- Home
- Standard 12
- Physics
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 $n_2$ capacitors, each of value $C_2$, is charged by a source of potential difference $V$, it has the same (total) energy store in it, as the first combination has. The value of $C_2$, in terms of $C_1$, is then
$\frac{{2{C_1}}}{{{n_1}{n_2}}}$
$16\frac{{{n_2}}}{{{n_1}}}{C_1}$
$2\frac{{{n_2}}}{{{n_1}}}{C_1}$
$\frac{{16{C_1}}}{{{n_1}{n_2}}}$
Solution
In series, the equivalent capacitance is$:$ $C s=\frac{C_{1}}{n_{1}}$ and
Energy stored, $E_{S}=\frac{1}{2} C_{S} V_{S}^{2}=\frac{1}{2}\left(C_{1} / n_{1}\right)(4 V)^{2}=\frac{8 C_{1} V^{2}}{n_{1}}$
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, $E_{S}=E_{P}$
$\frac{8 C_{1} V^{2}}{n_{1}}=\frac{n_{2} C_{2} V^{2}}{2}$
$\therefore C_{2}=\frac{16 C_{1}}{n_{1} n_{2}}$
