A capacitor of capacitance 150.0 μF is connected to an alternating source of emf given by E = 36 sin(120πt) V. The maximum value of current in the circuit is approximatively equal to: 1
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Question

A capacitor of capacitance 150.0 μF is connected to an alternating source of emf given by E = 36 sin(120πt) V. The maximum value of current in the circuit is approximatively equal to: 1

G1 Physics · Accepted Answer

For a capacitor connected to V = V₀ sinωt, current amplitude is I₀ = ωCV₀ Given: C = 150 μF = 150×10⁻⁶ F ω = 120π rad s⁻¹ V₀ = 36 V Therefore, I₀ = 120π × 150×10⁻⁶ × 36 ≈ 2.04 A So the nearest option is 2 A. Hence option D follows.