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- 2072 Chaitra · 8 marks
Parallel RL and RC branches
Short answer
Branch 1: 6.29 A lagging 38.15°, branch 2: 6.38 A leading 57.86°. Total current 8.48 A at 10.32° leading, pf 0.984 leading. Power consumed 2003 W.
A circuit consisting of a resistance of 30 Ω in series with an inductance of 75 mH is connected in parallel with a circuit consisting of a resistance of 20 Ω in series with a capacitance of 100 μF. If the parallel combination is connected to a 240 V, 50 Hz single phase supply, calculate (i) The current in each branch (ii) The total current and power factor and (iii) Power consumed. 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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