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Q. 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 :

JEE MainJEE Main 2016Alternating Current

Solution:

For $dc$
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$R = \frac{80}{10} = 8 \omega$
For $ac$
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$10 = \frac{220}{\sqrt{R^{2} + \omega^{2}L^{2}}} $
$R^{2} + \omega^{2}L^{2} = \left(\frac{220}{10}\right)^{2}$
$ L^{2} = \frac{22^{2} - 8^{2}}{\omega^{2}} $
$\therefore L = \frac{\sqrt{30 \times14}}{2 \pi\times50} = \frac{\sqrt{420}}{100 \pi} = 0.065\, H $

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