JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 21)
As shown in the figure an inductor of inductance 200 mH is connected to an AC 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 $${{\sqrt a } \over \pi }$$ A. The value of $$a$$ is ___________.
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Answer
242
Explanation
$${I_{rms}} = {{{V_{rms}}} \over z}$$
$$z = {X_2} = {\omega _2}$$
$$ = 2\pi \times 50 \times {{200} \over {1000}}$$
$$ = 20\,\pi $$
$$\therefore$$ $${I_{rms}} = {{220} \over {20\pi }} = {{11} \over \pi }$$
$$\therefore$$ $${I_{peak}} = \sqrt 2 \times {{11} \over \pi }$$
$$ = {{\sqrt {2 \times 121} } \over \pi }$$
$$ = {{\sqrt {242} } \over \pi }$$
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