# ch25_p15 - 15(a First the equivalent capacitance of the two...

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15. (a) First, the equivalent capacitance of the two 4.00 µ F capacitors connected in series is given by 4.00 F/2 = 2.00 F. This combination is then connected in parallel with two other 2.00- F capacitors (one on each side), resulting in an equivalent capacitance C = 3(2.00 F) = 6.00 F. This is now seen to be in series with another combination, which consists of the two 3.0- F capacitors connected in parallel (which are themselves equivalent to C' = 2(3.00 F) = 6.00 F). Thus, the equivalent capacitance of the circuit is () eq 600 F 600 F 300 F. .. CC C. CC . . µµ == = ++ (b) Let V = 20.0 V be the potential difference supplied by the battery. Then q = C eq V = (3.00 F)(20.0 V) = 6.00 × 10 –5 C. (c) The potential difference across C 1 is given by
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## This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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