When a current of $4 A$ at $50 Hz$ is passed through a coil,the power consumed by the coil is $240 W$. If the voltage of the $AC$ source is $100 V$,what is the inductance of the coil?

$A$ circuit contains an inductance of $\frac{1}{6 \pi} \text{ H}$ and a resistance of $15 \text{ } \Omega$ in series. If an $AC$ voltage of $100 \text{ V}$ and $60 \text{ Hz}$ is applied to the circuit,then the current in the circuit and the phase difference between voltage and current are,respectively:

In an $ac$ circuit,the reactance of a coil is $\sqrt{3}$ times its resistance. The phase difference between the voltage across the coil and the current through the coil will be:

When a coil is connected to an $AC$ supply of frequency $50 \, Hz$, a current of $4 \, A$ flows in it and it consumes $240 \, W$ power. If the potential difference across the coil is $100 \, V$, then the inductance value of the coil is

$A$ light bulb and an inductor coil are connected to an $AC$ source through a key as shown in the figure below. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the light bulb

An $AC$ circuit having supply voltage $E = 10 \sin \omega t$ consists of a resistor of resistance $R = 3 \Omega$ and an inductor of reactance $X_L = 4 \Omega$ as shown in the figure. The voltage across the inductor at $t = \pi / \omega$ is: (in $\text{ V}$)