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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 $\left(\tan 60^{\circ}=\sqrt{3}\right)$
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Verified Answer
The correct answer is:
$15 \Omega$
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 $Z=\frac{E}{I}=\frac{15}{0.5}=30 \Omega$
$$
\begin{aligned}
& \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \therefore \quad \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}
\end{aligned}
$$
The formula for impedance is
$$
\begin{array}{rl}
Z & =\sqrt{R^2+X_L^2} \\
Z & =\sqrt{R^2+(\sqrt{3} R)^2} \\
Z & =\sqrt{4 R^2} \\
2 & R=Z \\
\therefore \quad R & =\frac{Z}{2}=\frac{30}{2}=15 \Omega
\end{array}
$$
Impedance is given as $Z=\frac{E}{I}=\frac{15}{0.5}=30 \Omega$
$$
\begin{aligned}
& \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\
& \therefore \quad \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}
\end{aligned}
$$
The formula for impedance is
$$
\begin{array}{rl}
Z & =\sqrt{R^2+X_L^2} \\
Z & =\sqrt{R^2+(\sqrt{3} R)^2} \\
Z & =\sqrt{4 R^2} \\
2 & R=Z \\
\therefore \quad R & =\frac{Z}{2}=\frac{30}{2}=15 \Omega
\end{array}
$$
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