In a series circuit $C = 2\mu F,\ L = 1mH$ and $R = 10\Omega$. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Question

Dr. Abubacker · Accepted Answer

(d) Current will be maximum in the condition of resonance so $i_{\max} = \frac{V}{R} = \frac{V}{10}A$
Energy stored in the coil $W_{L} = \frac{1}{2}Li_{\max}^{2} = \frac{1}{2}L\left( \frac{E}{10} ight)^{2}$
$= \frac{1}{2} 	imes 10^{- 3}\left( \frac{E^{2}}{100} ight) = \frac{1}{2} 	imes 10^{- 5}E^{2}\ joule$
$	herefore$ Energy stored in the capacitor
$W_{C} = \frac{1}{2}CE^{2} = \frac{1}{2} 	imes 2 	imes 10^{- 6}E^{2} = 10^{- 6}E^{2}\ joule$
$	herefore\frac{W_{C}}{W_{L}} = \frac{1}{5}$

Q. In a series circuit $C = 2\mu F,\ L = 1mH$ and $R = 10\Omega$. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Explanation

Reviewed by Dr. Abubacker

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