This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Reilly (mar3978) – hw15 – turner – (56725) 1 This printout should have 16 questions. Multiplechoice 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. 88 V 4 V X Y 3 . 5 Ω 2 . 7 Ω 2 . 3 Ω R 7 A Find the resistance R . Correct answer: 3 . 5 Ω. Explanation: E 1 E 2 X Y R 1 R 3 R 2 R I Let : E 1 = 88 V , E 2 = 4 V , R 1 = 3 . 5 Ω , R 2 = 2 . 3 Ω , R 3 = 2 . 7 Ω , and I = 7 A . From Ohm’s law, the total resistance of the circuit is R total = V I = E 1 − E 2 I = (88 V) − (4 V) (7 A) = 12 Ω . Therefore the resistance R is R = R total − R 1 − R 2 − R 3 = (12 Ω) − (3 . 5 Ω) − (2 . 3 Ω) − (2 . 7 Ω) = 3 . 5 Ω . 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: 47 . 4 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 . 7 Ω + 3 . 5 Ω) (7 A) = 47 . 4 V , or = E 1 − R 1 I − R 2 I = (88 V) − (3 . 5 Ω + 2 . 3 Ω) (7 A) = 47 . 4 V . 003 (part 3 of 3) 10.0 points How much energy U E is dissipated by the 2 . 3 Ω resistor in 26 s? Correct answer: 2930 . 2 J. Explanation: Let : t = 26 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 = (7 A) 2 (2 . 3 Ω) (26 s) = 2930 . 2 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 3 μ F 2 1 μ F 3 2 μ F 4 1 μ F 53 V S 2 S 1 a b c d Reilly (mar3978) – hw15 – turner – (56725) 2 Find the charge on the 13 μ F upperleft capacitor between points a and c . Correct answer: 489 . 956 μ C. Explanation: Let : C 1 = 13 μ F , C 2 = 21 μ F , C 3 = 32 μ F , C 4 = 41 μ F , and E B = 53 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 = (13 μ F) (32 μ F) 13 μ F + 32 μ F = 9 . 24444 μ 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 = (21 μ F) (41 μ F) 21 μ F + 41 μ F = 13 . 8871 μ 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 = 9 . 24444 μ F + 13 . 8871 μ F = 23 . 1315 μ 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 = (9 . 24444 μ F) (53 V) = 489 . 956 μ C . C 2 and C 4 are in series, so Q 2 = Q 4 = Q l = C 24 E B = (13 . 8871 μ F) (53 V) = 736 . 016 μ C . C 1 and C 3 are in series, so Q 3 = Q 1 = 489 . 956 μ C ....
View
Full Document
 Spring '10
 TURNER,J
 SEPTA Regional Rail, Correct Answer, Jaguar Racing, Ω, Highways in Slovakia, Reilly

Click to edit the document details