---
title: "Branch currents in series-parallel AC"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:branch-currents-8ohm-0-6mh-220v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/branch-currents-8ohm-0-6mh-220v
site: BCT Engineering
---

# Branch currents in series-parallel AC

Basic Electrical Engineering · Chapter 6: Single-phase AC Circuits

- Asked in: 2074 Chaitra, 6 marks
- Video: [AC Analysis: Series/Parallel RLC Circuit](https://www.youtube.com/watch?v=s3Daf4GC_u4) — ElectronX Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/branch-currents-8ohm-0-6mh-220v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Find current flowing in each branches of the following circuit: [Figure: 8 Ω resistor in series with 0.6 mH inductor, followed by two parallel branches: a 4 Ω resistor, and a 6 Ω resistor in series with 8 μF capacitor; supply 220 V, 50 Hz.]

## Short answer

Total impedance 12.00 Ω, total current 18.33 A. Parallel voltage 73.3 V. Branch currents: I₁ (4 Ω) = 18.33 A, I₂ (6 Ω, 8 μF) = 0.184 A leading ~89°. Most current flows through the 4 Ω resistor due to large capacitive reactance.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is X_C for 8 μF at 50 Hz?
   A. 397.9 Ω
   B. 25 Ω
   C. 6 Ω
   D. 0.1885 Ω
   Answer: A. X_C = 1/(314.16 × 8×10^-6) = 397.9 Ω. (Key point: Capacitor reactance)

2. How is the current in each parallel branch found?
   A. Divide supply voltage by each branch impedance
   B. Add the branch resistances
   C. Total current divided equally
   D. Find Z_T, I = V/Z_T, V_p = I Z_p, then divide V_p by each branch impedance
   Answer: D. The parallel part receives V_p, not the full supply, because of the series 8 Ω and 0.6 mH. (Key point: Method)

3. Why does nearly all current flow through the 4 Ω branch?
   A. The capacitor branch has a very large reactance (about 398 Ω)
   B. 4 Ω branch has capacitance
   C. Supply is DC
   D. The inductor blocks current
   Answer: A. The branch with 6 Ω + 8 μF has |Z| ≈ 398 Ω, so I2 is only 0.184 A against 18.33 A. (Key point: Why current mostly in resistor)

4. What is the overall power factor?
   A. About 0.8 lagging
   B. Nearly unity (0.9999 lagging)
   C. About 0.5 leading
   D. Zero
   Answer: B. Z_T = 12.00 + j0.148 Ω is almost purely resistive. (Key point: Power factor)

## Related in Basic Electrical Engineering

- [Series-parallel AC circuit analysis](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-r1-15-l1-0-05-200v) — Basic Electrical Engineering, chapter 6, asked 2×
- [Series-parallel circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-z1-6-0-01h-100v) — Basic Electrical Engineering, chapter 6
- [Source current from parallel branch currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-branch-currents-time-expression) — Basic Electrical Engineering, chapter 5
- [Total power of series-parallel circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-total-power-265uf-19mh) — Basic Electrical Engineering, chapter 6
- [Parallel RLC supply current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-r-l-c-200v-supply-current) — Basic Electrical Engineering, chapter 6
- [Series path with parallel capacitor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-rl-parallel-capacitor-240v) — Basic Electrical Engineering, chapter 6

## Related in other subjects

- [Series-parallel AC circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Aseries-parallel-ac-circuit-power) — Fundamental of Electrical and Electronics Engineering, chapter 3
- [Laplace: current i(t) with two capacitor branches](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-current-50v-10ohm-two-rc) — Electrical Circuit Theory, chapter 4
- [Laplace: inductor and capacitor currents, 12 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-inductor-capacitor-current-12v-2ohm) — Electrical Circuit Theory, chapter 4, asked 3×
- [Parallel RL and RC branch currents, power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aparallel-impedance-total-current) — Basic Electrical and Electronics Engineering, chapter 1