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.