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Unformatted text preview: marout (ssm979) – 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. 48 V S 13 μ F 7 Ω 9 Ω 1 Ω 4 7 Ω What is the magnitude of the electric po tential across the capacitor? Correct answer: 20 V. Explanation: Let : R 1 = 7 Ω , R 2 = 9 Ω , R 3 = 1 Ω , R 4 = 47 Ω , and C = 13 μ F = 1 . 3 × 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 = 7 Ω + 9 Ω = 16 Ω and R b = R 3 + R 4 = 1 Ω + 47 Ω = 48 Ω , so I t = E R t = 48 V 16 Ω = 3 A and I b = E R b = 48 V 48 Ω = 1 A . Across R 1 , E 1 = I t R 1 = (3 A) (7 Ω) = 21 V and across R 3 E 3 = I b R 3 = (1 A) (1 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 − 21 V = − 20 V E C  = 20 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: 91 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 marout (ssm979) – hw16 – turner – (56725) 2 C R eq I eq where R ℓ = R 1 + R 3 = 7 Ω + 1 Ω = 8 Ω , R r = R 2 + R 4 = 9 Ω + 47 Ω = 56 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 8 Ω + 1 56 Ω parenrightbigg − 1 = 7 Ω , so the time constant is τ ≡ R eq C = (7 Ω) (13 μ F) = 91 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 ) = − (91 μ s) ( − 1) = 91 μ s . 003 10.0 points In the figure below the battery has an emf of 21 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 5 μ F 5 Ω 9 Ω 1 Ω 21 V Find the charge on the 5 μ F capacitor. Correct answer: 63 μ C. Explanation: Let : R 1 = 5 Ω , R 2 = 9 Ω , r in = 1 Ω , V = 21 V , and C = 5 μ F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = 5 Ω + 9 Ω + 1 Ω = 15 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 = 9 Ω 15 Ω (21 V) = 12 . 6 V . Since R 2 and C are parallel, the potential difference across each is the same, and the charge on the capacitor is Q = C V 2 = (5 μ F) (12 . 6 V) = 63 μ C ....
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 Fall '10
 TURNER,J
 Electron, Energy, Magnetic Field, Correct Answer, Electric charge

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