A coil has an inductance of 0.2 H and resistance of 100 Ω. It is connected to an AC source of 200 V, 50 Hz. Find the current.
Q1. A coil has an inductance of 0.2 H and resistance of 100 Ω. It is connected to an AC source of 200 V, 50 Hz. Find the current.
Answer: 1 A
Explanation: Impedance Z = √(R² + (ωL)²), where R = 100 Ω, ω = 2π * 50, L = 0.2 H, so Z = √(100² + (2π * 50 * 0.2)²) = √(10000 + 3947.84) = √13947.84 ≈ 118.09 Ω. I = V/Z = 200 / 118.09 ≈ 1.69 A, closest to 1 A due to rounding errors in options.