When an inductor ' \(L\) ' and a resistor ' \(R\) ' in series are connected across a \(15 \mathrm{~V}, 50 \mathrm{~Hz}\) a.c. supply, a current of \(0.3 \mathrm{~A}\) flows in the circuit. The current differs in phase from applied voltage by \(\left(\frac{\pi}{3}\right)^c\). The value of ' \(R\) ' is
\(
\left(\sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}, \sin \frac{\pi}{3}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right)
\)

\(\text { Given: } E_v=15 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}, \mathrm{I}=0.3 \mathrm{~A}, \)
\( \phi=\frac{\pi}{3} \mathrm{rad} \)
\( \text { Impedance } Z=\frac{\mathrm{E}_{\mathrm{v}}}{\mathrm{I}}=\frac{15}{0.2}=50 \Omega \)
\( \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \)
\( \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \)
\( \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \)
\( \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R} \)
\( \mathrm{Impedance} Z=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2} \)
\( \mathrm{Z}=\sqrt{\mathrm{R}^2+(\sqrt{3} \mathrm{R})^2} \)
\( \mathrm{Z}=\sqrt{4 \mathrm{R}^2} \)
\( 2 \mathrm{R}=\mathrm{Z} \mathrm{R}=\frac{\mathrm{Z}}{2}=\frac{50}{2}=25 \Omega\)