When a DC voltage of 200, V is applied to a c | vedclass

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(D) For $DC$ supply,the inductor acts as a simple resistor because $X_L = 2\pi fL$ and $f = 0$. Thus,$R = \frac{V}{I_1} = \frac{200}{1} = 200\,\Omega$

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$50$

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(D) For $DC$ supply,the inductor acts as a simple resistor because $X_L = 2\pi fL$ and $f = 0$. Thus,$R = \frac{V}{I_1} = \frac{200}{1} = 200\,\Omega$.For $AC$ supply,the impedance $Z$ is given by $Z = \frac{V}{I_2} = \frac{200}{0.5} = 400\,\Omega$.The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$.Substituting the values: $400 = \sqrt{200^2 + X_L^2}$.Squaring both sides: $160000 = 40000 + X_L^2$,which gives $X_L^2 = 120000$.$X_L = \sqrt{120000} = 200\sqrt{3}\,\Omega$.Since $X_L = 2\pi fL$,we have $200\sqrt{3} = 2\pi f \left( \frac{2\sqrt{3}}{\pi} \right)$.$200\sqrt{3} = 4\sqrt{3} f$.$f = \frac{200\sqrt{3}}{4\sqrt{3}} = 50\,Hz$.

When a $DC$ voltage of $200\, V$ is applied to a coil of self-inductance $\frac{2\sqrt{3}}{\pi}\,H$,a current of $1\, A$ flows through it. When the $DC$ source is replaced by an $AC$ source of $200\, V$,the current in the coil is reduced to $0.5\, A$. The frequency of the $AC$ supply is......$Hz$.

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