JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2022
- Medium
- Numerical
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 \(\frac{\sqrt{a}}{\pi} A\). The value of a is..........

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(242\)
Step-by-step Solution
Detailed explanation
\(f =50 Hz\)
\(X _{ L }=2 \pi fL\)
\(=2 \pi(50)\left(200 \times 10^{-3}\right)\)
\(=20 \pi \Omega\)
\(i _{0}=\frac{ V _{0}}{ X _{ L }} \Rightarrow \frac{ V _{ rms } \sqrt{2}}{ X _{ L }}\)
\(=\frac{(220) \sqrt{2}}{20 \pi}=\frac{11 \sqrt{2}}{\pi}\)
\(i _{ o }=\frac{\sqrt{242}}{\pi}\)
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