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Question

A \( 220 \, \text{V} \) (rms), \( 50 \, \text{Hz} \) AC source is connected to a \( 44 \, \text{mH} \)
inductor. Calculate the inductive reactance.

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Explanation

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. Inductive reactance: X_L = ω L , where ω = 2π f . Given: f = 50 Hz , L = 44 mH = 0.044 H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.044 = 13.816 Ω ≈ 13.82 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

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