Question

$n$ 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

• A. $nC$
• B. $n^{2}C/(n^2+1)$
• C. $nC/(n^2+1)$
• D. $(n^2+1)C/n$

Answer 1

Answer: (C)

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,
$C_N=\frac{C_S\times C_P}{C_S+C_P}=\frac{\frac{C}{n}\times nC}{\frac{C}{n}+nC}$ or $C_N=\frac{nC}{(1+n^2)}$