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Parallel RL and RC branches, total current

Short answer

Total current is 8.48 A at a leading angle of 10.32°. Power factor 0.984 leading. Active power 2003 W, reactive power -365 VAR (net capacitive).

A Circuit consisting of a resistance of 30Ω in series with an inductance of 75mH is connected in parallel with a circuit consisting of a resistance of 20Ω in series with a capacitance of 100μF, if the parallel combination is connected to a 240V, 50Hz, single-phase supply. Calculate (i) The total current (ii) Power factor (iii) Active and reactive power. Also draw a neat phasor diagram.

Circuit: Branch 1 is R₁ = 30 Ω in series with L = 75 mH. Branch 2 is R₂ = 20 Ω in series with C = 100 μF. Supply V = 240∠0° V, 50 Hz (reference). ω = 314.16 rad/s.

Reactances and branch impedances

  • X_L = ωL = 314.16 × 0.075 = 23.56 Ω
  • X_C = 1/(ωC) = 1/(314.16 × 100 × 10⁻⁶) = 31.83 Ω
  • Z₁ = 30 + j23.56 = 38.15∠38.15° Ω
  • Z₂ = 20 − j31.83 = 37.59∠−57.86° Ω

Branch currents

  • I₁ = 240/38.15 = 6.29 A, ∠−38.15° (lagging) = 4.95 − j3.89 A
  • I₂ = 240/37.59 = 6.38 A, ∠+57.86° (leading) = 3.40 + j5.41 A

Total current

I = I₁ + I₂ = 8.34 + j1.52 A

|I| = 8.48 A, at 10.32° (leading the voltage)

Power factor

cos φ = cos(10.32°) = 0.984 leading

Power

  • Active power: P = VI cos φ = 240 × 8.48 × 0.984 = 2003 W (check: I₁²R₁ + I₂²R₂ = 6.29² × 30 + 6.38² × 20 = 1187 + 815 = 2002 W)
  • Reactive power: Q = VI sin φ = −365 VAR (net capacitive; inductive branch 933 VAR lagging, capacitive branch 1297 VAR leading)
  • Apparent power: S = VI = 240 × 8.48 = 2036 VA

Phasor diagram

                 I2 (+57.9°, 6.38 A)
                ↗
              I (+10.3°, 8.48 A)
   ─────────────────────────► V (0°)
                ↘
                 I1 (−38.2°, 6.29 A)

I₁ lags V, I₂ leads V, and their sum I is slightly ahead of V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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