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JEE Main202229 Jul 2022Morning ShiftPhysicsAlternating CurrentActual

An alternating emf E = 440 sin 100 π t is applied to a circuit containing an inductance of 2 π H . If an a.c. ammeter is connected in the circuit, its reading will be :

Options

  1. A4 . 4   A
  2. B1 . 55   A
  3. C2 . 2   A
  4. D3 . 11   A

Correct answer

C. 2 . 2   A

Step-by-step solution

Given that E = 440 sin 100 π t ,   L = 2 π H Angular frequency of the source is ω = 100 π   rad   s - 1 . Now the reactance of the inductor will be, X L = ω L = 100 π 2 π = 100 2   Ω Therefore, the peak current I 0 = E 0 X L = 440 100 2 = 2 . 2 2   A A C ammeter reads RMS value therefore reading will be I rms   I rms = I 0 2 = 2 . 2   A

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