Question

A series combination of $N_1$ capacitors (each of capacity $C_1$) is charged to potential difference ${}^{\prime }3V^{\prime }$. Another parallel combination of $N_2$ capacitors (each of capacity $C_2$) is charged to potential difference 'V'. The total energy stored in both the combinations is same. The value of $C_1$ in terms of $C_2$ is

• A. $\frac{C_2{N}_1{N}_2}{9}$
• B. $\frac{C_2{N}_1^2{N}_2^2}{9}$
• C. $\frac{C_2{N}_1}{9N_2}$
• D. $\frac{C_2{N}_1}{9N_1}$

Answer 1

The correct option is A $\frac{C_2{N}_1{N}_2}{9}$

For series combination:-

equivalence capacitance=$\frac{C_1}{N_1}$

Hence, charge on each capacitor, $q=\left(3V\right)\frac{C_1}{N_1}$

Hence, total energy, $U=N_1\times \frac{q^2}{2C_1}=\frac{9C_1{V}^2}{2N_1}$

For parallel combination:-

Potential across each capacitor is $V$.

Hence, total energy, $U=N_2\times \frac{1}{2}{C}_2{V}^2=\frac{N_2{C}_2{V}^2}{2}$

$\frac{9C_1{V}^2}{2N_1}=\frac{N_2{C}_2{V}^2}{2}$

$⟹C_1=\frac{N_1{N}_2{C}_2}{9}$

Hence, answer is option-(A).