A coil having an inductance of $\frac{1}{\pi} \mathrm{H}$ is connected in series with a resistance of $300…
A coil having an inductance of $\frac{1}{\pi} \mathrm{H}$ is connected in series with a resistance of $300 \Omega$. If $20 \mathrm{~V}$ from a $200 \mathrm{~Hz}$ source are impressed across the combination, the value of the phase angle between the voltage and the current is
$\tan ^{-1}\left(\frac{5}{4}\right)$
$\tan ^{-1}\left(\frac{4}{5}\right)$
$\tan ^{-1}\left(\frac{3}{4}\right)$
$\tan ^{-1}\left(\frac{4}{3}\right)$
Solution
$\begin{aligned}
& \mathrm{X}_{\mathrm{L}}=\mathrm{L} \omega=\mathrm{L} \times 2 \pi \mathrm{f} \\
\therefore & \mathrm{X}_{\mathrm{L}}=\frac{1}{\pi} \times 2 \pi \times 200 \\
\therefore & \mathrm{X}_{\mathrm{L}}=400 \Omega
\end{aligned}$
Now, the phase angle' between voltage and current is given by, $\tan \phi \stackrel{X_L}{R}=\frac{400}{300}$
$\therefore \quad \phi=\tan ^{-1}\left(\frac{4}{3}\right)$