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

Question

As shown in the figure an inductor of inductance 200mH is connected to an AC source of emf 220V and frequency 50Hz .

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Correct answer: 0

The reactance of the given inductor is,

${X}_{L} = \omegaL = 2\pifL = 2\pi \times 50 \times 0 . 2 = 20\piΩ$ .

The maximum current in the given circuit will be,

${i}_{0} = \dfrac{\sqrt{2}{V}_{rms}}{{X}_{L}} = \dfrac{\sqrt{2} \times 220}{20\pi} = \dfrac{\sqrt{242}}{\pi}A$

Therefore, $a = 242$

As shown in the figure an inductor of inductance 200mH is connected...

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