Question

A series combination of $n_1$ capacitors, each of value $C_1$, is charged by a source of potential difference 4 V. When another parallel combination of $n_2$ capacitors, each of value $C_2$, 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. $\frac{2C_1}{n_1{n}_2}$
• B. $16\frac{n_2}{n_1}{C}_1$
• C. $2\frac{n_2}{n_1}{C}_1$
• D. $\frac{16C_1}{n_1{n}_2}$

Answer 1

Answer: (D)

$\frac{16C_1}{n_1{n}_2}$

In series, the equivalent capacitance is: $C_S=\frac{C_1}{n_1}$ and
Energy stored, $E_S=\frac{1}{2}{C}_S{V}_S^2=\frac{1}{2}\left(C_1/n_1\right)(4V{)}^2=\frac{8C_1{V}^2}{n_1}$

In parallel, the equivalent capacitance is $C_P=n_2{C}_2$ and
Energy stored , $E_P=\frac{1}{2}{C}_P{V}_P^2=\frac{1}{2}\left(n_2{C}_2\right)(V{)}^2=\frac{n_2{C}_2{V}^2}{2}$

Here, $E_S=E_P$

$\frac{8C_1{V}^2}{n_1}=\frac{n_2{C}_2{V}^2}{2}$

$\therefore C_2=\frac{16C_1}{n_1{n}_2}$