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quiz2 - Daniels Matthew Quiz 2 Due 11:00 pm Inst Ditmire...

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Daniels, Matthew – Quiz 2 – Due: Oct 19 2006, 11:00 pm – Inst: Ditmire 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A capacitor that is connected to a 40 V source contains 95 μ C of charge. What is its capacitance? Correct answer: 2 . 375 μ F. Explanation: Let : V = 40 V and q = 95 μ C . The capacitance is C = q V = 95 μ C 40 V = 2 . 375 μ F keywords: 002 (part 1 of 1) 10 points Find the equivalent resistance between points a and b in the figure. 4 Ω 2 . 8 Ω 4 . 9 Ω 5 . 1 Ω 4 . 6 Ω a b Correct answer: 11 . 0728 Ω. Explanation: r 1 r 2 r 3 r 4 r 5 a b Let : r 1 = 4 Ω , r 2 = 2 . 8 Ω , r 3 = 4 . 9 Ω , r 4 = 5 . 1 Ω , and r 5 = 4 . 6 Ω . r 4 and r 5 are in series, so r 45 = r 4 + r 5 r 1 r 2 r 3 r 45 a b r 2 and r 45 are in parallel, so 1 r 245 = 1 r 2 + 1 r 45 = r 2 + r 45 r 2 r 45 r 245 = r 2 r 45 r 2 + r 45 = r 2 ( r 4 + r 5 ) r 2 + r 4 + r 5 r 1 r 245 r 3 a b r 1 , r 245 and r 3 are in series, so R ab = r 1 + r 245 + r 3 = r 1 + r 2 ( r 4 + r 5 ) r 2 + r 4 + r 5 + r 3 = 4 Ω + (2 . 8 Ω)(5 . 1 Ω + 4 . 6 Ω) 2 . 8 Ω + 5 . 1 Ω + 4 . 6 Ω + 4 . 9 Ω = 11 . 0728 Ω .
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Daniels, Matthew – Quiz 2 – Due: Oct 19 2006, 11:00 pm – Inst: Ditmire 2 keywords: 003 (part 1 of 2) 10 points A coaxial cable with length has an inner conductor that has a radius a and carries a charge of Q . The surrounding conductor has an inner radius b and a charge of - Q . Assume the region between the conductors is air. The linear charge density λ Q . radius = a + Q radius = b - Q What is the electric field halfway between the conductors? 1. E = Q 2 π ² 0 r 2 2. E = λ 4 π ² 0 r 3. E = λ 2 π ² 0 r correct 4. E = Q 2 π ² 0 r 5. E = λ 2 π ² 0 r 2 6. E = Q π ² 0 r 2 7. E = λ π ² 0 r 2 8. E = Q 4 π ² 0 r 9. E = Q π ² 0 r 10. E = λ π ² 0 r Explanation: The electric field of a cylindrical capacitor is given by E = λ 2 π r ² 0 . 004 (part 2 of 2) 10 points What is the capacitance C of this coaxial cable? 1. C = k e ln b a 2. C = ‘ a 2 k e b 3. C = k e 4. C = 2 k e ln b a correct 5. C = k e ln b a 6. C = 2 k e ln b a 7. C = 2 k e 8. C = k e ln a b · 9. C = k e 2 ln b a 10. C = 2 k e ln a b · Explanation: C = 2 k e ln b a . keywords: 005 (part 1 of 3) 10 points See the circuit below.
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Daniels, Matthew – Quiz 2 – Due: Oct 19 2006, 11:00 pm – Inst: Ditmire 3 75 V 3 V X Y 3 . 5 Ω 1 . 3 Ω 3 . 1 Ω R 6 A Find the resistance R . Correct answer: 4 . 1 Ω. Explanation: E 1 E 2 X Y R 1 R 3 R 2 R I Let : E 1 = 75 V , E 2 = 3 V , R 1 = 3 . 5 Ω , R 2 = 3 . 1 Ω , R 3 = 1 . 3 Ω , and I = 6 A . From Ohm’s law, the total resistance of the circuit is R total = V I = E 1 - E 2 I = (75 V) - (3 V) (6 A) = 12 Ω . Therefore the resistance R is R = R total - R 1 - R 2 - R 3 = (12 Ω) - (3 . 5 Ω) - (3 . 1 Ω) - (1 . 3 Ω) = 4 . 1 Ω . 006 (part 2 of 3) 10 points Find the potential difference V XY = V Y - V X between points X and Y . Correct answer: 35 . 4 V. Explanation: The current in the circuit goes counter- clockwise, so the potential difference between Y and X is V XY = E 2 + R 3 I + R I = (3 V) + (1 . 3 Ω + 4 . 1 Ω) (6 A) = 35 . 4 V , or = E 1 - R 1 I - R 2 I = (75 V) - (3 . 5 Ω + 3 . 1 Ω) (6 A) = 35 . 4 V .
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