Discussion Forum

1)

A $2 \mu F$  capacitor i scharge as shown in the figure . The percentage of its stored energy dissipated after the switch  S is turned to position 2 , is 

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