Each of the capacitors shown in 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.
Solution
There are three rows of capacitors connected in parallel in the given system. In each row, three capacitors of capacitance 2 μF are connected in series.
For each row, the equivalent capacitance is given by
`1/C_r = 1/2 + 1/2 + 1/2`
`⇒ C_r = 2/3 "uF"`
As three rows are connected in parallel, their equivalent capacitance is given by
`C_(eq) = C_r + C_r + C_r = 2/3 + 2/3 + 2/3 = 2 "uF"`
The voltage across each row is the same and is equal to 60 V.
As all capacitors have the same capacitance in each row, the potential difference across their plates is the same.
∴ Potential difference across each capacitor = 20 V