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Parallel RLC supply current

Short answer

Nearly equal reactances ~10 Ω. Branch currents: I_R=20 A, I_L=20.02 A lagging, I_C=19.98 A leading. Supply current is 20.0 A, power factor 1.0 (parallel resonance).

A 10 ohm resistor, a 31.8 mH inductor and 318μF capacitor are connected in parallel and supplied from a 200V, 50Hz supply source. Calculate the supply current and power factor and also calculate current in each branch.

R = 10 Ω, L = 31.8 mH, C = 318 μF in parallel across 200 V, 50 Hz. ω = 314.16 rad/s.

Reactances

  • X_L = ωL = 314.16 × 0.0318 = 9.99 Ω
  • X_C = 1/(ωC) = 1/(314.16 × 318 × 10⁻⁶) = 10.01 Ω

Branch currents (V = 200∠0° as reference)

  • I_R = 200/10 = 20.0 A (in phase with V)
  • I_L = 200/9.99 = 20.02 A (lagging by 90°)
  • I_C = 200/10.01 = 19.98 A (leading by 90°)

Supply current

The reactive currents are in opposite directions:

I = √(I_R² + (I_C − I_L)²) = √(20² + (19.98 − 20.02)²) = 20.0 A

Phase angle φ = tan⁻¹((I_C − I_L)/I_R) = tan⁻¹(−0.04/20) ≈ −0.1° (practically zero).

Power factor

cos φ = I_R/I = 20/20 = 1.0 (unity)

The circuit is practically at parallel resonance (X_L ≈ X_C, f ≈ 50 Hz), so the supply current is only the resistor current, even though each reactive branch carries about 20 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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