An inductor of self-inductance 0.20 H is connected across an alternating current voltage source of frequency 50 Hz. Calculate the inductive reactance of this inductor. (Take π ≈ 3.142)
Answer and explanation
Correct answer: A. 62.8 Ω
- Step 1: Inductive reactance formula: X_L = 2π f L. Step 2. Substitute values: X_L = 2 × 3.142 × 50 × 0.20 = 100 × 3.142 × 0.20 = 20 × 3.142 = 62.84 Ω ≈ 62.8 Ω. Answer: 62.8 Ω. Why not the others: 31.4 Ω uses L = 0.10 H; 125.6 Ω doubles the frequency (100 Hz); 10.0 Ω is the product f × L.
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