JEE Mains · Physics · STD 12 -7. Alternating current
An alternating \(emf\) \(E =440 \sin 100 \pi t\) is applited to a circuit containing an inductance of \(\frac{\sqrt{2}}{\pi} H\). If an a.c. ammeter is connected in the circuit, its reading will be \(.......A\)
- A \(4.4\)
- B \(1.55\)
- C \(2.2\)
- D \(3.1\)
Answer & Solution
Correct Answer
(C) \(2.2\)
Step-by-step Solution
Detailed explanation
\(E =440 \sin 100 \pi t , L =-\frac{\sqrt{2}}{\pi} H\) \(X _{ L }=\omega L =100 \pi \frac{\sqrt{2}}{\pi}=100 \sqrt{2} \Omega\) Peak current \(I _{0}=\frac{ E _{0}}{ X _{ L }}=\frac{440}{100 \sqrt{2}}=2.2 \sqrt{2} A\) \(AC\) ammeter reads RMS value therefore reading will be…
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