Question-20 If maximum energy is stored in a capacitor at t=0 then the time after which, current in the circuit will be maximum is
(A) \pi\times10^{-3}\ s
(B) 2\pi\times10^{-3}\ s
(C) 2\pi\times10^{-4}\ s
(D) \pi\times10^{-4}\ s
Answer : (B)
Explanation:
At t = 0, the energy and hence the charge stored in the capacitor is maximum.
Since, q=q_0\cos\omega_0 t
∴ i=\frac{dq}{dt}=q_0\omega_0\sin\omega_0 t=i_0\sin\omega_0 t
When i=i_0, i_0=i_0\sin\omega_0 t
∴ \sin\omega_0 t=1
∴ \omega_0 t=\frac{\pi}{2}
∴ t=\frac{\pi}{2\omega_0}
Now, \omega_0=\sqrt{\frac{1}{LC}}=\frac{1}{\sqrt{16\times10^{-3}\times10^{-5}}}
=\frac{1}{\sqrt{16\times10^{-8}}}=\frac{1}{4\times10^{-4}}=2500\ rad/s∴ t=\frac{\pi}{2\times2500}=\frac{\pi}{5000}=2\pi\times10^{-4}\ s