oldhw14_pdf - mcconnell (kam2342) oldhomework 14 Turner...

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Unformatted text preview: mcconnell (kam2342) oldhomework 14 Turner (60230) 1 This print-out should have 14 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. 24 V S 14 F 5 7 1 2 3 What is the magnitude of the electric po- tential across the capacitor? Correct answer: 9 V. Explanation: E S 1 C t b a b I t R 1 I t R 2 I b R 3 I b R 4 Let : R 1 = 5 , R 2 = 7 , R 3 = 1 , R 4 = 23 , and C = 14 F . After a long time implies that the ca- pacitor C is fully charged and therefore the capacitor acts as an open circuit with no cur- rent 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 = 5 + 7 = 12 R b = R 3 + R 4 = 1 + 23 = 24 I t = E R t = 24 V 12 = 2 A I b = E R b = 24 V 24 = 1 A Across R 1 E 1 = I t R 1 = (2 A) (5 ) = 10 V . 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- 10 V =- 9 V |E C | = 9 V . 002 (part 2 of 2) 10.0 points If the battery is disconnected, how long does it take for the capacitor to discharge to E t E = 1 e of its initial voltage? Correct answer: 70 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 mcconnell (kam2342) oldhomework 14 Turner (60230) 2 C R eq I eq where R = R 1 + R 3 = 5 + 1 = 6 , R r = R 2 + R 4 = 7 + 23 = 30 and R eq = parenleftbigg 1 R + 1 R r parenrightbigg- 1 = parenleftbigg 1 6 + 1 30 parenrightbigg- 1 = 5 . Therefore the time constant is R eq C = (5 ) (14 F) = 70 s . The equation for discharge of the capacitor is Q t Q = e- t/ , or E t E = e- t/ = 1 e . Taking the logarithm of both sides, we have- t = ln parenleftbigg 1 e parenrightbigg t =- (- ln e ) =- (70 s)(- 1) = 70 s . 003 (part 1 of 2) 10.0 points In this problem assume 1) the batteries have zero internal resistance, 2) the currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. 14 . 1 21 . 8 3 . 6 V 12 . 9 V I 1 I 2 I 3 Using the direction of the arrow as posi- tive (as shown in the figure), find the current through the 14 . 1 resistor and the 3 . 6 V bat- tery at the top of the circuit....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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oldhw14_pdf - mcconnell (kam2342) oldhomework 14 Turner...

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