Three identical capacitors C_{1}, C_{2} and C_{3} of capacitance 6 μF each are connected to a 12 V battery as shown.

Find

(i) charge on each capacitor

(ii) equivalent capacitance of the network

(iii) energy stored in the network of capacitors

#### Solution

The 12 V battery is in parallel with *C*_{1}*, C*_{2}, and *C*_{3}.* C*_{1} and *C*_{2}are in series with each other while *C*_{3 }is in parallel with the combination formed by *C*_{1}and *C*_{2}.

Total voltage drop across *C*_{3} = 12 V

*q*_{3} = *CV*

Where, *q* = Charge on the capacitor

*C*_{1}*, C*_{2}*, C*_{3} = 6 μF (Given in the question)

*q*_{3} = 6 × 12 = 72 μC

Voltage drop across *C*_{1} and *C*_{2} combined will be 12 V.

Let the voltage drop at *C*_{1}* = V*_{1}

Let the voltage drop at *C*_{2}* = V*_{2}

Then,

*V = V*_{1}* + V*_{2}

`V_1 = q_1/C`

`V_2 = q_2/C`

`q_1/6+q_2/6 = 12`

As both the capacitors are in series,

`q_1 =q_2 =q`

Then,

`q{1/6 +1/6} =12`

`q xx 1/3 =12`

Or,

*q* = 36 micro coulombs

Thus, charge on each of *C*_{1} and *C*_{2} is 36 coulombs.