PHY303LEXAM2

# PHY303LEXAM2 - Version 162 – Exam 02 – Chiu –(58295 1...

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Unformatted text preview: Version 162 – Exam 02 – Chiu – (58295) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 13 . 1 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. 1.52222 2. 1.43333 3. 1.32222 4. 1.62222 5. 1.47778 6. 1.24444 7. 1.38889 8. 1.5 9. 1.35556 10. 1.45556 Correct answer: 1 . 45556 Ω. Explanation: Let : R = 13 . 1 Ω , ℓ = 1 3 ℓ , and A = 3 A . The original resistance was R = ρ ℓ A and the new resistance is R = ρ ℓ 3 3 A = 1 9 R = 1 9 (13 . 1 Ω) = 1 . 45556 Ω . 002 10.0 points A current of 5 A flows in a copper wire 8 mm in diameter. The density of valence electrons in copper is roughly 9 × 10 28 m − 3 . Find the drift speed of these electrons. 1. 7.36828e-06 2. 8.84194e-05 3. 6.63146e-05 4. 5.52621e-06 5. 4.14466e-06 6. 0.000110524 7. 6.90777e-06 8. 9.82438e-06 9. 2.21049e-05 10. 1.76839e-06 Correct answer: 6 . 90777 × 10 − 6 m / s. Explanation: Let : I = 5 A , r = 8 mm 2 = 0 . 004 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 = 5 A (1 . 6 × 10 − 19 C)(9 × 10 28 m − 3 ) × 1 π (0 . 004 m) 2 = 6 . 90777 × 10 − 6 m / s = 2 . 4868 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 twice as brightly as before and bulb B will burn as brightly as before. Version 162 – Exam 02 – Chiu – (58295) 2 2. bulb A will go out and bulb B will burn as brightly as before. 3. bulb A will burn as brightly as before and bulb B will burn as brightly as before. 4. bulb A will burn twice as brightly as before and bulb B will go out. 5. bulb A will burn as brightly as before and bulb B will go out. 6. bulb A will burn four times as brightly as before and bulb B will go out. correct 7. bulb A will go out and bulb B will go out. 8. bulb A will burn as brightly as before and bulb B will burn half as brightly as before. 9. bulb A will burn twice as brightly as before and bulb B will burn half as brightly as before. 10. bulb A will go out and bulb B will burn half 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 difference between the two ends of bulb B , it goes off....
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PHY303LEXAM2 - Version 162 – Exam 02 – Chiu –(58295 1...

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