question_answer

In   the   figure shown, after the switch S is turned from position A to position B the energy dissipated in the circuit in terms of capacitance C and total charge Q is-                                             [JEE Main 12-Jan-2019 Morning]

A) \[\frac{5}{8}\frac{{{Q}^{2}}}{C}\]

B) \[\frac{1}{8}\frac{{{Q}^{2}}}{C}\]

C) \[\frac{3}{8}\frac{{{Q}^{2}}}{C}\]

D) \[\frac{3}{4}\frac{{{Q}^{2}}}{C}\]

Correct Answer: C

Solution :

[c] Initially the energy stored in the circuit is

When the switch S is turned into position B the net capacitance becomes and total charge Q remains the same.

So the energy stored will be

So the difference of energy is dissipated in the given situation i.e.