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(D) When a d.c. voltage is applied, the coil acts as a pure resistor because the inductive reactance $X_L = 2 \pi f L = 0$ for $f = 0$.$R = \frac{V}{I

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$50$

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(D) When a d.c. voltage is applied, the coil acts as a pure resistor because the inductive reactance $X_L = 2 \pi f L = 0$ for $f = 0$.$R = \frac{V}{I_{dc}} = \frac{200}{1} = 200 \, \Omega$When an a.c. voltage is applied, the impedance $Z$ of the $LR$ circuit is given by:$Z = \frac{V}{I_{ac}} = \frac{200}{0.5} = 400 \, \Omega$The impedance is related to resistance and inductive reactance by $Z^2 = R^2 + X_L^2$.$(400)^2 = (200)^2 + X_L^2$$160000 = 40000 + X_L^2$$X_L^2 = 120000$$X_L = \sqrt{120000} = 200 \sqrt{3} \, \Omega$Since $X_L = 2 \pi f L$, we have:$2 \pi f \left( \frac{2 \sqrt{3}}{\pi} \right) = 200 \sqrt{3}$$4 \sqrt{3} f = 200 \sqrt{3}$$f = \frac{200}{4} = 50 \, Hz$

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