In $LR$ circuit the current increases to one fourth of its maximum value in $4\,sec$, then the time constant of the circuit is

$AC$ alternating emf $\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$ volt is applied to a capacitor of $2 \mu \mathrm{F}$, the rms value of current in the circuit is.. . . . . .$\mathrm{mA}$.

Obtain an equation of current for $AC$ voltage applied to an inductor and draw a graph of $V$ and $I$.

An arc lamp requires a direct current of $10\ A$ at $80\ V$ to function. If it is connected to a $ 220\ V (rms ), 50\ Hz\ AC$ supply, the series inductor needed for it to work is close to :

At time $t =0$, terminal $A$ in the circuit shown in the figure is connected to $B$ by a key and alternating current $I ( t )= I _0 \cos (\omega t$,$) , with I _0=1 A$ and $\omega=500 \ rad s ^{-1}$ starts flowing in it with the initial direction shown in the figure.

At $t=\frac{7 \pi}{6 \omega}$, the key is switched from $B$ to $D$. Now onwards on ly $A$ and $D$ are connected. $A$ total charge $Q$ flows from the battery to charge the capacitor fully. If $C =20 \mu, R =10 \Omega$ and the battery is ideal with emf of $50 \ V$, identify the correct statement $(s)$

$(A)$ Magnitude of the maximum charge on the capacitor before $t=\frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3} C$.

$(B)$ The current in the left part of the circuit just before $t=\frac{7 \pi}{6 \omega}$ is clockwise

$(C)$ Immediately after $A$ is connected to $D$. the current in $R$ is $10 A$.

$(D)$ $Q =2 \times 10^{-3} C$.