8 - roofner(bar784 Homework 8 Weathers(22202 Explanation...

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roofner (bar784) – Homework 8 – Weathers – (22202) 1 This print-out should have 11 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 Three capacitors of 3 . 2 μ ±, 1 . 6 μ ±, and 16 . 8 μ ± are connected to the terminals of a 16 . 1 V battery. How much energy does the battery supply if the capacitors are connected in series? Correct answer: 0 . 129992 mJ. Explanation: The equivalent capacitance when the three capacitors are connected in series is equal to C s = 1 1 C 1 + 1 C 2 + 1 C 3 = 1 1 3 . 2 μ ± + 1 1 . 6 μ ± + 1 16 . 8 μ ± = 1 . 00299 μ ± . The energy supplied by the battery is then U = 1 2 C s V 2 = 1 2 1 . 00299 μ ± (16 . 1 V) 2 = 0 . 129992 mJ . 002 (part 2 of 2) 10.0 points How much energy does the battery supply if the capacitors are connected parallel? Correct answer: 2 . 79947 mJ. Explanation: The equivalent capacitance when the three capacitors are connected parallel is C p = C 1 + C 2 + C 3 = 3 . 2 μ ± + 1 . 6 μ ± + 16 . 8 μ ± = 21 . 6 μ ± . The energy supplied by the battery is U = 1 2 C p V 2 = 1 2 21 . 6 μ ± (16 . 1 V) 2 = 2 . 79947 mJ . 003 (part 1 of 2) 10.0 points A capacitor network is shown below. 64 V 34 μ ± 40 What is the e²ective capacitance of the circuit? Correct answer: 37 μ ±. Explanation: Let : C 1 = 34 μ ± , C 2 = 40 μ ± , C 3 = 34 μ ± , C 4 = 40 μ ± and E B = 64 V . ±or capacitors in series, 1 C series = s 1 C i V series = s V i , and the individual charges are the same. ±or parallel capacitors, C parallel = s C i Q parallel = s Q i , and the individual voltages are the same. E B C 1 C 2 C 3 C 4
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roofner (bar784) – Homework 8 – Weathers – (22202) 2 Parallel connections: C parallel = C 1 + C 2 + C 3 + · · · Series connections: 1 C series = 1 C 1 + 1 C 2 + 1 C 3 + · · · C 1 and C 2 are in parallel, so C 12 = C 1 + C 2 = 34 μ F + 40 μ F = 74 μ F C 3 and C 4 are in parallel, so C 34 = C 3 + C 4 = 34 μ F + 40 μ F = 74 μ F . E B C 12 C 34 C 12 and C 34 are in series with the battery, so C 1234 = p 1 C 12 + 1 C 34 P 1 = p C 12 + C 34 C 12 C 34 P 1 = C 12 C 34 C 12 + C 34 = (74 μ F) (74 μ F) 74 μ F + 74
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8 - roofner(bar784 Homework 8 Weathers(22202 Explanation...

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