Question 428: AC fundamentals

Question

When a d.c. voltage of $$200 \mathrm{~V}$$ is applied to a coil of self-inductance $$\left(\frac{2 \sqrt{3}}{\pi}\right) \mathrm{H}$$, a current of $$1 \mathrm{~A}$$ flows through it. But by replacing d.c. source with a.c. source of $$200 \mathrm{~V}$$, the current in the coil is reduced to $$0.5 \mathrm{~A}$$. Then the frequency of a.c. supply is

Accepted Answer

When a d.c. voltage is applied only the resistance of the coil comes into play, its inductive reactance is zero.
$$\mathrm{R=\frac{V}{I}=\frac{200}{1}=200~\Omega}$$
When a.c. voltage is applied, the resistance and inductive reactance come into play and the coil has an impedance Z.
The impedance
$$\mathrm{Z=\frac{V}{I}=\frac{200}{0.5}=400~\Omega}$$
$$\mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2$$
$$	herefore (400)^2=(200)^2+\mathrm{X}_{\mathrm{L}}^2$$
$$	herefore \mathrm{X}_{\mathrm{L}}^2=(400)^2-(200)^2=12 	imes 10^4$$
$$	herefore \mathrm{X}_{\mathrm{L}}=2 \sqrt{3} 	imes 10^2=200 \sqrt{3} \Omega$$ ..... (1)
Also $$\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \mathrm{f} 	imes \frac{2 \sqrt{3}}{\pi}=4 \sqrt{3} \mathrm{f}$$ ..... (2)
By (1) and (2) : $$\quad 4 \sqrt{3} \mathrm{f}=200 \sqrt{3}$$
$$	herefore \quad \mathrm{f}=50 \mathrm{~Hz}$$

Answer

100 Hz

Answer

60 Hz

Answer

75 Hz

Answer

50 Hz

AC fundamentals

Choose the correct answer: