As shown in the figure an inductor of inductance $200 mH$ is connected to an $A C$ source of emf $220 V$ and frequency $50 Hz$ . The instantaneous voltage of the source is $0 V$ when the peak value of current is $\frac{a}{π} A$ . The value of $a$ is _____ .

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TestHub · Accepted Answer

The reactance of the given inductor is,

$X_{L} = ω L = 2 π f L = 2 π \times 50 \times 0.2 = 20 π Ω$ . The maximum current in the given circuit will be,

$i_{0} = \frac{2 V _{r m s}}{X _{L}} = \frac{2 \times 220}{20 π} = \frac{242}{π} A$

Therefore, $a = 242$

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