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 . . . . . . .

$A$ capacitor of reactance $4 \sqrt{3} \Omega$ and a resistor of resistance $4 \Omega$ are connected in series with an $AC$ source of peak value $8 \sqrt{2} \text{ V}$. The power dissipation in the circuit is . . . . . $\text{W}$.

The adjoining figure shows an $AC$ circuit with resistance $R$,inductance $L$ and source voltage $V_s$. Then

$A$ $400 \Omega$ resistor, a $250 \text{ mH}$ inductor, and a $2.5 \mu \text{F}$ capacitor are connected in series with an $AC$ source of peak voltage $5 \text{ V}$ and angular frequency $\omega = 2000 \text{ rad/s}$. What is the peak value of the electrostatic energy of the capacitor (in $\mu \text{J}$)?

$A$ series $L, R$ circuit connected with an ac source $E = (25 \sin 1000 t) \ V$ has a power factor of $\frac{1}{\sqrt{2}}$. If the source of emf is changed to $E = (20 \sin 2000 t) \ V$,the new power factor of the circuit will be:

When alternating current is passed through an $L-R$ series circuit,the power factor is $\frac{\sqrt{3}}{2}$ and $R=50 \ \Omega$. If the frequency of the source is $50 \ Hz$,then the value of $L$ is (Assume $\pi \approx 3.14$):
$\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$