Here’s How to Crack It
Once the charging currents stop, the voltage across C 3 is equal to the voltage across
the battery, so V 3 = 180 V. This gives us Q 3 = C 3 V 3 = (90 μF)(180 V) = 16.2 mC.
Since C 1 and C 2 are in series, they must store identical amounts of charge, and,
from the diagram, the sum of their voltages must equal the voltage of the battery. So
if we let Q be the charge on each of these two capacitors, then Q = C 1 V 1 = C 2 V 2
and V 1 + V 2 = 180 V. The equation C 1 V 1 = C 2 V 2 becomes (30 μF)V 1 = (60 μF)V 2 ,
so V 1 = 2V 2. Substituting this into V 1 + V 2 = 180 V gives us V 1 = 120 V and V 2 =
60 V.
The charge stored on each of these capacitors is
(30 μF)(120 V) = C 1 V 1 = C 2 V 2 = (60 μF)(60 V) = 3.6 mC- In the following diagram, C 1 = 2 mF and C 2 = 4 mF. When
switch S is open, a battery (which is not shown) is connected
between points a and b and charges capacitor C 1 so that Vab = 12 V.
The battery is then disconnected.