Exam_02-solutions - Version 150 Exam 02 Chiu(58295 This...

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Version 150 – Exam 02 – Chiu – (58295) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A 14 . 6 Ω metal wire is cut into three equal pieces that are then connected side by side to Form a new wire the length oF which is equal to one-third the original length. What is the resistance oF this new wire? 1. 0.733333 2. 0.655556 3. 0.844444 4. 1.81111 5. 0.988889 6. 0.822222 7. 1.62222 8. 1.35556 9. 1.43333 10. 1.02222 Correct answer: 1 . 62222 Ω. Explanation: Let : R 0 = 14 . 6 Ω , = 1 3 0 , and A = 3 A 0 . The original resistance was R 0 = ρ 0 A 0 and the new resistance is R = ρ 0 3 3 A 0 = 1 9 R 0 = 1 9 (14 . 6 Ω) = 1 . 62222 Ω . 002 10.0 points A current oF 4 A ±ows in a copper wire 4 mm in diameter. The density oF valence electrons in copper is roughly 9 × 10 28 m 3 . ²ind the driFt speed oF these electrons. 1. 1.65786e-05 2. 1.38155e-06 3. 8.84194e-05 4. 8.84194e-07 5. 1.76839e-06 6. 7.36828e-06 7. 5.52621e-06 8. 2.21049e-05 9. 4.42097e-05 10. 2.45609e-06 Correct answer: 2 . 21049 × 10 5 m / s. Explanation: Let : I = 4 A , r = 4 mm 2 = 0 . 002 m , and n e = 9 × 10 28 m 3 . I = j ( π r 2 ) = e n e v d ( π r 2 ) v d = I e n e π r 2 = 4 A (1 . 6 × 10 19 C) (9 × 10 28 m 3 ) × 1 π (0 . 002 m) 2 = 2 . 21049 × 10 5 m / s = 7 . 95775 cm / hours . 003 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 AFter the wire is connected across B , 1. bulb A will burn as brightly as beFore and bulb B will go out. 2. bulb A will burn twice as brightly as
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Version 150 – Exam 02 – Chiu – (58295) 2 before and bulb B will burn as brightly as before. 3. bulb A will go out and bulb B will burn as brightly as before. 4. bulb A will burn twice as brightly as before and bulb B will burn half as brightly as before. 5. bulb A will burn twice as brightly as before and bulb B will go out. 6. bulb A will go out and bulb B will go out. 7. bulb A will burn four times as brightly as before and bulb B will go out. correct 8. bulb A will burn as brightly as before and bulb B will burn half as brightly as before. 9. bulb A will go out and bulb B will burn half as brightly as before. 10. bulb A will burn as brightly as before and bulb B will burn as brightly as before. Explanation: The electric power is given by P = I 2 R = V 2 R . Before the wire is connected, I A = I B = V 2 R P A = V 2 4 R . After the wire is connected, only bulb A is in the circuit, so I A = V R I B = 0 P A = V 2 R = 4 P A , therefore bulb A will burn four times as brightly as before. Since there is no potential diFerence between the two ends of bulb B , it goes oF. 004 10.0 points E A B C D E Rank the brightness of the identical bulbs in the following circuit. Hint: You may ±nd it helpful to work out the currents through the bulbs for the case V = 1 V , and R = 1 Ω for all the bulbs, then compare the currents. 1. A = B = C > D = E 2. A = C > B > D = E 3. E = D > A > B = C 4. A = B > C > D = E 5. A = B = C = D = E 6. A = B = C > D > E 7. C > B > A > D > E 8. A = D = E > B > C 9. A = D = E > B = C correct 10. B = C > A = D = E Explanation: ²rom the ±gure, with all the bulbs having the resistance R , I A = I D = I E = V R
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Version 150 – Exam 02 – Chiu – (58295) 3 I B = I C = V 2 R The rankings of the currents shown here are
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Exam_02-solutions - Version 150 Exam 02 Chiu(58295 This...

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