In branch 1 (8 Ω, 120 μF, 0.1 H in series), the net reactance is:

X_L − X_C1 = +4.89 Ω (inductive)

X_C1 − X_L = 4.89 Ω capacitive

Which are the branch currents?

I1 = 25.60 A lagging 31.4°, I2 = 13.57 A leading 90°

Why is the overall pf nearly unity?

Capacitor branch supplies nearly the lagging reactive power of branch 1

Which element dissipates the active power?

Only the 8 Ω resistor: P = I1^2 R = 5242 W

2072 Chaitra · 8 marks

Series path with parallel capacitor

Short answer

Branch 1 current 25.60 A lagging 31.4°, branch 2 13.57 A leading 90°. Total current 21.84 A nearly in phase, pf ≈1. Active power 5242 W, reactive power negligible.

For a series path with a resistance of 8 Ω, capacitor of 120μF and an inductance of 0.1 H, a capacitor 180μF is kept in parallel. Then the combination is fed by 240V, 50Hz, 1-φ supply. Calculate branch currents, total current from supply, power factor of whole circuit, active power and reactive power consumed by the circuit. Also show phasor diagram.

Circuit: Branch 1 is R = 8 Ω, C = 120 μF and L = 0.1 H in series. Branch 2 is C = 180 μF alone, in parallel. Supply 240 V, 50 Hz (reference). ω = 314.16 rad/s.

Reactances

X_L = ωL = 31.42 Ω
X_C1 = 1/(ω × 120 μF) = 26.53 Ω
X_C2 = 1/(ω × 180 μF) = 17.68 Ω

Branch impedances

Z₁ = 8 + j(31.42 − 26.53) = 8 + j4.89 = 9.38∠31.4° Ω
Z₂ = −j17.68 Ω

Branch currents

I₁ = 240/9.38 = 25.60 A, lagging V by 31.4° = 21.84 − j13.35 A
I₂ = 240/17.68 = 13.57 A, leading V by 90° = j13.57 A

Total current

I = I₁ + I₂ = 21.84 + j0.22 A

|I| = 21.84 A at +0.58° (practically in phase with V)

Power factor

cos φ = cos 0.58° = 0.99995 ≈ 1 (slightly leading)

Active and reactive power

P = I₁²R = 25.60² × 8 = 5242 W (= VI cos φ = 240 × 21.84 × 1.0)
Q = VI sin φ ≈ −53 VAR (net capacitive, essentially zero)
S = VI = 240 × 21.84 = 5242 VA

The capacitor of branch 2 supplies almost exactly the lagging reactive power drawn by branch 1 (about 3.2 kVAR lag against 3.26 kVAR lead), so the circuit behaves as nearly resistive.

Phasor diagram

        I2 (+90°, 13.57 A)
        ↑
        │
 ───────┼────────────► V, I (21.84 A)
        │
        ↘ I1 (−31.4°, 25.60 A)

✓ Answer checked by Claude Opus 5.5 (high effort)

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