Each of the capacitors shown in the figure has a capacitance of 2 μF. Find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.
Given
Capacitance C= 2μF
The equalent capacitance of the first row is calculated as
The capacitance of each row is the same, and it is equal to 
All the three rows are arranged in parallel. So, Voltage or potential difference across each row is the same and is equal to 60V.
So, Voltage across each capacitor is =20V.
The equalent capacitance between A and B is 
Potential difference across each capacitor is 20V.
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