Question

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

• A. 100 Hz
• B. 75 Hz
• C. 60 Hz
• D. 30 Hz
• E. 50 Hz

Answer 1

Answer: (E)

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 $ or{{\left( 2\pi f\times \frac{2\sqrt{3}}{\pi } \right)}^{2}}={{(400)}^{2}}-{{(200)}^{2}} $
$ =120000 $
Or $ 4f\sqrt{3}=200\sqrt{3} $
Or $ f=50\,Hz $