$n_1$ identical capacitors, each of capacitance $C_1$ are connected in series and charged by a source of potential difference 3 V. $n_2$ identical capacitors, each of capacitance $C_2$ are connected in parallel and charged by a source of potential difference 2 V. Total energy stored in both the combinations is the same. What is the value of capacitance $C_2$ in terms of $C_1$?

$\frac{9 C_1}{4 n_1 n_2}$

The correct answer is Option (2) → $\frac{9 C_1}{4 n_1 n_2}$

Energy stored in capacitor, $U=\frac{1}{2}CV^2$

where,

C = Equivalent capacitance

V = Voltage Across the applied system

Case 1: for $n_1$ identical capacitors each of capacitance $C_1$ connected in series -

then,

$C_{eq}=\frac{C_1}{n_1}$

$U_{series}=\frac{1}{2}(\frac{C_1}{n_1})3^2=\frac{9C_1}{2n_1}$ ($V=3V$ [given])

Case 2: for $n_2$ identical capacitor each of capacitance $C_1$ connected in parallel -

$C_{eq}=n_2C_2$

$U_{parallel}=\frac{1}{2}n_2C_2(2)^2=2n_2C_2$ ($V=2V$ [given])

$U_{series}=U_{parallel}$ [given]

$⇒\frac{9C_1}{2n_1}=2n_2C_2$

$⇒C_2=\frac{9C_1}{4n_1n_2}$