- 2073 Shrawan · 8 marks
Series-parallel circuit currents and power
Short answer
Total impedance 17.29 Ω, supply current I=5.79 A lagging 30.76°. Parallel voltage 60.9 V gives I₁=8.18 A (lag), I₂=3.80 A (lead). Power factor 0.859 lagging, total power 497 W.
Determine the value of current I1, I2 and I and overall power factor of the circuit shown in figure below for series and parallel circuit. Also draw the phasor diagram and find the total power consumed by the circuit. [Figure: Z1 = 6 Ω in series with 0.01 H, followed by two parallel branches: Z2 = 4 Ω in series with 0.02 H (current I1) and a branch 2 Ω in series with 200 μF capacitor (current I2); supply 100 V, 50 Hz.]
Circuit: Z₁ = 6 Ω + 0.01 H in series with two parallel branches. Branch 1 (current I₁): Z₂ = 4 Ω + 0.02 H. Branch 2 (current I₂): 2 Ω + 200 μF. Supply 100 V, 50 Hz as reference. ω = 314.16 rad/s.
Reactances
- X_L1 = 3.142 Ω; X_L2 = 6.283 Ω; X_C = 15.92 Ω
Impedances
- Z₁ = 6 + j3.142 Ω
- Z_a = 4 + j6.283 = 7.45∠57.5° Ω
- Z_b = 2 − j15.92 = 16.04∠−82.8° Ω
Parallel part
Z_p = Z_a Z_b/(Z_a + Z_b) = 8.854 + j5.697 = 10.53∠32.8° Ω
Total impedance
Z_T = Z₁ + Z_p = 14.85 + j8.84 = 17.29∠30.76° Ω
Currents
- I = V/Z_T = 100/17.29 = 5.79 A, lagging V by 30.76°
- Parallel voltage V_p = I Z_p = 5.79 × 10.53 = 60.9 V
- I₁ = 60.9/7.45 = 8.18 A (∠−55.5°)
- I₂ = 60.9/16.04 = 3.80 A (∠+84.8°)
Check: I₁ + I₂ = (4.63 − j6.74) + (0.35 + j3.78) = 4.97 − j2.96 → 5.79∠−30.8° A.
Power factor and power
- cos φ = 14.85/17.29 = 0.859 lagging
- P = VI cos φ = 100 × 5.79 × 0.859 = 497 W (check: 5.79² × 6 + 8.18² × 4 + 3.80² × 2 = 201 + 268 + 29 = 498 W)
- Q = VI sin φ = 296 VAR, S = 579 VA
Phasor diagram
I2 (+84.8°, 3.80 A)
↑
│
────────────┼──────────────► V (0°)
│ ↘ I (−30.8°, 5.79 A)
│ ↘ I1 (−55.5°, 8.18 A)
I₁ lags V (inductive), I₂ leads V (capacitive), and I = I₁ + I₂.
✓ Answer checked by Claude Opus 5.5 (high effort)
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