Electrostatics 5 Question 23
25. A $2 \mu \mathrm{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 is
(2011)
(a) $0 %$
(b) $20 %$
(c) $75 %$
(d) $80 %$
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Answer:
Correct Answer: 25. (d)
Solution:
- $q_{i}=C_{i} V=2 V=q$
This charge will remain constant after switch is shifted from position 1 to position 2 .
$$ \begin{aligned} U_{i} & =\frac{1}{2} \frac{q^{2}}{C_{i}}=\frac{q^{2}}{2 \times 2}=\frac{q^{2}}{4} \\ U_{f} & =\frac{1}{2} \frac{q^{2}}{C_{f}}=\frac{q^{2}}{2 \times 10}=\frac{q^{2}}{20} \end{aligned} $$
$\therefore$ Energy dissipated $=U_{i}-U_{f}=\frac{q^{2}}{5}$
This energy dissipated $=\frac{q^{2}}{5}$ is $80 %$ of the initial stored energy $=\frac{q^{2}}{4}$.
$\therefore$ Correct option is (d).
