An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

First note what the arc-lamp itself needs. On direct current it draws $$I = 10 \text{ A}$$ when the potential difference across it is $$V = 80 \text{ V}$$. For a purely resistive device the resistance is defined by Ohm’s law

$$R = \dfrac{V}{I}.$$

Substituting the given values, we have

$$R = \dfrac{80}{10} = 8 \;\Omega.$$

Now the lamp must be operated from a sinusoidal mains supply of $$220 \text{ V(rms)},\; 50 \text{ Hz}$$, yet we still want the current through the lamp to be only $$10 \text{ A(rms)}$$. Because the mains voltage is higher than 80 V, an additional impedance must be placed in series to limit the current. The problem asks for an inductor, so that extra impedance will be the inductive reactance $$X_L$$ of the coil.

For a series circuit containing a resistance $$R$$ and an inductive reactance $$X_L$$, the magnitude of the total impedance $$Z$$ is given by

$$Z = \sqrt{R^{2} + X_L^{2}}.$$

But on alternating current we also have, from Ohm’s law extended to impedances,

$$Z = \dfrac{V_{\text{rms}}}{I_{\text{rms}}}.$$

Substituting the required rms values,

$$Z = \dfrac{220}{10} = 22 \;\Omega.$$

Equating the two expressions for $$Z$$ gives

$$\sqrt{R^{2} + X_L^{2}} = 22.$$

We already found $$R = 8 \;\Omega$$, so

$$\sqrt{8^{2} + X_L^{2}} = 22,$$

$$\sqrt{64 + X_L^{2}} = 22.$$

Now square both sides to remove the square root:

$$64 + X_L^{2} = 22^{2} = 484.$$

Solving for $$X_L^{2}$$, we obtain

$$X_L^{2} = 484 - 64 = 420,$$

so

$$X_L = \sqrt{420} \approx 20.49 \;\Omega.$$

The inductive reactance is related to the inductance $$L$$ by the formula

$$X_L = \omega L,$$

where $$\omega$$ is the angular frequency $$\omega = 2\pi f$$. For a supply frequency of $$f = 50 \text{ Hz}$$, we have

$$\omega = 2\pi \times 50 = 100\pi \;\text{rad s}^{-1} \approx 314.16 \;\text{rad s}^{-1}.$$

Substituting into the reactance formula gives

$$L = \dfrac{X_L}{\omega} = \dfrac{20.49}{314.16} \;\text{H}.$$

Carrying out the division,

$$L \approx 0.0652 \;\text{H}.$$

This value rounds to $$0.065 \text{ H}$$, matching option B.

Hence, the correct answer is Option B.