$A$ coil of resistance $450 \Omega$ and self-inductance $1.5 \text{ H}$ is connected to an $A$.$C$. source of frequency $\frac{150}{\pi} \text{ Hz}$. The phase difference between voltage and current is

$A$ series combination of resistor $R$ and capacitor $C$ is connected to an $A$.$C$. source of angular frequency $\omega$. Keeping the voltage same,if the frequency is changed to $\frac{\omega}{3}$,the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

In the given circuit, the r.m.s. value of the current through the resistor $R$ is:

When an inductor $L$ and a resistor $R$ in series are connected across a $15 \, V, 50 \, Hz$ a.c. supply, a current of $0.3 \, A$ flows in the circuit. The current differs in phase from the applied voltage by $(\frac{\pi}{3})^c$. The value of $R$ is:
$(\sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}, \sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2})$

An $a.c.$ source is applied to a series $LR$ circuit with $X_L = 3R$ and power factor is $X_1$. Now a capacitor with $X_c = R$ is added in series to the $LR$ circuit and the power factor is $X_2$. The ratio $X_1$ to $X_2$ is

$A$ lamp is connected in series with a capacitor and an $AC$ source. What happens if the capacitance of the capacitor is reduced?