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Q.
$n$ identical capacitors are grouped in series. $n$ such capacitors are grouped in parallel. These two groups are connected in series. The effective capacitance of the combination is
COMEDKCOMEDK 2006Electrostatic Potential and Capacitance
Solution:
Let capacitance of each capacitor be $C$.
If n capacitors are connected .in series then net capacitance of this combination will be,
$ C_S = \frac{C}{n}$
If n capacitors are connected in parallel then net capacitance of this combination will be, $C_p = nC$
Now, $C_S$ and $C_P$ are connected in series so equivalent capacitance,
$ C_N = \frac{C_S \times C_P}{C_S + C_P} = \frac{\frac{C}{n} \times n C}{\frac{C}{n} + nC} $ or $C_N = \frac{nC}{( 1 + n^2)}$