---
title: "Series path with parallel capacitor"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-rl-parallel-capacitor-240v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rl-parallel-capacitor-240v
site: BCT Engineering
---

# Series path with parallel capacitor

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

- Asked in: 2072 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rl-parallel-capacitor-240v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

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.

## 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.

## Answer

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

## Check yourself

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

1. In branch 1 (8 Ω, 120 μF, 0.1 H in series), the net reactance is:
   A. X_L − X_C1 = +4.89 Ω (inductive)
   B. 0
   C. X_C1 − X_L = 4.89 Ω capacitive
   D. X_L + X_C1 = 57.9 Ω
   Answer: A. 31.42 − 26.53 = +4.89 Ω, so Z1 = 8 + j4.89. (Key point: Branch 1 reactance)

2. Which are the branch currents?
   A. I1 = 25.60 A lagging 31.4°, I2 = 13.57 A leading 90°
   B. I1 = 25.6 A leading
   C. I1 = 30 A, I2 = 13.5 A lagging
   D. I1 = 13.57 A, I2 = 25.60 A
   Answer: A. I1 = 240/9.38, I2 = 240/17.68 with a pure capacitor leading by 90°. (Key point: Branch currents)

3. Why is the overall pf nearly unity?
   A. Both branches inductive
   B. Capacitor branch supplies nearly the lagging reactive power of branch 1
   C. Supply frequency is zero
   D. Resistance is zero
   Answer: B. Branch 2 gives about 3.26 kVAR leading against 3.2 kVAR lagging, giving I = 21.84 + j0.22 A. (Key point: Total current and pf)

4. Which element dissipates the active power?
   A. Capacitor 180 μF
   B. Both capacitors
   C. Inductor
   D. Only the 8 Ω resistor: P = I1^2 R = 5242 W
   Answer: D. Ideal L and C consume no real power. (Key point: Active power)

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