In the circuit shown, switch S is placed in position 1 till the capacitor is charged to half of the maximum possible charge in this situation. Now, the switch S is placed in position 2. The maximum energy lost by the circuit after switch S is placed in position 2 is

Options

A. `(1)/(2) CE^(2)`
B. `(1)/(8)CE^(2)`
C. `(7)/(8)CE^(2)`
D. `(9)/(8)CE^(2)`

Answer

D

Explanation

Initially, capacitor is charged to Q//2.
Energy stored in the capacitor=`(1)/(2)((Q)/(2))^(2)xx(1)/(C)=(1)/(8)(Q^(2))/(2)implies(1)/(8)CE^(2)=U_(i)` Finally, energy stored in the capacitor,`U_(f)=(1)/(2)CE^(2)`
therefore Work done by the battery =`((3)/(2)CE)xxE=(3)//(2)CE^(2)`
Heat energy lost = `=(3)/(2)CE^(2)-(U_(f)-U_(i))=(3)/(2)CE^(2)-((1)/(2)CE^(2)-(1)/(8)CE^(2))=(9)/(8)CE^(2)`
Hence, (D) is correct.