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Power factor and parameters from v and i

Short answer

Power factor 0.643 leading. Circuit parameters: R=1.07 Ω, C=1566 μF (series RC). Apparent power 30 VA, active power 19.3 W, reactive power 23.0 VAR leading. The circuit is capacitive because current leads voltage.

When a voltage v= 10sin(500t–60°)v is applied to a series ac circuit, the current is i=6sin(500t–10°). Find (i) Power factor (ii) Circuit parameters (iii) Apparent, active and reactive power (iv) Also indicate whether the circuit is capacitive or inductive and why?

Given v = 10 sin(500t − 60°) V and i = 6 sin(500t − 10°) A. So Vm = 10 V, Im = 6 A, ω = 500 rad/s.

(i) Power factor

Phase angle φ = (−60°) − (−10°) = −50°, so the current leads the voltage by 50°.

Power factor = cos 50° = 0.643 leading

(ii) Circuit parameters

  • |Z| = Vm/Im = 10/6 = 1.667 Ω
  • R = Z cos φ = 1.667 × 0.643 = 1.07 Ω
  • X = Z sin φ = 1.667 × 0.766 = 1.277 Ω (capacitive)
  • X_C = 1/(ωC) ⟹ C = 1/(500 × 1.277) = 1.566 × 10⁻³ F = 1566 μF

The circuit is a series R–C: R = 1.07 Ω, C ≈ 1566 μF.

(iii) Powers

V = 10/√2 = 7.07 V, I = 6/√2 = 4.24 A

  • Apparent: S = VI = 30 VA
  • Active: P = VI cos φ = 30 × 0.643 = 19.3 W
  • Reactive: Q = VI sin φ = 30 × 0.766 = 23.0 VAR (leading)

(iv) Nature of the circuit

The circuit is capacitive, because the current leads the voltage (the current wave reaches its peak 50° earlier than the voltage wave). Inductive circuits give lagging current.

✓ Answer checked by Claude Opus 5.5 (high effort)

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