Exam II - Version 147/ACBAD – midterm 02 – Turner –...

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Unformatted text preview: Version 147/ACBAD – midterm 02 – Turner – (60230) 1 This print-out should have 18 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 11 μ F 1 4 Ω 1 8 Ω 2 Ω 3 Ω What is the magnitude of the electric po- tential across the capacitor? 1. 20.0 2. 19.0 3. 3.0 4. 12.0 5. 16.0 6. 40.0 7. 1.0 8. 2.0 9. 14.0 10. 6.0 Correct answer: 12 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 = 14 Ω , R 2 = 18 Ω , R 3 = 2 Ω , R 4 = 30 Ω , and C = 11 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 = 14 Ω + 18 Ω = 32 Ω R b = R 3 + R 4 = 2 Ω + 30 Ω = 32 Ω I t = E R t = 32 V 32 Ω = 1 A I b = E R b = 32 V 32 Ω = 1 A Across R 1 E 1 = I t R 1 = (1 A) (14 Ω) = 14 V . Across R 3 E 3 = I b R 3 = (1 A) (2 Ω) = 2 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 = 2 V − 14 V = − 12 V |E C | = 12 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. 91.0 Version 147/ACBAD – midterm 02 – Turner – (60230) 2 2. 132.0 3. 147.0 4. 504.0 5. 1020.0 6. 456.0 7. 510.0 8. 462.0 9. 196.0 10. 374.0 Correct answer: 132 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 = 14 Ω + 2 Ω = 16 Ω , R r = R 2 + R 4 = 18 Ω + 30 Ω = 48 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 16 Ω + 1 48 Ω parenrightbigg − 1 = 12 Ω . Therefore the time constant τ is τ ≡ R eq C = (12 Ω) (11 μ F) = 132 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 ) = − (132 μ s)( − 1) = 132 μ s . 003 10.0 points In the figure below the battery has an emf of 8 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 2 μ F 11 Ω 4 Ω 1 Ω 8 V Find the charge on the 2 μ F capacitor. 1. 15.5556 2. 9.88235 3. 4.0 4. 32.4 5. 11.375 6. 7.22222 7. 27.6923 8. 40.5 9. 33.3333 10. 7.35294 Correct answer: 4 μ C. Explanation: Let : R 1 = 11 Ω , R 2 = 4 Ω , r in = 1 Ω , V = 8 V , and C = 2 μ F ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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Exam II - Version 147/ACBAD – midterm 02 – Turner –...

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