A $20\, Henry$ inductor coil is connected to a $10\, ohm$ resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, 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.. . . . . .$\mathrm{mA}$.

A coil of inductance $0.50\;H$ and resistance $100 \;\Omega$ is connected to a high frequency supply $(240\; V, 10 \;k\,Hz)$

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

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

Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a $dc$ circuit after the steady state?

An ac-circuit having supply voltage $E$ consists of a resistor of resistance $3\Omega$ and an inductor of reactance $4\Omega$ as shown in the figure. The voltage across the inductor at $t = \pi /\omega$ is…….$volts$

There is no resistance in the inductive circuit. Kirchhoff 's voltage law for the circuit is