Skip to main content

Power factor and series-parallel circuit calculation

Short answer

A low power factor increases current, losses, and size of equipment for a given real power. For the circuit, total impedance Z = 12.33 + j6.89 Ω (14.13∠29.19° Ω), circuit current I = 14.16∠−29.19° A, power factor = 0.873 lagging, and power consumed P = 2472 W.

What is the significance of the power factor? Calculate the total impedance, circuit current, power factor, and power consumed for the following circuit. [Figure: a 200 V, 50 Hz AC source supplies 3 Ω + j6 Ω in series with a parallel combination of two branches: branch 1 = 4 Ω + j8 Ω, branch 2 = 5 Ω − j8 Ω.]

Significance of power factor

Power factor (cos φ = P/S) shows what fraction of the apparent power does useful work. For a given power at a given voltage, I = P/(V cos φ), so a low power factor means a larger current. This causes higher copper losses (I²R), more voltage drop, poorer efficiency, and needs bigger cables, transformers and generators (rated in kVA). Utilities therefore charge penalties for low pf and consumers improve it with capacitors.

Calculation

V = 200∠0° V. Series part Z1 = 3 + j6 Ω; branches Z2 = 4 + j8 Ω, Z3 = 5 − j8 Ω.

Parallel combination:

Z2 × Z3 = (4 + j8)(5 − j8) = 20 − j32 + j40 + 64
        = 84 + j8
Z2 + Z3 = 9 + j0
Zp = (84 + j8)/9 = 9.333 + j0.889 Ω = 9.376∠5.44° Ω

Total impedance:

Z = Z1 + Zp = (3 + j6) + (9.333 + j0.889)
  = 12.333 + j6.889 Ω
|Z| = √(12.333² + 6.889²) = 14.127 Ω
φ  = tan⁻¹(6.889/12.333) = 29.19°
Z  = 14.127∠29.19° Ω

Circuit current:

I = V/Z = 200∠0° / 14.127∠29.19°
  = 14.157∠−29.19° A

Branch currents (for completeness): Vp = I × Zp = 132.73∠−23.75° V, I2 = Vp/Z2 = 14.84∠−87.18° A, I3 = Vp/Z3 = 14.07∠34.25° A.

Power factor:

pf = cos 29.19° = R/|Z| = 12.333/14.127 = 0.873 (lagging)

Power consumed:

P = V I cos φ = 200 × 14.157 × 0.873 = 2472 W
Check: P = I² R = 14.157² × 12.333 = 2472 W ✓

Answer: Z = 12.33 + j6.89 Ω = 14.13∠29.19° Ω, I = 14.16∠−29.19° A, pf = 0.873 lagging, P ≈ 2472 W (Q ≈ 1381 VAR, S ≈ 2831 VA).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Electrical and Electronics Engineering chapter 3 questions →