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(D) For $DC$ voltage, the inductor acts as a pure resistor because $X_L = 0$ for $DC$.$R = \frac{V}{I} = \frac{100 \, V}{5 \, A} = 20 \, \Omega$.For $

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$250$

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(D) For $DC$ voltage, the inductor acts as a pure resistor because $X_L = 0$ for $DC$.$R = \frac{V}{I} = \frac{100 \, V}{5 \, A} = 20 \, \Omega$.For $AC$ voltage, the circuit is an $LR$ series circuit.Given $X_L = 20\sqrt{3} \, \Omega$ and $R = 20 \, \Omega$.The impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = \sqrt{1600} = 40 \, \Omega$.The $RMS$ voltage is $V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{200}{\sqrt{2}} \, V$.The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200 / \sqrt{2}}{40} = \frac{5}{\sqrt{2}} \, A$.The power dissipated in the circuit is $P = I_{rms}^2 R$.$P = \left( \frac{5}{\sqrt{2}} ight)^2 	imes 20 = \frac{25}{2} 	imes 20 = 250 \, W$.

When a $DC$ voltage of $100 \, V$ is applied to an inductor, a $DC$ current of $5 \, A$ flows through it. When an $AC$ voltage of $200 \, V$ peak value is connected to the inductor, its inductive reactance is found to be $20\sqrt{3} \, \Omega$. The power dissipated in the circuit is . . . . . . $W$.

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