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

Question

Dr. Abubacker · Accepted Answer

(d) Resistance of coil$(R) = \frac{200}{1} = 200\ \Omega$
Current$,\ I = \frac{200}{\sqrt{R^{2} + X_{L}^{2}}}$
or $0.5 = \frac{200}{\sqrt{R^{2} + X_{L}^{2}}}$
or $R^{2} + (2\pi fL)^{2} = (400)^{2}$
or $\left( 2\pi f 	imes \frac{2\sqrt{3}}{\pi} ight)^{2} = (400)^{2} - (200)^{2}$
= 120000
or $4f\sqrt{3} = 200\sqrt{3}$
or $f = 50\ Hz$

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

Explanation

Reviewed by Dr. Abubacker

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