---
title: "Parallel RL and RC branches"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:parallel-rl-rc-240v-30ohm-75mh"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-rl-rc-240v-30ohm-75mh
site: BCT Engineering
---

# Parallel RL and RC branches

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

- Asked in: 2072 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-240v-30ohm-75mh
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

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.

## 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.

## 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. Which are Z1 and Z2?
   A. 30 + j23.56 Ω and 20 − j31.83 Ω
   B. 53.56 Ω and 51.83 Ω
   C. 30 + j75 and 20 − j100
   D. 30 − j23.56 and 20 + j31.83
   Answer: A. X_L = 23.56 Ω (inductive) and X_C = 31.83 Ω (capacitive). (Key point: Branch impedances)

2. Which describes the branch currents?
   A. I1 = 38.15 A, I2 = 37.59 A
   B. Both 8.48 A
   C. I1 = 6.29 A leading, I2 lagging
   D. I1 = 6.29 A lagging 38.15°, I2 = 6.38 A leading 57.86°
   Answer: D. I = V/Z for each branch; the RL branch lags, the RC branch leads. (Key point: Branch currents)

3. Why is the total current 8.48 A rather than 6.29 + 6.38 = 12.67 A?
   A. Mistake in values
   B. Supply voltage drops
   C. Branch currents have different phase angles and add as phasors
   D. Resistors cancel
   Answer: C. I = I1 + I2 = (4.95 − j3.89) + (3.40 + j5.41) = 8.34 + j1.52 A. (Key point: Total current)

4. How can active power be checked?
   A. P = I1^2 R1 + I2^2 R2 = 2002 W
   B. P = V × I = 2036 W
   C. P = Q cos φ
   D. P = I^2 X
   Answer: A. Real power is dissipated only in resistors; reactive powers of the branches partly cancel (net −365 VAR). (Key point: Power)

## Close variants set in other exams

- [Parallel RL and RC branches, total current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-rl-rc-total-current-pf-power) — Basic Electrical Engineering, chapter 6, asked 2×

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