Question

A series combination of $n_1$ capacitors, each of capacity $C_1$ is charged by source of potential difference $4\, V$. When another parallel combination of $n_2$ capacitors each of capacity $C_2$ is charged by a source of potential difference $V$, it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is then

• A. $16 \frac{n_{2}}{n_{1}}C_{1}$
• B. $\frac{2 C_{1}}{n_{1}n_{2}}$
• C. $2 \frac{n_{1}}{n_{2}}C_{2}$
• D. $\frac{16 C_{1}}{n_{1}n_{2}}$

Answer 1

Answer: (D)

$\frac{16 C_{1}}{n_{1}n_{2}}$

Answer 2

Equivalent capacitance of $n_2$ number of capacitors each of capacitance $C_2$ in parallel $= n_2C_2$ Equivalent capacitance of $n_1$ number of capacitors each of capacitances $C_1$ in series. Capacitance of each is $C_{1} \frac{C_{1}}{n_{1}}$ According to question, total energy stored in both the combinations are same $i.e, \frac{1}{2}\left(\frac{C_{1}}{n_{1}}\right)\left(4V\right)^{2} = \frac{1}{2}\left(n_{2}C_{2}\quad\right)^{2}$ $\therefore\,C_{2} = \frac{16C_{1}}{n_{1}n_{2}}$