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Unformatted text preview: malsam (wgm329) hw15 turner (56705) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points See the circuit below. 112 V 4 V X Y 3 . 8 2 . 9 1 . 8 R 9 A Find the resistance R . Correct answer: 3 . 5 . Explanation: E 1 E 2 X Y R 1 R 3 R 2 R I Let : E 1 = 112 V , E 2 = 4 V , R 1 = 3 . 8 , R 2 = 1 . 8 , R 3 = 2 . 9 , and I = 9 A . From Ohms law, the total resistance of the circuit is R total = V I = E 1 E 2 I = (112 V) (4 V) (9 A) = 12 . Therefore the resistance R is R = R total R 1 R 2 R 3 = (12 ) (3 . 8 ) (1 . 8 ) (2 . 9 ) = 3 . 5 . 002 (part 2 of 3) 10.0 points Find the potential difference V XY = V Y V X between points X and Y . Correct answer: 61 . 6 V. Explanation: The current in the circuit goes counter- clockwise, so the potential difference between Y and X is V XY = E 2 + R 3 I + RI = (4 V) + (2 . 9 + 3 . 5 ) (9 A) = 61 . 6 V , or = E 1 R 1 I R 2 I = (112 V) (3 . 8 + 1 . 8 ) (9 A) = 61 . 6 V . 003 (part 3 of 3) 10.0 points How much energy U E is dissipated by the 1 . 8 resistor in 37 s? Correct answer: 5394 . 6 J. Explanation: Let : t = 37 s . The work done is W = P t = V I t = I 2 R t, so the energy dissipated is W = I 2 R 2 t = (9 A) 2 (1 . 8 ) (37 s) = 5394 . 6 J . 004 (part 1 of 2) 10.0 points In the figure below consider the case where switch S 1 is closed and switch S 2 is open. 1 6 F 2 4 F 3 9 F 4 4 F 58 V S 2 S 1 a b c d malsam (wgm329) hw15 turner (56705) 2 Find the charge on the 16 F upper-left capacitor between points a and c . Correct answer: 658 . 036 C. Explanation: Let : C 1 = 16 F , C 2 = 24 F , C 3 = 39 F , C 4 = 44 F , and E B = 58 V . C 1 C 2 C 3 C 4 E B S 2 S 1 a b c d Redrawing the figure, we have E B C 1 C 2 C 3 C 4 b a c d C 1 and C 3 are in series, so C 13 = parenleftbigg 1 C 1 + 1 C 3 parenrightbigg 1 = C 1 C 3 C 1 + C 3 = (16 F) (39 F) 16 F + 39 F = 11 . 3455 F . C 2 and C 4 are in series, so C 24 = parenleftbigg 1 C 2 + 1 C 4 parenrightbigg 1 = C 2 C 4 C 2 + C 4 = (24 F) (44 F) 24 F + 44 F = 15 . 5294 F . Simplifying the circuit, we have E B C 13 C 24 a b C 13 and C 24 are parallel, so C ab = C 13 + C 24 = 11 . 3455 F + 15 . 5294 F = 26 . 8749 F . E B C ab a b C 1 and C 3 are in series, so Q 1 = Q 3 = Q 13 = C 13 E B = (11 . 3455 F) (58 V) = 658 . 036 C . C 2 and C 4 are in series, so Q 2 = Q 4 = Q l = C 24 E B = (15 . 5294 F) (58 V) = 900 . 706 C . C 1 and C 3 are in series, so Q 3 = Q 1 = 658 . 036 C ....
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This note was uploaded on 11/22/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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