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A charged capacitor of unknown capacitance is connected in series with 100kOmega resistance and an ideal ammeter. The initial current in the circuit is found to be 0.2mA and the current after 7 sec is found to be 0.1mA. The potential energy that was stored in the capacitor before it was connected in the circuit is ______mJ. [Take log_(2)2=0.7] |
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Answer» Since initial current `i_(0)=(V)/(R), V=20`Volts Time after which current reduces to half `T_(1//2)=RC log_(e)2` Therefore, `C=100muF` so, initial potential energy STORED, `U_(i)=(1)/(2)CV^(2)=(1)/(2)(10^(-4))(20)^(2)=20mJ`. |
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