If the power factor is $1/2$ in a series $RL$ circuit with $R = 100\,\Omega$ connected to an $ac$ mains $(50\,Hz)$,find the value of $L$.

$A$ resistor of $200 \; \Omega$ and a capacitor of $15.0 \; \mu F$ are connected in series to a $220 \; V, 50 \; Hz$ $ac$ source.
$(a)$ Calculate the current in the circuit.
$(b)$ Calculate the voltage $(rms)$ across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,resolve the paradox.

Calculate the capacitive reactance of a capacitor in order to run a bulb rated at $10 \, W, 60 \, V$ when connected in series to an $a.c.$ source of $100 \, V$.

$A$ 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

$A$ coil having an inductance of $\frac{1}{\pi} \text{ H}$ is connected in series with a resistance of $300 \text{ } \Omega$. If an $A$.$C$. source $(20 \text{ V}, 200 \text{ Hz})$ is connected across the combination,the phase angle between voltage and current is

In an $AC$ circuit,a resistance of $R$ $\Omega$ is connected in series with an inductor of self-inductance $L$. If the phase angle between voltage and current is $45^{\circ}$,the value of inductive reactance $(X_{L})$ will be equal to . . . . . . .