P26_054

# P26_054 - 6 = 4 . 0 V across the group of capacitors in the...

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

54. (a) The potential across capacitor 1 is 10 V, so the charge on it is q 1 = C 1 V 1 =(10 µ F)(10 V) = 100 µ C . (b) Reducing the right portion of the circuit produces an equivalence equal to 6 . 0 µ F, with 10 V across it. Thus, a charge of 60 µ C is on it – and consequently also on the bottom right capacitor. The bottom right capacitor has, as a result, a potential across it equal to V = q C = 60 µ C 10 µ F =6 . 0V , which leaves 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 = 4 . 0 V across the group of capacitors in the upper right portion of the circuit. Inspection of the arrangement (and capacitance values) of that group reveals that this 4 . 0 V must be equally divided by C 2 and the capacitor directly below it (in series with it). Therefore, with 2 . 0 V across capacitor 2, we nd q 2 = C 2 V 2 = (10 F)(2 . 0 V) = 20 C ....
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online