An a.c. source of \(15 \mathrm{~V}, 50 \mathrm{~Hz}\) is connected across an inductor (L) and resistance (R) in series R.M.S. current of \(0.5 \mathrm{~A}\) flows in the circuit. The phase difference between applied voltage and current is \(\left(\frac{\pi}{3}\right)\) radian. The value of resistance \((R)\) is \(\left(\tan 60^{\circ}=\sqrt{3}\right)\)

Given data: \(\mathrm{E}=15 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}, \mathrm{I}=0.5 \mathrm{~A}\), \(\phi=\frac{\pi}{3} \mathrm{rad}\)
Impedance is given as \(Z=\frac{E}{I}=\frac{15}{0.5}=30 \Omega\)
\(
\begin{aligned}
& \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \therefore \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}
\end{aligned}
\)
The formula for impedance is
\(Z =\sqrt{R^2+X_L^2} \)
\( Z =\sqrt{R^2+(\sqrt{3} R)^2} \)
\( Z =\sqrt{4 R^2} \)
\( 2 R=Z \)
\( \therefore \quad R =\frac{Z}{2}=\frac{30}{2}=15 \Omega\)