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Unformatted text preview: Version 079 – TEST02 – Tsoi – (59090) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The circuit has been connected as shown in the figure for a “long” time. 32 V S 17 μ F 3 Ω 3 4 Ω 2 Ω 6 2 Ω What is the magnitude of the electric po tential across the capacitor? 1. 3.0 2. 17.0 3. 33.0 4. 8.0 5. 40.0 6. 18.0 7. 28.0 8. 14.0 9. 2.0 10. 20.0 Correct answer: 14 V. Explanation: E S 1 C t b a b I t R 1 I t R 2 I b R 3 I b R 4 Let : R 1 = 30 Ω , R 2 = 34 Ω , R 3 = 2 Ω , R 4 = 62 Ω , and C = 17 μ F . 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 flowing to it. The equivalent circuit is I t R 1 I t R 2 R 3 I b I b R 4 a b R t = R 1 + R 2 = 30 Ω + 34 Ω = 64 Ω R b = R 3 + R 4 = 2 Ω + 62 Ω = 64 Ω I t = E R t = 32 V 64 Ω = 0 . 5 A I b = E R b = 32 V 64 Ω = 0 . 5 A Across R 1 E 1 = I t R 1 = (0 . 5 A) (30 Ω) = 15 V . Across R 3 E 3 = I b R 3 = (0 . 5 A) (2 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 15 V = 14 V E C  = 14 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 = 1 e of its initial voltage? 1. 684.0 Version 079 – TEST02 – Tsoi – (59090) 2 2. 570.0 3. 504.0 4. 198.0 5. 196.0 6. 850.0 7. 408.0 8. 91.0 9. 352.0 10. 330.0 Correct answer: 408 μ s. Explanation: With the battery removed, the circuit is C I ℓ R 1 I r R 2 R 3 I ℓ I r R 4 ℓ r C R eq I eq where R ℓ = R 1 + R 3 = 30 Ω + 2 Ω = 32 Ω , R r = R 2 + R 4 = 34 Ω + 62 Ω = 96 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 32 Ω + 1 96 Ω parenrightbigg − 1 = 24 Ω . Therefore the time constant τ is τ ≡ R eq C = (24 Ω) (17 μ F) = 408 μ s . The equation for discharge of the capacitor is Q t Q = e t/τ , or E t E = e t/τ = 1 e . Taking the logarithm of both sides, we have t τ = ln parenleftbigg 1 e parenrightbigg t = τ ( ln e ) = (408 μ s)( 1) = 408 μ s . 003 10.0 points The equivalent resistance of the circuit in the figure is R eq = 69 . 0 Ω . E R S 91 Ω 15 Ω 15 Ω 91 Ω Find the value of R . 1. 21.5 2. 18.0 3. 17.5 4. 13.0 5. 23.5 6. 16.0 7. 15.0 8. 11.0 9. 8.0 10. 5.5 Correct answer: 16 Ω. Explanation: E R S R 1 R 2 R 3 R 4 Let : R 1 = 91 . 0 Ω , R 2 = 15 . 0 Ω , R 3 = 15 . 0 Ω , R 4 = 91 . 0 Ω , and R eq = 69 . 0 Ω . Version 079 – TEST02 – Tsoi – (59090) 3 Basic Concepts: For resistors in parallel, 1 R eq,p = 1 R a + 1 R b For resistors in series, R eq,s = R a + R b R 12 = R 1 + R 2 = 91 Ω + 15 Ω = 106 Ω , and R 34 = R 3 + R 4 = 15 Ω + 91 Ω = 106 Ω , and R 1234 = parenleftbigg 1 R 12 + 1 R 34 parenrightbigg − 1 = parenleftbigg 1 106 Ω + 1 106 Ω parenrightbigg − 1 = 53 Ω , and R = R...
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This note was uploaded on 10/21/2009 for the course PHY 59090 taught by Professor Tsoi during the Fall '09 term at University of Texas at Austin.
 Fall '09
 TSOI
 Physics

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