JEE MAIN - Physics (2011 - No. 8)
A fully charged capacitor $$C$$ with initial charge $${q_0}$$ is connected to a coil of self inductance $$L$$ at $$t=0.$$ The time at which the energy is stored equally between the electric and the magnetic fields is :
$${\pi \over 4}\sqrt {LC} $$
$$2\pi \sqrt {LC} $$
$$\sqrt {LC} $$
$$\pi \sqrt {LC} $$
Explanation
Energy stored in magnetic field $$ = {1 \over 2}L{i^2}$$
Energy stored in electric field $$ = {1 \over 2}{{{q^2}} \over C}$$
$$\therefore$$ $${1 \over 2}L{i^2} = {1 \over 2}{{{q^2}} \over C}$$
Also $$q = {q_0}\,\cos \,\omega t$$ and $$\omega = {1 \over {\sqrt {LC} }}$$
On solving $$t = {\pi \over 4}\sqrt {LC} $$
Energy stored in electric field $$ = {1 \over 2}{{{q^2}} \over C}$$
$$\therefore$$ $${1 \over 2}L{i^2} = {1 \over 2}{{{q^2}} \over C}$$
Also $$q = {q_0}\,\cos \,\omega t$$ and $$\omega = {1 \over {\sqrt {LC} }}$$
On solving $$t = {\pi \over 4}\sqrt {LC} $$
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