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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.
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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=1/2CV2
=12×4×10−6×(6)2
=12×4×36×10−6
=72×10−6 J
=72 μJ
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