chapter25-017 - 17. (a) First, the equivalent capacitance...

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3(2.00 µ 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. 600 F 600 F .. CC C. CC . . µµ µ µµ ++ (b) Let V = 20.0 V be the potential difference supplied by the battery. Then q = C eq V = (3.00 µ F)(20.0 V) = 6.00 × 10 –5 (c) The potential difference across C 1 is given by ( ) ( ) 1 600 F 200V 10 0V 600 F 600 F .. CV V. . CC . . µ µµ == = ++ (d) The charge carried by C q 1 = C 1 V 1 = (3.00 µ F)(10.0 V) = 3.00 × 10 –5 C. (e) The potential difference across C 2 is given by 2 = V – V 1 = 20.0 V – 10.0 V = 10.0 V. (f) The charge carried by C 2 is q 2 = C 2 V 2
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