# Exam 2 - Version 079 – TEST02 – Tsoi –(59090 1 This print-out should have 14 questions Multiple-choice questions may continue on the next

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Unformatted text preview: Version 079 – TEST02 – Tsoi – (59090) 1 This print-out should have 14 questions. Multiple-choice 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.

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Exam 2 - Version 079 – TEST02 – Tsoi –(59090 1 This print-out should have 14 questions Multiple-choice questions may continue on the next

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