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A series LCR circuit with resistance $R=500 \mathrm{ohm}$ is connected to an a.c. source of $250 \mathrm{~V}$. When only the capacitance is removed, the current lags behind the voltage by $60^{\circ}$. When only the inductance is removed, the current leads the voltage by $60^{\circ}$. The impedance of the circuit is
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Verified Answer
The correct answer is:
$500 \Omega$
When capacitance is removed
$$
\begin{aligned}
& \tan \phi=\tan 60^{\circ}=\frac{X_L}{R} \\
& \therefore \frac{X_L}{R}=\sqrt{3} \text { or } X_L=R \sqrt{3}
\end{aligned}
$$
When inductance is removed,
$$
\begin{aligned}
& \frac{X_C}{R}=\tan 60^{\circ}=\sqrt{3} \\
& \therefore X_C=R \sqrt{3} \\
& \therefore X_C=X_L
\end{aligned}
$$
Impedance, $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)^2}=\mathrm{R}=500 \Omega$
$$
\begin{aligned}
& \tan \phi=\tan 60^{\circ}=\frac{X_L}{R} \\
& \therefore \frac{X_L}{R}=\sqrt{3} \text { or } X_L=R \sqrt{3}
\end{aligned}
$$
When inductance is removed,
$$
\begin{aligned}
& \frac{X_C}{R}=\tan 60^{\circ}=\sqrt{3} \\
& \therefore X_C=R \sqrt{3} \\
& \therefore X_C=X_L
\end{aligned}
$$
Impedance, $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)^2}=\mathrm{R}=500 \Omega$
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