Answer:
Option D
Explanation:
$q_{i}=C_{i}V=2V=q(say)$
This charge will remain constant after switch is shifted from position 1 to position 2.
$ 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}$
$\therefore$ Energy Dissipated = $U_{i}-U_{f}= \frac{q^{2}}{5}$
This energy dissipated $\left(=\frac{q^{2}}{5}\right)$ is 80%
of the initial stored energy $\left(=\frac{q^{2}}{4}\right)$
Analysis of Question
(i) This question is moderately tough'
(ii) In a capacitor circuit, redistribution of charge takes Place under the following three conditions'
(a) A switch is closed.
(b) A closed switch is opened.
(c) A switch is shifted from one Position to another Position.
In the redistribution of charge, energy is dissipated
