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Coil inductance and resistance from two frequencies

Short answer

From Z = V/I, at 127 Hz Z=60Ω, at 456 Hz Z=120Ω. Solving R²+(2πfL)² gives L≈37.8 mH, R≈51.9 Ω.

At 127 Hz frequency, an emf source of 6V sends a current of 100mA through an inductive coil. If the frequency changes to 456Hz, the current drawn by the circuit becomes 50mA. Calculate the inductance and resistance of coil.

Let the coil have resistance R and inductance L. Its impedance is Z = √(R² + (2πfL)²).

Data

  • At f₁ = 127 Hz: Z₁ = V/I = 6/0.1 = 60 Ω
  • At f₂ = 456 Hz: Z₂ = 6/0.05 = 120 Ω

Equations

R² + (2π × 127 × L)² = 60² = 3600 …(1) R² + (2π × 456 × L)² = 120² = 14400 …(2)

Subtract (1) from (2):

L² [(2π × 456)² − (2π × 127)²] = 10800

(2π)² (456² − 127²) L² = 10800 39.478 × (207 936 − 16 129) L² = 10800 39.478 × 191 807 L² = 10800 7.572 × 10⁶ L² = 10800 ⟹ L² = 1.426 × 10⁻³

L = 0.0378 H ≈ 37.8 mH

Resistance

X_L at 127 Hz = 2π × 127 × 0.03777 = 30.14 Ω

R = √(3600 − 30.14²) = √(3600 − 908.4) = 51.9 Ω

Check at 456 Hz: X_L = 108.2 Ω, √(51.9² + 108.2²) = 120 Ω.

Answer: L ≈ 37.8 mH, R ≈ 51.9 Ω.

✓ Answer checked by Claude Opus 5.5 (high effort)

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