Branch currents in series-parallel AC

Circuit: 8 Ω and 0.6 mH in series with two parallel branches: a 4 Ω resistor, and a 6 Ω resistor in series with an 8 μF capacitor. Supply 220 V, 50 Hz (reference). ω = 314.16 rad/s.

Reactances

• X_L = ωL = 314.16 × 0.6 × 10⁻³ = 0.1885 Ω
• X_C = 1/(ωC) = 1/(314.16 × 8 × 10⁻⁶) = 397.9 Ω

Impedances

• Z_series = 8 + j0.1885 Ω
• Z₁ = 4 Ω (resistor branch)
• Z₂ = 6 − j397.9 Ω = 397.9∠−89.1° Ω

Parallel combination

Z_p = Z₁Z₂/(Z₁ + Z₂) = 4(6 − j397.9)/(10 − j397.9) = 3.999 − j0.040 Ω

Total impedance and current

Z_T = (8 + 3.999) + j(0.1885 − 0.040) = 12.00 + j0.148 = 12.00∠0.71° Ω

Total current I = 220/12.00 = 18.33 A (lagging 0.71°)

Voltage across the parallel part

V_p = I × Z_p = 18.33 × 3.999 = 73.3 V

Branch currents

• I₁ (4 Ω branch) = 73.3/4 = 18.33 A
• I₂ (6 Ω + 8 μF branch) = 73.3/397.9 = 0.184 A (leading by about 89°)

Because the capacitor has a very large reactance at 50 Hz, almost all the current passes through the 4 Ω resistor. The overall power factor is nearly 1 (0.9999 lagging) and P = I²R_T ≈ 4.03 kW.