Question 286: AC fundamentals

Question

If maximum energy is stored in a capacitor at $\mathrm{t}=0$ then the time after which, current in the circuit will be maximum is

Accepted Answer

At $t=0$, the energy and hence the charge stored in the capacitor is maximum.
$$ \begin{array}{ll}  & 	ext { Since, } \mathrm{q}=\mathrm{q}_0 \cos \omega_0 \mathrm{t} \ 	herefore & \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{q}_0 \omega_0 \sin \omega_0 \mathrm{t}=\mathrm{i}_0 \sin \omega_0 \mathrm{t} \ & 	ext { When } \mathrm{i}=\mathrm{i}_0, \mathrm{i}_0=\mathrm{i}_0 \sin \omega_0 \mathrm{t} \ 	herefore & \sin \omega_0 \mathrm{t}=1 \ 	herefore & \omega_0 \mathrm{t}=\frac{\pi}{2} \ 	herefore & \mathrm{t}=\frac{\pi}{2 \omega_0} \end{array} $$
$$ \begin{aligned} 	ext { Now, } \omega_0 & =\sqrt{\frac{1}{\mathrm{LC}}}=\frac{1}{\sqrt{16 	imes 10^{-3} 	imes 10^{-5}}} \ & =\frac{1}{\sqrt{16 	imes 10^{-8}}}=\frac{1}{4 	imes 10^{-4}}=2500 \mathrm{rad} / \mathrm{s} \ 	herefore \quad t & =\frac{\pi}{2 	imes 2500}=\frac{\pi}{5000}=2 \pi 	imes 10^{-4} \mathrm{~s} \end{aligned} $$

Answer

$\pi 	imes 10^{-3} \mathrm{~s}$

Answer

$2 \pi 	imes 10^{-3} \mathrm{~s}$

Answer

$2 \pi 	imes 10^{-4} \mathrm{~s}$

Answer

$\pi 	imes 10^{-4} \mathrm{~s}$

AC fundamentals

If maximum energy is stored in a capacitor at \mathrm{t}=0 then the time after which, current in the circuit will be maximum is

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