Total power of series-parallel circuit

Circuit: 4 Ω in series with L = 19.1 mH, then two parallel branches: Branch A is 10 Ω in series with C = 265.15 μF, Branch B is 6 Ω in series with L = 31.8 mH. Supply 200 V, 50 Hz (reference). ω = 314.16 rad/s.

Reactances

• X_L (series) = 314.16 × 0.0191 = 6.00 Ω
• X_C = 1/(314.16 × 265.15 × 10⁻⁶) = 12.0 Ω
• X_L (branch B) = 314.16 × 0.0318 = 9.99 ≈ 10.0 Ω

Impedances

• Z_s = 4 + j6.0 Ω
• Z_A = 10 − j12.0 = 15.62∠−50.2° Ω
• Z_B = 6 + j10.0 = 11.66∠59.0° Ω

Parallel combination

Z_p = Z_A Z_B/(Z_A + Z_B) = 10.85 + j3.11 Ω

Total impedance

Z_T = Z_s + Z_p = 14.85 + j9.11 = 17.43∠31.5° Ω

Total current

I = 200/17.43 = 11.48 A, lagging V by 31.5°

Branch currents

V_p = I Z_p = 11.48 × 11.29 = 129.6 V

• I_A = 129.6/15.62 = 8.29 A (leading V by 34.7°)
• I_B = 129.6/11.66 = 11.12 A (lagging V by 74.6°)

Total power

P = I²R_T = I²(Re Z_T) = 11.48² × 14.85 = 1957 W (≈ 1.96 kW)

Check by summing resistor losses:

• 4 Ω: 11.48² × 4 = 527 W
• 10 Ω: 8.29² × 10 = 688 W
• 6 Ω: 11.12² × 6 = 742 W
• Total = 1957 W

Also Q = 1200 VAR (lagging), S = 2296 VA, power factor = 0.852 lagging.

Phasor diagram

         I_A (+34.7°, 8.29 A)
        ↗
  ─────────────────────────► V (0°)
        ↘ I (−31.5°, 11.48 A)
          ↘ I_B (−74.6°, 11.12 A)

I = I_A + I_B. The drop across the series part (4 Ω, 6 Ω reactance) is I Z_s and V = I Z_s + V_p.