A capacitor is charged to potential difference of 12V it delivers 40% of its stored energy to lamp? what is final potential difference across capacitor? a. 4.3V b. 4.8V c. 7.2V d. 9.3V

A capacitor is charged to potential difference of 12V. It delivers 40% of its stored to lamp.The final potential difference across capacitor is 9.295volts.
Initially:
We know that
Energy stored in capacitor=##E_i=1/2CV^2##
Energy stored in capacitor=##E_i=1/2C(12)^2## [since initial voltage=12 v]
Energy stored in capacitor=##E_i=72C## joulseAfter consumption of 40% energy:
Remaining energy in capacitor is 60% of initial energy.
##E_f=0.6*E_i## joulse
##E_f=0.6*(72C)## joulse
##E_f=43.2C## joulse__(1)
but ##E_f=1/2C(V_f)^2##___(2)
From equation(1) and equation(2);
##1/2C(V_f)^2=43.2C##
##1/2cancel(C)(V_f)^2=43.2cancel(C)##
##V_f=9.2951## volts
The final potential difference across capacitor if is constant(didn’t mentioned in question) is 9.295volts.