Electrostatic Potential and Capacitance - Result Question 31
33. 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
[2010]
(a) $\frac{2 C_1}{n_1 n_2}$
(b) $16 \frac{n_2}{n_1} C_1$
(c) $2 \frac{n_2}{n_1} C_1$
(d) $\frac{16 C_1}{n_1 n_2}$
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Answer:
Correct Answer: 33. (d)
Solution:
- (d) In series, $C _{eff}=\frac{C_1}{n_1}$
$\therefore \quad$ Energy stored,
$E_S=\frac{1}{2} C _{e f V_S} V_S^{2}=\frac{1}{2} \frac{C_1}{n_1} 16 V^{2}$
$=8 V^{2} \frac{C_1}{n_1}$
In parallel, $C _{eff}=n_2 C_2$ $\therefore \quad$ Energy stored, $E_p=\frac{1}{2} n_2 C_2 V^{2}$
According to question $E_s=E_p$
$\therefore \quad \frac{8 V^{2} C_1}{n_1}=\frac{1}{2} n_2 C_2 V^{2}$
$\Rightarrow C_2=\frac{16 C_1}{n_1 n_2}$
