ch25-p057

# ch25-p057 - 57. Initially the capacitors C1, C2, and C3...

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57. Initially the capacitors C 1 , C 2 , and C 3 form a series combination equivalent to a single capacitor which we denote C 123 . Solving the equation 12 23 13 123 1 2 3 1 2 3 11 1 1 CC C C C C C + + =++= , we obtain C 123 = 2.40 µ F. With V = 12.0 V, we then obtain q = C 123 V = 28.8 µ C. In the final situation, C 2 and C 4 are in parallel and are thus effectively equivalent to 24 12.0 F C µ = . Similar to the previous computation, we use 4 2 43 1234 1 24 3 1 24 3 1 1 C C C C C C C + + =+ += and find C 1234 = 3.00 µ F. Therefore, the final charge is q = C 1234 V = 36.0 µ C. (a) This represents a change (relative to the initial charge) of q = 7.20 µ C. (b) The capacitor C 24 which we imagined to replace the parallel pair C 2 and C 4 is in series with
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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