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An R-L load is connected to a 250 V, 400 Hz step down dc converter. The average load current is 100 A...

 

RL Circuit Inductor Calculation

RL Circuit Inductor Calculation

Given Data

  • Supply voltage V = 250 V
  • Frequency f = 400 Hz
  • Load current IDC = 100 A
  • Load resistance R = 0.5 Ω
  • Allowed ripple = 15%

Step 1: Ripple Current

Iripple = 0.15 × 100 = 15 A

Step 2: Ripple Voltage

Approximation:

Vr ≈ 250 V

Step 3: RL Ripple Formula

Iripple = Vr / √(R² + (ωL)²)

Rearranging:

√(R² + (ωL)²) = Vr / Iripple
R² + (ωL)² = (250 / 15)²

Step 4: Solve

250 / 15 = 16.67
R² = (0.5)² = 0.25
(ωL)² = 16.67² − 0.25
(ωL)² ≈ 277.8 − 0.25 = 277.55
ωL ≈ √277.55 ≈ 16.66

Step 5: Find Inductance

ω = 2Ï€f = 2Ï€ × 400 ≈ 2513
L = 16.66 / 2513
L ≈ 0.00663 H

Final Answer

L ≈ 6.6 mH option B
  • High frequency → smaller inductance needed
  • Low resistance → inductor dominates ripple control
  • 15% ripple → moderate inductance value

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