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.
The equivalent circuit is given below.
There are two capacitors in one branch in series. So, the equivalent capacitance of one branch will be
The arrangement will be further reduced to the form given below.
Now, both the capacitors are in parallel, so the equivalent capacitance will be
Therefore, the equivalent capacitance is 4 μF.
Voltage, V = 6 V
The charge in the network is given by
Here, C is the equivalent capacitance.
=24×10−6 C=24 μC
The energy stored in the network is given by