ch25_p15 - 15. (a) First, the equivalent capacitance of the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online