JEE Mains · Physics · STD 12 -7. Alternating current
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..........
- A \(282\)
- B \(242\)
- C \(247\)
- D \(867\)
Answer & Solution
Correct Answer
(B) \(242\)
Step-by-step Solution
Detailed explanation
\(f =50 Hz\) \(X _{ L }=2 \pi fL\) \(=2 \pi(50)\left(200 \times 10^{-3}\right)\) \(=20 \pi \Omega\) \(i _{0}=\frac{ V _{0}}{ X _{ L }} \Rightarrow \frac{ V _{ rms } \sqrt{2}}{ X _{ L }}\) \(=\frac{(220) \sqrt{2}}{20 \pi}=\frac{11 \sqrt{2}}{\pi}\)…
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