---
title: "Parallel RL and RC branches, total current"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:parallel-rl-rc-total-current-pf-power"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-rl-rc-total-current-pf-power
site: BCT Engineering
---

# Parallel RL and RC branches, total current

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

- Asked in: 2070 Chaitra, 8 marks
- Video: [Parallel RLC Circuit Example Problem](https://www.youtube.com/watch?v=8MMzeeHNjIw) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-rl-rc-total-current-pf-power
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

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.

## 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).

## Answer

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
```text
                 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.

## Check yourself

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

1. How are I1 and I2 obtained?
   A. Both equal
   B. I = V/Z for each branch: 6.29 A lagging and 6.38 A leading
   C. Each is V/X
   D. Each is V/R
   Answer: B. Z1 = 38.15 Ω and Z2 = 37.59 Ω with the RL branch lagging and RC branch leading. (Key point: Branch currents)

2. Total current is found by:
   A. Phasor addition: 8.34 + j1.52 = 8.48 A
   B. Divide V by sum of impedances
   C. Subtract the magnitudes
   D. 6.29 + 6.38
   Answer: A. Phase angles differ, so use complex addition. (Key point: Total current)

3. The power factor is cos(10.32°) = 0.984. It is:
   A. Lagging
   B. Zero
   C. Unity
   D. Leading, as the capacitive branch dominates
   Answer: D. Total current has a positive imaginary part, so it leads the voltage. (Key point: Power factor)

4. Q = −365 VAR arises because:
   A. Inductive branch 933 VAR lag is less than capacitive branch 1297 VAR lead
   B. Resistive branch
   C. No reactive power
   D. Both lag
   Answer: A. Net reactive power = 933 lag − 1297 lead = −365 VAR (capacitive). (Key point: Reactive power sign)

## Close variants set in other exams

- [Parallel RL and RC branches](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-rl-rc-240v-30ohm-75mh) — Basic Electrical Engineering, chapter 6, asked 2×

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