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Unformatted text preview: Reilly (mar3978) – hw16 – turner – (56725) 1 This printout should have 17 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. 40 V S 22 μ F 1 7 Ω 2 3 Ω 3 Ω 5 7 Ω What is the magnitude of the electric po tential across the capacitor? Correct answer: 15 V. Explanation: Let : R 1 = 17 Ω , R 2 = 23 Ω , R 3 = 3 Ω , R 4 = 57 Ω , and C = 22 μ F = 2 . 2 × 10 − 5 F . E S C t b a b I t R 1 I t R 2 I b R 3 I b R 4 “After a long time” implies that the capac itor C is fully charged, so it acts as an open circuit with no current 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 = 17 Ω + 23 Ω = 40 Ω and R b = R 3 + R 4 = 3 Ω + 57 Ω = 60 Ω , so I t = E R t = 40 V 40 Ω = 1 A and I b = E R b = 40 V 60 Ω = 0 . 666667 A . Across R 1 , E 1 = I t R 1 = (1 A) (17 Ω) = 17 V and across R 3 E 3 = I b R 3 = (0 . 666667 A) (3 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 − 17 V = − 15 V E C  = 15 V . 002 (part 2 of 2) 10.0 points If the battery is disconnected, how long does it take for the voltage across the capacitor to drop to a value of V ( t ) = E e , where E is the initial voltage across the capacitor? Correct answer: 352 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 Reilly (mar3978) – hw16 – turner – (56725) 2 C R eq I eq where R ℓ = R 1 + R 3 = 17 Ω + 3 Ω = 20 Ω , R r = R 2 + R 4 = 23 Ω + 57 Ω = 80 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 20 Ω + 1 80 Ω parenrightbigg − 1 = 16 Ω , so the time constant is τ ≡ R eq C = (16 Ω) (22 μ F) = 352 μ s . The capacitor discharges according to Q t Q = e − t/τ V ( t ) E = e − t/τ = 1 e − t τ = ln parenleftbigg 1 e parenrightbigg = − ln e t = τ (ln e ) = − (352 μ s) ( − 1) = 352 μ s . 003 10.0 points In the figure below the battery has an emf of 10 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 6 μ F 5 Ω 7 Ω 1 Ω 10 V Find the charge on the 6 μ F capacitor. Correct answer: 32 . 3077 μ C. Explanation: Let : R 1 = 5 Ω , R 2 = 7 Ω , r in = 1 Ω , V = 10 V , and C = 6 μ F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = 5 Ω + 7 Ω + 1 Ω = 13 Ω , so the current in the circuit is I = V R eq , and the voltage across R 2 is V 2 = I R 2 = R 2 R eq V = 7 Ω 13 Ω (10 V) = 5 . 38462 V ....
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This note was uploaded on 11/30/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas.
 Spring '10
 TURNER,J

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