An arc lamp requires a direct current of $$10\,A$$  at $$80\,V$$  to function. If it is connected to a $$200\,V\left( {rms} \right),50\,Hz\,AC$$     supply, the series inductor needed for it to work is close to:

Solution :
Here
$$\eqalign{ & i = \frac{e}{{\sqrt {{R^2} + X_L^2} }} = \frac{e}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} = \frac{e}{{\sqrt {{R^2} + 4{\pi ^2}{v^2}{L^2}} }} \cr & 10 = \frac{{220}}{{\sqrt {64 + 4{\pi ^2}{{\left( {50} \right)}^2}L} }}\,\,\left[ {\because R = \frac{V}{I} = \frac{{80}}{{10}} = 8} \right] \cr} $$
On solving we get $$L= 0.065\,H$$