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: (in $H$)

When a coil is connected to a $d.c.$ source of $e.m.f.$ $12 \ V$,the current of $4 \ A$ flows in it. If the same coil is connected to a $12 \ V, 50 \ Hz$ $a.c.$ source,the current flowing in it is $2.4 \ A$. The self-inductance of the coil will be:

The figure given below shows an $LCR$ series circuit with two switches $S_1$ and $S_2$. When switch $S_1$ is closed keeping $S_2$ open,the phase difference $(\phi)$ between the current and source voltage is $30^\circ$ and the phase difference is $60^\circ$ when $S_2$ is closed keeping $S_1$ open. The value of $(3L_1 - L_2)$ is . . . . . . $H$.

An inductor and a resistor are connected in series to an $AC$ source of $10 \ V$. If the potential difference across the inductor is $6 \ V$,then the potential difference across the resistor is: (in $V$)

An $AC$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current registered is $I$. If the frequency of the source is changed to $\omega/3$ (maintaining the same voltage),the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$.

An $A.C.$ source is connected to a series $LCR$ circuit. If the voltage across $R$ is $40 \,V$, the voltage across $L$ is $80 \,V$, and the voltage across $C$ is $40 \,V$, then the e.m.f. '$e$' of the $A.C.$ source is: