Question

A series combination of $n_1$ capacitors, each of value $C_1$ , is charged by a source of potential different 4 V. When another parallel combination of $n_2$ capacitors, each of value $C_2$, is charged y a source of potential difference V, it has the same(total) energy stored in it, as the first combination has. The value $C_2$, 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. $\frac{{16}_{C1}}{{n2}_{n1}}$

Answer 1

The correct option is D $\frac{{16}_{C1}}{{n2}_{n1}}$

1st case:-

Energy $=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{C_1}{n_1}\right)(4V{)}^2=\frac{8C_1{V}^2}{n_1}$

2nd case:-

Energy$=\frac{1}{2}c{V}^2=\frac{1}{2}\left(n_2{C}_2\right)V^2$

According to question-

$\frac{8C_1{V}^2}{n_1}=\frac{n_2{C}_2{V}^2}{2}\Rightarrow \frac{8C_1}{n_1}=\frac{n_2{C}_2}{2}\Rightarrow \frac{16C_1}{n_1{n}_2}=C_2$