Physics - Capacitor Question with Solution | TestHub

Let $n$ capacitors be connected in parallel and the remaining $7-n$ capacitors be connected in series with this parallel combination.

The equivalent capacitance of the $n$ capacitors in parallel is:
$C_p=nC$

The equivalent capacitance $C_{eq}$ of the entire circuit is given by:
$\frac{1}{C_{eq}}=\frac{1}{C_p}+\frac{7-n}{C}$

$\frac{1}{C_{eq}}=\frac{1}{nC}+\frac{7-n}{C}=\frac{1+7n-n^2}{nC}$

Given $C_{eq}=\frac{6}{13}\mu \mathrm{F}$ and $C=2\mu \mathrm{F}$, we substitute these values:
$\frac{13}{6}=\frac{1+7n-n^2}{2n}$

Cross-multiplying yields:
$26n=6(1+7n-n^2)$

$26n=6+42n-6n^2$

$6n^2-16n-6=0$

Dividing by 2:
$3n^2-8n-3=0$

Factoring the quadratic equation:
$3n^2-9n+n-3=0$

$3n(n-3)+1(n-3)=0$

$(3n+1)(n-3)=0$

Since $n$ must be a positive integer, $n=3$.

Therefore, 3 capacitors are in parallel and the remaining 4 capacitors are in series with this combination.

Answer: 3 capacitors in parallel and then 4 capacitors in series.