hw16-solutions

# hw16-solutions - howard (cah3459) hw16 turner (56705) This...

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howard (cah3459) – hw16 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points The circuit has been connected as shown in the fgure For a “long” time. 32 V S 24 μ ± 14 Ω 18 2 30 What is the magnitude oF the electric po- tential across the capacitor? Correct answer: 12 V. Explanation: Let : R 1 = 14 Ω , R 2 = 18 Ω , R 3 = 2 Ω , R 4 = 30 Ω , and C = 24 μ ± = 2 . 4 × 10 - 5 ± . E S C t b a b I t R 1 b 3 4 “AFter a long time” implies that the capac- itor C is Fully charged, so it acts as an open circuit with no current ²owing to it. The equivalent circuit is a b R t = R 1 + R 2 = 14 Ω + 18 Ω = 32 Ω and R b = R 3 + R 4 = 2 Ω + 30 Ω = 32 Ω , so I t = E R t = 32 V 32 Ω = 1 A and 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 and 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 voltage across the capacitor to drop to a value oF V ( t )= E 0 e , where E 0 is the initial voltage across the capacitor? Correct answer: 288 μ s. Explanation: With the battery removed, the circuit is r ( r

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howard (cah3459) – hw16 – turner – (56705) 2 C R eq I where R ± = 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 = ± 1 R ± + 1 R r ² - 1 = ± 1 16 Ω + 1 48 Ω ² - 1 = 12 Ω , so the time constant is τ R eq C = (12 Ω) (24 μ F) = 288 μ s . The capacitor discharges according to Q t Q 0 = e - t/τ V ( t ) E 0 = e - t/τ = 1 e - t τ = ln ± 1 e ² = - ln e t = τ (ln e )= - (288 μ s) ( - 1) = 288 μ s . 003 10.0 points In the fgure below the battery has an em± o± 17 V and an internal resistance o± 1 Ω . Assume there is a steady current ²owing in the circuit. 5 μ F 8 Ω 7 Ω 1 Ω 17 V Find the charge on the 5 μ F capacitor. Correct answer: 37 . 1875 μ C. Explanation: Let : R 1 = 8 Ω , R 2 = 7 Ω , r in = 1 Ω , V = 17 V , and C =5 μ F . The equivalent resistance o± the three resistors in series is R eq = R 1 + R 2 + r in = 8 Ω + 7 Ω + 1 Ω = 16 Ω , so the current in the circuit is I = V R eq , and the voltage across R 2 is V 2 = IR 2 = R 2 R eq V = 16 Ω (17 V) =7 . 4375 V . Since R 2 and C are parallel, the potential di³erence across each is the same, and the charge on the capacitor is Q = CV 2 = (5 μ F) (7 . 4375 V) = 37 . 1875 μ C . 004 10.0 points A0 . 29 μ F capacitor is given a charge Q 0 . A±ter 4 s, the capacitor’s charge is 0 . 5 Q 0 . What is the e³ective resistance across this capacitor? Correct answer: 19 . 8992 MΩ. Explanation: Let : C =0 . 29 μ F = 2 . 9 × 10 - 7 F and t = 4 s .
howard (cah3459) – hw16 – turner – (56705) 3 The charge on the capacitor is Q ( t )= Q 0 e - t/τ . e t/τ = Q 0 Q t τ = ln ± Q 0 Q ² τ = t ln ± Q 0 Q ² . The efective resistance is R = τ C = t C ln ± Q 0 Q ² = 4s (2 . 9 × 10 - 7 F) ln ± Q 0 0 . 5 Q 0 ² · 1 MΩ 10 6 Ω = 19 . 8992 MΩ .

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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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hw16-solutions - howard (cah3459) hw16 turner (56705) This...

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