#### Question

Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 μF. When the ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network.

#### Solution

The equivalent circuit is given below.

There are two capacitors in one branch in series. So, the equivalent capacitance of one branch will be

1/1+1/1=2muF

The arrangement will be further reduced to the form given below.

Now, both the capacitors are in parallel, so the equivalent capacitance will be

2+2=4μF

Therefore, the equivalent capacitance is 4 μF.

(i)

Voltage, *V =* 6 V

The charge in the network is given by

q=CV

Here, C is the equivalent capacitance.

Now,q=4×10^{−6}×6

=24×10^{−6} C=24 μC

(ii)

The energy stored in the network is given by

E=1/2CV^{2}

=12×4×10^{−6}×(6)^{2}

=12×4×36×10^{−6}

=72×10^{−6} J

=72 μJ