We have two capacitors `C_1 and C_2` of capacitance `12 muF` and `6 muF` respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch `S_1` is closed but switch `S_2` is open. After some time, the switch `S_1` is opened and simultaneously switch `S_2` is closed.
Now answer the following questions :
The fractional loss in electrostatic energy of the arrangement after closing the switch `S_2` is

Options

A. `1/4`
B. `1/2`
C. `1/3`
D. `3/4`

Answer

C

Explanation

Initial value of electrostatic energy of the arrangement
`u_i= 1/2 C_1 V_1^2 = 1/2 xx (12 mu F) xx (6 V)^2 = 126muJ`
and final value of electrostatic energy of the arrangement
`u_f = 1/2 (C_1 + C_2) V^2 = 1/2 (12 muF + 6 mu F) xx (4V)^2 = 144 muJ`
`:.` Loss in electrostatic energy `Delta u_i = u_i = (216 - 144) = 72 mu J`
`rArr (Delta u) /u_1 = 72/216 = 1/3`