P26_069.PDF

69. (a) The voltage across C 1 is 12 V, so the charge is

q 1 = C 1 V 1 =24 µ C .

(b) We reduce the circuit, starting with C 4 and C 3 (in parallel) which are equivalent to 4 µ F. This is

then in series with C 2 , resulting in an equivalence equal to 3 µ F which would have 12 V across it.

The charge on this 3 µ F capacitor (and therefore on C 2 )is( 3 µ F)(12V) = 16 µ C. Consequently,

the voltage across C 2 is

C 2 = 16 µ C

2 µ F

=8V .

This leaves 12

−

8=4Vacross C 4 (similarly for C 3 ).

V 2 = q 2