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JEE Main20236 Apr 2023Evening ShiftPhysicsAlternating CurrentActual

A capacitor of capacitance 150 . 0 μ F is connected to an alternating source of emf given by E = 36 sin 120 π t V . The maximum value of current in the circuit is approximately equal to:

Options

  1. A2   A
  2. B2   A
  3. C2 2   A
  4. D1 2   A

Correct answer

A. 2   A

Step-by-step solution

The formula for maximum current is given by I m a x = ω C V m       . . . ( i ) The given data is ω = 120 π V m = 36   V C = 150   μF Substituting the values in equation (i) I max = 120 π × 150 × 10 - 6 × 36 = 2 . 036   A ≈ 2   A

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