NEET 2020 Model Question Paper 3

Initially, the energy stored in the circuit is

Q2

2C

When the switch S is turned into position B, the net capacitance becomes C + 3C = 4C and total charge Q remains the same
So, the energy stored will be

Q2

2(4C)

=

Q2

8C

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

Q2

8C

=

Q2

2C

=−

3

8

Q2

C