# ch25_p52 - with C 1 and C 3 and thus also has the final...

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

52. 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 1 C 1 + 1 C 2 + 1 C 3 = 1 C 123 , 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 1 C 1 + 1 C 24 + 1 C 3 = 1 C 1234 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: with C 1 and C 3 and thus also has the final charge q =36.0 µ C found above. The voltage across C 24 would be V 24 = q / C 24 = 36.0/12.0 = 3.00 V. This is the same voltage across each of the parallel pair. In particular, V 4 = 3.00 V implies that q 4 = C 4 V 4 = 18.0 µ C. (c) The battery supplies charges only to the plates where it is connected. The charges on the rest of the plates are due to electron transfers between them, in accord with the new distribution of voltages across the capacitors. So, the battery does not directly supply the charge on capacitor 4....
View Full Document

## 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.

Ask a homework question - tutors are online