hw16-solutions - marout (ssm979) hw16 turner (56725) 1 This...

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Unformatted text preview: marout (ssm979) hw16 turner (56725) 1 This print-out should have 17 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. 48 V S 13 F 7 9 1 4 7 What is the magnitude of the electric po- tential across the capacitor? Correct answer: 20 V. Explanation: Let : R 1 = 7 , R 2 = 9 , R 3 = 1 , R 4 = 47 , and C = 13 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 + 9 = 16 and R b = R 3 + R 4 = 1 + 47 = 48 , so I t = E R t = 48 V 16 = 3 A and I b = E R b = 48 V 48 = 1 A . Across R 1 , E 1 = I t R 1 = (3 A) (7 ) = 21 V and across R 3 E 3 = I b R 3 = (1 A) (1 ) = 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 s. Explanation: With the battery removed, the circuit is C I R 1 I r R 2 R 3 I I r R 4 r marout (ssm979) hw16 turner (56725) 2 C R eq I eq where R = R 1 + R 3 = 7 + 1 = 8 , R r = R 2 + R 4 = 9 + 47 = 56 and R eq = parenleftbigg 1 R + 1 R r parenrightbigg 1 = parenleftbigg 1 8 + 1 56 parenrightbigg 1 = 7 , so the time constant is R eq C = (7 ) (13 F) = 91 s . The capacitor discharges according to Q t Q = e t/ V ( t ) E = e t/ = 1 e t = ln parenleftbigg 1 e parenrightbigg = ln e t = (ln e ) = (91 s) ( 1) = 91 s . 003 10.0 points In the figure below the battery has an emf of 21 V and an internal resistance of 1 . Assume there is a steady current flowing in the circuit. 5 F 5 9 1 21 V Find the charge on the 5 F capacitor. Correct answer: 63 C. Explanation: Let : R 1 = 5 , R 2 = 9 , r in = 1 , V = 21 V , and C = 5 F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = 5 + 9 + 1 = 15 , 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 15 (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 F) (12 . 6 V) = 63 C ....
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hw16-solutions - marout (ssm979) hw16 turner (56725) 1 This...

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