Question

A series combination of $N_1$ capacitors (each of capacity $C_1$) is charged to potential difference '3V'. 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. $\frac{C_{2}N_{1}}{9 N_2}$
• D. $\frac{C_{2}N_{2}}{9 N_1}$

Answer 1

Answer: (A)

In the first condition,
$C_{ eq }=\frac{C_{ G }}{N_{1}}$
Potential difference $(V)=3 V$
$\therefore \,\,\,\,\,\,$ Energy stored $\left(E_{1}\right)=\frac{1}{2} C V^{2}$
$=\frac{1}{2}\left(\frac{C_{1}}{N_{1}}\right)(3 V)^{2}$
$=\frac{9}{2} \frac{C_{1}}{N_{1}} V^{2} \,\,\,\,\,\,\,\,\, ....(i)$
In the second condition,
$C_{ eq }=N_{2} C_{2},$ potential difference $=V$
Energy stored $\left(E_{2}\right)=\frac{1}{2} C V^{2} $
$=\frac{1}{2} N_{2} C_{2} V^{2} \,\,\,\,\,\,\,\, ...(ii)$
According to the question,
$E_{1}=E_{2}$
From Eqs. (i) and (ii), we get
$\frac{9}{2} \frac{C_{1}}{N_{1}} V^{2} =\frac{1}{2} N_{2} C_{2} V^{2} $
$C_{1} =C_{2} \frac{N_{2} N_{1}}{9}$