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A capacitor is charged to potential difference of 12V

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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.