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A series combination of $n_1$ capacitors, each of value $C_1$, is charged by a source of potential difference 4V. 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
$\frac{16C_1}{n_1n_2}$
B
$\frac{2C_1}{n_1n_2}$
C
$16\frac{n_2}{n_1}C_1$
D
$2\frac{n_2}{n_1}C_1$
Detailed Solution
Series combination of $n_1$ capacitors each of $C_1$: $C_s = \frac{C_1}{n_1}$
Energy stored at potential difference 4V: $U_1 = \frac{1}{2}C_s(4V)^2 = \frac{1}{2}\frac{C_1}{n_1}\times16V^2 = \frac{8C_1V^2}{n_1}$
Parallel combination of $n_2$ capacitors each of $C_2$: $C_p = n_2C_2$
Energy stored at potential difference V: $U_2 = \frac{1}{2}n_2C_2V^2$
Given $U_1 = U_2$: $\frac{8C_1V^2}{n_1} = \frac{1}{2}n_2C_2V^2$
$C_2 = \frac{16C_1}{n_1n_2}$
Energy stored at potential difference 4V: $U_1 = \frac{1}{2}C_s(4V)^2 = \frac{1}{2}\frac{C_1}{n_1}\times16V^2 = \frac{8C_1V^2}{n_1}$
Parallel combination of $n_2$ capacitors each of $C_2$: $C_p = n_2C_2$
Energy stored at potential difference V: $U_2 = \frac{1}{2}n_2C_2V^2$
Given $U_1 = U_2$: $\frac{8C_1V^2}{n_1} = \frac{1}{2}n_2C_2V^2$
$C_2 = \frac{16C_1}{n_1n_2}$