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:

Question

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:, 2 A, 2 A, 2 2 A, 1 2 A

MARKS App · Accepted Answer

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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