The equation of current in a purely inductive circuit is $5 \sin \left(49 \pi t-30^{\circ}\right)$. If the inductance is $30\,mH$ then the equation for the voltage across the inductor, will be.$\left\{\right.$ Let $\left.\pi=\frac{22}{7}\right\}$

A $100 \;\mu \,F$ capacitor in series with a $40\; \Omega$ resistance is connected to a $110\; V , 12\; k\,Hz$ supply.

$(a)$ What is the maximum current in the circuit?

$(b)$ What is the time lag between the current maximum and the voltage maximum?

Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a $dc$ circuit after the steady state.

At a moment $(t = 0)$ when charge on capacitor $C_1$ is zero, the switch is closed. If $I_0$ be the current through inductor at that instant, for $t > 0,$

One $10\, \,\,V, 60\, W$ bulb is to be connected to $100\, V$ line. The required induction coil has self inductance of value $(f = 50\,Hz)$

A series $R-C$ combination is connected to an $AC$ voltage of angular frequency $\omega=500 \ radian / s$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is