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Asked by jmorakhia | 08 May, 2019, 02:23: PM
answered-by-expert Expert Answer
Total work done by the battery in fully charging the capacitor is given by
W equals Q V equals C V squared
Energy stored in the capacitor is
U equals 1 half C V squared
The final energy stored in the capacitor is half of the work done by the battery.The other half is dissipated in the resistor in the form of heat.
Energy stored in capacitor
equals 1 half cross times 10 cross times 10 to the power of negative 6 end exponent cross times 20 squared
equals 2 m J
Hence the enrgy supplied by the battery is greater than 2mJ.
Answered by Utkarsh Lokhande | 08 May, 2019, 02:51: PM
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