Homework 8 Solutions

# Homework 8 Solutions - lim (kl9356) – Homework 8 –...

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Unformatted text preview: lim (kl9356) – Homework 8 – Weathers – (22202) 1 This print-out should have 11 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 Three capacitors of 6 . 8 μ F, 1 . 6 μ F, and 16 . 3 μ F are connected to the terminals of a 9 V battery. How much energy does the battery supply if the capacitors are connected in series? Correct answer: 0 . 0485956 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 6 . 8 μ F + 1 1 . 6 μ F + 1 16 . 3 μ F = 1 . 19989 μ F . The energy supplied by the battery is then U = 1 2 C s V 2 = 1 2 1 . 19989 μ F (9 V) 2 = . 0485956 mJ . 002 (part 2 of 2) 10.0 points How much energy does the battery supply if the capacitors are connected parallel? Correct answer: 1 . 00035 mJ. Explanation: The equivalent capacitance when the three capacitors are connected parallel is C p = C 1 + C 2 + C 3 = 6 . 8 μ F + 1 . 6 μ F + 16 . 3 μ F = 24 . 7 μ F . The energy supplied by the battery is U = 1 2 C p V 2 = 1 2 24 . 7 μ F (9 V) 2 = 1 . 00035 mJ . 003 (part 1 of 2) 10.0 points A capacitor network is shown below. 91 V 43 μ F 25 μ F 47 μ F 21 μ F What is the effective capacitance of the circuit? Correct answer: 34 μ F. Explanation: Let : C 1 = 43 μ F , C 2 = 25 μ F , C 3 = 47 μ F , C 4 = 21 μ F and E B = 91 V . For capacitors in series, 1 C series = summationdisplay 1 C i V series = summationdisplay V i , and the individual charges are the same. For parallel capacitors, C parallel = summationdisplay C i Q parallel = summationdisplay Q i , and the individual voltages are the same. E B C 1 C 2 C 3 C 4 lim (kl9356) – 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 = 43 μ F + 25 μ F = 68 μ F C 3 and C 4 are in parallel, so C 34 = C 3 + C 4 = 47 μ F + 21 μ F = 68 μ F . E B C 12 C 34 C 12 and C 34 are in series with the battery, so C 1234 = parenleftbigg 1 C 12 + 1 C 34 parenrightbigg − 1 = parenleftbigg C 12 + C 34 C 12 C 34 parenrightbigg − 1 = C 12 C 34 C 12 + C 34 = (68 μ F) (68 μ F) 68 μ F + 68...
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## This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework 8 Solutions - lim (kl9356) – Homework 8 –...

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