If maximum energy is stored in a capacitor at $t=0$ then find the time after which current in the circuit will be maximum.

\[\begin{align}
  & A.\dfrac{\pi }{2}ms \\
 & B.\dfrac{\pi }{4}ms \\
 & C.\pi ms \\
 & D.2ms \\
\end{align}\]

Hint: In a LCR circuit, the inductor $L$, capacitance $C$ and resistance $R$ are connected to a AC source. Here the LCR is connected in series circuit; it can also be connected in parallel circuit. Then the phase difference between the current and the voltage is $\theta$ generally.

Formula used:
 $T=2\pi\sqrt{LC}$

Complete step by step answer:
We know that the source of an AC circuit is sinusoidal. Then there is a phase difference between the voltage and current. If the phase difference between the current and voltage is zero, then both are said to be in phase, and if the phase difference is not equal to zero, then both are said to be out of phase. Resonance is a special condition, which is observed when RLC is connected in series; here there is no phase difference between the current and the voltage.
We know that the frequency of the circuit is given as,$\omega=\dfrac{1}{\sqrt{LC}}$
$\implies T=2\pi\sqrt{LC}$, where $T$ is the time-period of one cycle.
Here, we have LC circuit, and given that, $L=25mH$ and $C=10\mu F$, substituting the values, we get,
$T=2\pi\sqrt{25\times 10^{-3}\times 10\times 10^{-6}}$
$\implies T=2\pi\sqrt{25\times 10^{-8}}$
$\implies T=2\pi\times 5\times 10^{-4}$
$\therefore T=\pi\times 10^{-3}s$
Since the AC is sinusoidal in nature and the time-period of the wave is $T=\pi\times 10^{-3}s$, we know that at $\dfrac{T}{4}$, the capacitance discharges the charge stored in it, and thus the current is maximum. Hence at $T_{max}=\dfrac{T}{4}=\dfrac{\pi\times 10^{-3}}{4}$
$\therefore T_{max}=\dfrac{\pi}{4}ms$

So, the correct answer is “Option B”.

Note:
$Z$ is the impedance of the circuit or the total resistance offered by the circuit, it is given as $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$. Here $X_{L},X_{C}$ are the inductive reactance and the capacitive reactance. Using the frequency of the circuit, we can solve this sum.