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Question

Physics Question on Capacitors and Capacitance

The capacity of a condenser is 4×106F4 \times 10^{-6} F and its potential is 100V100\, V. The energy released on discharging it fully will be

A

0.04 J

B

0.02 J

C

0.025 J

D

0.05 J

Answer

0.02 J

Explanation

Solution

Energy released on discharging the capacitor fully is given by
E=12CV2E=\frac{1}{2} C V^{2}
Given, C=4×106F,V=100C=4 \times 10^{-6} F,\,\, V=100 volt
E=12×4×106×(100)2\therefore E=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^{2}
E=2×102\Rightarrow E=2 \times 10^{-2}
=0.02J=0.02\, J