Question 368: AC fundamentals

Question

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 $$\left(\sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}, \sin \frac{\pi}{3}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right)$$

Accepted Answer

$$\begin{array}{ll} 
& 	ext { Given: } E_v=15 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}, \mathrm{I}=0.3 \mathrm{~A}, \
& \phi=\frac{\pi}{3} \mathrm{rad} \
& 	ext { Impedance } Z=\frac{\mathrm{E}_{\mathrm{v}}}{\mathrm{I}}=\frac{15}{0.2}=50 \Omega \
& 	an \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
& 	an \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
& \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
	herefore \quad & \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R} \
& \mathrm{Impedance} \mathrm{~Z}=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2} \
& \mathrm{Z}=\sqrt{\mathrm{R}^2+(\sqrt{3} \mathrm{R})^2} \
& \mathrm{Z}=\sqrt{4 \mathrm{R}^2} \
	herefore & 2 \mathrm{R}= \mathrm{Z} \
	herefore \quad & \mathrm{R}=\frac{\mathrm{Z}}{2}=\frac{50}{2}=25 \Omega
\end{array}$$

Answer

$$10 \Omega$$

Answer

$$15 \Omega$$

Answer

$$20 \Omega$$

Answer

$$25 \Omega$$

AC fundamentals

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