Solveeit Logo

Question

Physics Question on Alternating current

A fully charged capacitor CC with initial charge Q0Q_0 is connected to a coil of self inductance LL at t=0t = 0. The time at which the energy is stored equally between the electric and the magnetic field is

A

π4LC\frac{\pi}{4}\sqrt{LC}

B

2πLC2\pi\sqrt{LC}

C

LC\sqrt{LC}

D

πLC\pi\sqrt{LC}

Answer

π4LC\frac{\pi}{4}\sqrt{LC}

Explanation

Solution

As ω2=1LC\omega^{2}=\frac{1}{LC} or ω=1LC\omega=\frac{1}{\sqrt{LC}} Maximum energy stored in capacitor =12Q02C=\frac{1}{2} \frac{Q^{2}_{0}}{C} Let at any instant tt, the energy be stored equally between electric and magnetic field. Then energy stored in electric field at instant tt is 12Q2C=12[12Q02C]\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \left[\frac{1}{2} \frac{Q^{2}_{0}}{C}\right] or Q2=Q022Q^{2}=\frac{Q^{2}_{0}}{2} or Q=Q02Q0cosωt=Q02Q=\frac{Q_{0}}{\sqrt{2}} \Rightarrow Q_{0}\,cos\,\omega t=\frac{Q_{0}}{\sqrt{2}} or ωt=π4\omega t =\frac{\pi}{4} or t=π4ωt=\frac{\pi}{4\omega} =π4×(1/LC)=\frac{\pi}{4\times\left(1/ \sqrt{LC} \right)} =πLC4=\frac{\pi\sqrt{LC}}{4}