Question 369: AC fundamentals

Question

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 $$\left(\frac{\pi}{3}ight)$$ radian. The value of resistance $$(\mathrm{R})$$ is $$\left(	an 60^{\circ}=\sqrt{3}ight)$$

Accepted Answer

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 $$Z=\frac{E}{I}=\frac{15}{0.5}=30 \Omega$$
$$\begin{gathered}
	an \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
	an \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
\sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \
	herefore \quad \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}
\end{gathered}$$
The formula for impedance is
$$\begin{aligned}
Z & =\sqrt{R^2+X_L^2} \
Z & =\sqrt{R^2+(\sqrt{3} R)^2} \
Z & =\sqrt{4 R^2} \
	herefore \quad 2 R & =Z \
	herefore \quad R & =\frac{Z}{2}=\frac{30}{2}=15 \Omega
\end{aligned}$$

Answer

$$10 \Omega$$

Answer

$$12 \Omega$$

Answer

$$15 \Omega$$

Answer

$$20 \Omega$$

AC fundamentals

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