$A$ resistance of $300 \ \Omega$ and an inductance of $\frac{1}{\pi} \ H$ are connected in series to an $AC$ voltage of $20 \ V$ and $200 \ Hz$ frequency. The phase angle between the voltage and current is

In a series $L-C-R$ circuit,the frequency of a $10 \, V$ $a.c.$ voltage source is adjusted in such a fashion that the reactance of the inductor measures $15 \, \Omega$ and that of the capacitor $11 \, \Omega$. If $R = 3 \, \Omega$,the potential difference across the series combination of $L$ and $C$ will be.....$V$.

An $AC$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit is found to be $I$. Now,the frequency of the source is changed to $\frac{\omega}{3}$ (maintaining the same voltage),and the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

$A$ resistor of $450 \Omega$ and an inductor are connected in series to an ac source of frequency $\frac{75}{\pi} \text{ Hz}$. If the power factor of the circuit is $0.6$,then the inductance connected in the circuit is:

In an $RL$ circuit,the resistance is $\pi \sqrt{3} \,\Omega$. The phase difference between the current and the voltage is $30^\circ$. The $ac$ frequency is $50 \,Hz$. The inductance is . . . . . . $Henry$.

In the given $AC$ circuit,the phase difference between $I_1$ and $I_2$ is: