Question 427: AC fundamentals

Question

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 $$\left(	an \frac{\pi}{3}=\sqrt{3}ight)$$

Accepted Answer

$$R=500 \Omega, \phi=60^{\circ}$$
When capacitance is removed
$$\begin{aligned}
& 	an \phi=	an 60^{\circ}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
	herefore & \frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=\sqrt{3} \quad 	ext { OR } \quad \mathrm{X}_{\mathrm{L}}=\mathrm{R} \sqrt{3}
\end{aligned}$$
When inductance is removed,
$$\begin{aligned}
& \frac{X_{\mathrm{C}}}{R}=	an 60^{\circ}=\sqrt{3} \
	herefore \quad X_C & =R \sqrt{3} \
	herefore \quad X_C & =X_L
\end{aligned}$$
Impedance $$\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight)^2}=\mathrm{R}=500 \Omega$$

Answer

$$\frac{500}{\sqrt{3}} \Omega$$

Answer

$$500 \sqrt{3} \Omega$$

Answer

$$250 \Omega$$

Answer

$$500 \Omega$$

AC fundamentals

Choose the correct answer: