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. $\frac{C_2{N}_1}{9N_2}$
• D. $\frac{C_2{N}_2}{9N_1}$

Answer 1

Answer: (A)

In the first condition,
$C_{eq}=\frac{C_G}{N_1}$
Potential difference $(V)=3V$
$\therefore$ Energy stored $\left(E_1\right)=\frac{1}{2}C{V}^2$
$=\frac{1}{2}\left(\frac{C_1}{N_1}\right)(3V{)}^2$
$=\frac{9}{2}\frac{C_1}{N_1}{V}^2....(i)$
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
$\frac{9}{2}\frac{C_1}{N_1}{V}^2=\frac{1}{2}{N}_2{C}_2{V}^2$
$C_1=C_2\frac{N_2{N}_1}{9}$