The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,behaves effectively as:

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 $H$)

An $a.c.$ 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 $\frac{\omega}{3}$ (while maintaining the same voltage),the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency $\omega$ is:

An inductor and a resistor are connected in series to an ac supply. If the potential differences across the inductor and the resistor are $180 \ V$ and $240 \ V$ respectively,then the voltage of the ac supply is (in $V$)

$A$ series $R-C$ circuit is connected to an $AC$ voltage source. Consider two cases: $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with a dielectric of constant $K = 4$. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following statements is/are true?
$(A)$ $I_R^A > I_R^B$
$(B)$ $I_R^B > I_R^A$
$(C)$ $V_C^A > V_C^B$
$(D)$ $V_C^B > V_C^A$

An inductor of self-inductance $1 \ H$ is connected in series with a resistor of $100 \pi \ \Omega$ and an $AC$ supply of $100 \pi \ V$,$50 \ Hz$. The maximum current flowing in the circuit is . . . . . . $A$.