Three capacitors having capacitances 20 µF, 30 µF and 40 µF are connected in series with a 12 V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?

#### Solution

When the capacitors are connected in series, the equivalent capacitance is given by `1/C_(eq) = 1/C_1 + 1/C_2 + 1/C_3`

`1/C_(eq) = (1/20 + 1/30 + 1/40) xx 1/10^-6`

⇒ `C_(eq) = 9.23 "uF"`

Because the capacitors are connected in series, the same charge will go to each of them and it is equal to the total charge given by the battery.

Now,

Let the charge at each capacitor be *q.*

`therefore q = CV = (9.23 xx 10^-6) xx 12`

`q = 110.76 "uC"`

The Work done by the battery (W) is given by

W = qV

⇒ `W = 12 xx 110.76 xx 10^-6`

⇒ `W = 1.33 xx 10^-3 J`