 (part 3 of 4) 10.0 points Bulb D is then removed from its socket. How does the brightness of bulb A change? 1. The brightness of bulb A cannot be de- termined. 2. The brightness of bulb A increases. 3. The brightness of bulb A remains the same. 4. The brightness of bulb A decreases. cor- rect Explanation: When bulb D is removed from its socket, the circuit becomes serial connection of bulb A , B , and C . R ABC = R A + R B + R C = 3 R I ′ A = E 3 R . I ′ A is less than I A = 3 E 5 R , so the brightness of bulb A decreases. 004 (part 4 of 4) 10.0 points How does the brightness of bulb B change when bulb D is removed from its socket? 1. The brightness of bulb B remains the same. 2. The brightness of bulb B cannot be de- termined. 3. The brightness of bulb B decreases. 4. The brightness of bulb B increases. cor- rect Explanation: Since bulbs A , B , and C are in serial con- nection, I ′ B = I ′ A = E 3 R ....
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This note was uploaded on 12/11/2011 for the course PHYSICS 101 taught by Professor Graves during the Fall '10 term at San Jose City College.

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16 - fryer(vdf96 – Circuits 2 – graves –(6 1 This...

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