solutiontes2_pdf

# solutiontes2_pdf - Version 120 – TEST02 – Tsoi –...

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Version 120 – TEST02 – Tsoi – (59090) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points The circuit has been connected as shown in the fgure For a “long” time. 7 V S 15 μ ± 18 Ω 24 Ω 10 Ω 60 Ω What is the magnitude oF the electric po- tential across the capacitor? 1. 2.0 2. 5.0 3. 18.0 4. 30.0 5. 16.0 6. 17.0 7. 8.0 8. 3.0 9. 1.0 10. 15.0 Correct answer: 2 V. Explanation: E S 1 C t b a b I t R 1 2 b 3 4 Let : R 1 = 18 Ω , R 2 = 24 Ω , R 3 = 10 Ω , R 4 = 60 Ω , and C = 15 μ ± . AFter a “long time” implies that the ca- pacitor C is Fully charged and thereFore the capacitor acts as an open circuit with no cur- rent ²owing to it. The equivalent circuit is a b R t = R 1 + R 2 = 18 Ω + 24 Ω = 42 Ω R b = R 3 + R 4 = 10 Ω + 60 Ω = 70 Ω I t = E R t = 7 V 42 Ω = 0 . 166667 A I b = E R b = 7 V 70 Ω = 0 . 1 A Across R 1 E 1 = I t R 1 = (0 . 166667 A) (18 Ω) = 3 V . Across R 3 E 3 = I b R 3 = (0 . 1 A) (10 Ω) = 1 V . Since E 1 and E 3 are “measured” From the same point “ a ”, the potential across C must be E C = E 3 - E 1 = 1 V - 3 V = - 2 V |E C | = 2 V . 002 (part 2 oF 2) 10.0 points IF the battery is disconnected, how long does it take For the capacitor to discharge to E t E 0 = 1 e oF its initial voltage? 1. 198.0

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Version 120 – TEST02 – Tsoi – (59090) 2 2. 308.0 3. 900.0 4. 147.0 5. 315.0 6. 91.0 7. 504.0 8. 153.0 9. 224.0 10. 1080.0 Correct answer: 315 μ s. Explanation: With the battery removed, the circuit is C I R 1 r 2 3 4 r eq where R = R 1 + R 3 = 18 Ω + 10 Ω = 28 Ω , R r = R 2 + R 4 = 24 Ω + 60 Ω = 84 Ω and R eq = p 1 R + 1 R r P 1 = p 1 28 Ω + 1 84 Ω P 1 = 21 Ω . Therefore the time constant τ is τ R eq C = (21 Ω) (15 μ F) = 315 μ s . The equation for discharge of the capacitor is Q t Q 0 = e - t/τ , or E t E 0 = e - t/τ = 1 e . Taking the logarithm of both sides, we have - t τ = ln p 1 e P t = - τ ( - ln e ) = - (315 μ s) ( - 1) = 315 μ s . 003 10.0 points Given a spherical capacitor with radius of the inner conducting sphere a and the outer shell b . The outer shell is grounded. The charges are + Q and - Q . A point C is located at r = R 2 , where R = a + b . a A B C + Q - Q b What is the capacitance of this spherical capacitor? 1. C = 1 k e ( a - b ) 2. C = b - a 2 k e ln p b a P 3. C = b k e 4. C = 1 k e p 1 a - 1 b P correct 5. C = a k e 6. C = k e b 7. C = k e a 8. C = a + b k e
Version 120 – TEST02 – Tsoi – (59090) 3 9. C = b 2 4 k e ( b - a ) 10. C = 1 k e ( a + b ) Explanation: Δ V = V a - V b = k e Q p 1 a - 1 b P - 0 since V b is grounded. The charge on the inside of the shell doesn’t aFect the grounded potential.

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## This note was uploaded on 06/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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solutiontes2_pdf - Version 120 – TEST02 – Tsoi –...

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