An arc lamp requires a direct current of 10 A at 80 V to function....

Given that the lamp draws a current of $10A$ upon a $80V$ D.C. supply, the resistance is,

$R = \dfrac{80}{10} = 8Ω$

When connected to an AC supply with an inductor in series, as shown in the figure above, the current,

$I = 10A$

and the voltage drop over the lamp is $80V$ and ${V}_{L}$ over the inductance.

Applying the rules of $L - R$ circuit,

${V}_{L}^{2} + {80}^{2} = {220}^{2}$

$⇒ {V}_{L} = \sqrt{48400 - 6400} = 210$ V

The Inductive reactance of the given inductor of inductance $L$, for the frequency of supply $f$,

${X}_{L} = 2\pi fL . . . ( 1 )$

The voltage drop across the inductance,

${V}_{L} = I {X}_{L} = 210$

$⇒ {X}_{L} = 2\pifL = \dfrac{{V}_{L}}{I} = \dfrac{210}{10} = 21Ω$

Substituting values in equation $( 1 )$

$L = \dfrac{21}{2\pif} = \dfrac{21}{2\pi \times 50} = \dfrac{21}{100\pi} = 0.065H$