When a coil is connected to a 200V dc supply, the current flowing through it is found to be 1A…

In a dc circuit, frequency is zero, so inductive reactance is zero ($X_L = 0$) and current is opposed only by ohmic resistance $R$ ($I = \frac{V}{R}$). In an ac circuit, the alternating current produces a changing magnetic flux, giving rise to an inductive reactance $X_L = 2\pi f L$. The total opposition is impedance $Z = \sqrt{R^2 + X_L^2}$. Since $Z > R$, the current in the ac circuit is less than in the dc circuit.

From the dc circuit data: Resistance $R = \frac{V_{\text{dc}}}{I_{\text{dc}}} = \frac{200\text{ V}}{1\text{ A}} = 200\,\Omega$ From the ac circuit data ($f = 50\text{ Hz}$): Impedance $Z = \frac{V_{\text{ac}}}{I_{\text{ac}}} = \frac{200\text{ V}}{0\cdot5\text{ A}} = 400\,\Omega$ Inductive reactance: $X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{400^{2} - 200^{2}} = \sqrt{120000} = 200\sqrt{3}\,\Omega \approx 346\cdot41\,\Omega$ Since $X_{L} = 2\pi f L$: $L = \frac{X_{L}}{2\pi f} = \frac{346\cdot41\,\Omega}{2 \times 3\cdot14 \times 50\text{ s}^{-1}} = \frac{346\cdot41}{314} \approx 1\cdot10\text{ H}$ Final answer: 1.10 H

Power factor of an ac circuit is given by: $\cos\phi = 0\cdot5$ Therefore, the phase difference $\phi$ between voltage and current is: $\phi = \cos^{-1}(0\cdot5) = 60^{\circ} = \frac{\pi}{3}\text{ rad}$ Final answer: 60^\circ