When a d.c. voltage of $200 \, V$ is applied to a coil of self-inductance $\left(\frac{2 \sqrt{3}}{\pi}\right) \, H$, a current of $1 \, A$ flows through it. But by replacing the d.c. source with an a.c. source of $200 \, V$, the current in the coil is reduced to $0.5 \, A$. Then the frequency of the a.c. supply is: (in $ \, Hz$)

$A$ coil of $200 \, \Omega$ resistance and $1.0 \, H$ inductance is connected to an $ac$ source of frequency $200/2\pi \, Hz.$ The phase angle between potential and current will be.....$^o$

$A$ $12\,\Omega$ resistor and a $0.21\,H$ inductor are connected in series to an $ac$ source operating at $20\,V$,$50\,Hz$. The phase angle between the current and the source voltage is.....$^o$

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 . . . . . . $Henry$.

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 $\pi \approx 3.14$):
$\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

In an $RL$ circuit,the current is $i = 100 \sin(314t) \ A$ and the voltage is $e = 200 \sin(314t + \pi/3) \ V$. If the resistance is $1 \ \Omega$,what is the reactance?