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Unformatted text preview: zhou (ejz224) – Homework 5 – shih – (57480) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A length of wire is cut into 7 equal pieces. The 7 pieces are then connected parallel, with the resulting resistance being 2 Ω. What was the resistance r of the original length of wire? Correct answer: 98 Ω. Explanation: Let : n = 7 and R p = 2 Ω . The resulting resistance R p of n equal piece resistors of resistance r connected parallel is R p = r n , but when they are in series, the total resis tance R s is R s = n r = n 2 R p = 7 2 (2 Ω) = 98 Ω . 002 (part 1 of 2) 10.0 points The circuit has been connected as shown in the figure for a “long” time. 9 V S 25 μ F 1 2 Ω 1 5 Ω 6 Ω 4 8 Ω What is the magnitude of the electric po tential across the capacitor? Correct answer: 3 V. Explanation: Let : R 1 = 12 Ω , R 2 = 15 Ω , R 3 = 6 Ω , R 4 = 48 Ω , and C = 25 μ F = 2 . 5 × 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 = 12 Ω + 15 Ω = 27 Ω and R b = R 3 + R 4 = 6 Ω + 48 Ω = 54 Ω , so I t = E R t = 9 V 27 Ω = 0 . 333333 A and I b = E R b = 9 V 54 Ω = 0 . 166667 A . Across R 1 , E 1 = I t R 1 = (0 . 333333 A) (12 Ω) = 4 V and across R 3 E 3 = I b R 3 = (0 . 166667 A) (6 Ω) = 1 V . zhou (ejz224) – Homework 5 – shih – (57480) 2 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 4 V = 3 V E C  = 3 V . 003 (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: 350 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 = 12 Ω + 6 Ω = 18 Ω , R r = R 2 + R 4 = 15 Ω + 48 Ω = 63 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 18 Ω + 1 63 Ω parenrightbigg − 1 = 14 Ω , so the time constant is τ ≡ R eq C = (14 Ω) (25 μ F) = 350 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 ) = (350 μ s) ( 1) = 350 μ s . 004 10.0 points The resistance to the right of A ′ B ′ is the same as the resistance to the right of AB; that is, R AB = R → A ′ B ′ ....
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This note was uploaded on 04/09/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas at Austin.
 Spring '08
 morrison
 Physics, Work

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