---
title: "Parallel RLC supply current"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:parallel-r-l-c-200v-supply-current"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-r-l-c-200v-supply-current
site: BCT Engineering
---

# Parallel RLC supply current

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

- Asked in: 2073 Chaitra, 6 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-r-l-c-200v-supply-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A 10 ohm resistor, a 31.8 mH inductor and 318μF capacitor are connected in parallel and supplied from a 200V, 50Hz supply source. Calculate the supply current and power factor and also calculate current in each branch.

## Short answer

Nearly equal reactances ~10 Ω. Branch currents: I_R=20 A, I_L=20.02 A lagging, I_C=19.98 A leading. Supply current is 20.0 A, power factor 1.0 (parallel resonance).

## Answer

R = 10 Ω, L = 31.8 mH, C = 318 μF in parallel across 200 V, 50 Hz. ω = 314.16 rad/s.

#### Reactances
- X_L = ωL = 314.16 × 0.0318 = 9.99 Ω
- X_C = 1/(ωC) = 1/(314.16 × 318 × 10⁻⁶) = 10.01 Ω

#### Branch currents (V = 200∠0° as reference)
- I_R = 200/10 = **20.0 A** (in phase with V)
- I_L = 200/9.99 = **20.02 A** (lagging by 90°)
- I_C = 200/10.01 = **19.98 A** (leading by 90°)

#### Supply current
The reactive currents are in opposite directions:

I = √(I_R² + (I_C − I_L)²) = √(20² + (19.98 − 20.02)²) = **20.0 A**

Phase angle φ = tan⁻¹((I_C − I_L)/I_R) = tan⁻¹(−0.04/20) ≈ −0.1° (practically zero).

#### Power factor
cos φ = I_R/I = 20/20 = **1.0 (unity)**

The circuit is practically at parallel resonance (X_L ≈ X_C, f ≈ 50 Hz), so the supply current is only the resistor current, even though each reactive branch carries about 20 A.

## Check yourself

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

1. In a parallel RLC circuit, how is each branch current found?
   A. Add the three component values
   B. Use series current
   C. Divide the supply voltage by each branch's R, X_L, X_C
   D. Divide by total impedance
   Answer: C. Each element sees the full 200 V, so I = V/R, V/X_L, V/X_C. (Key point: Parallel branch currents)

2. Relative to V, I_R, I_L and I_C are respectively:
   A. lagging, lagging, leading
   B. in phase, lagging 90°, leading 90°
   C. in phase, leading 90°, lagging 90°
   D. all in phase
   Answer: B. Inductor current lags, capacitor current leads, so they oppose. (Key point: Phase of branch currents)

3. How is the supply current found?
   A. I = I_L − I_C
   B. I = √(I_R^2 + (I_C − I_L)^2) = 20 A
   C. I = I_R + I_L + I_C = 60 A
   D. I = I_L
   Answer: B. Reactive currents are in opposite directions, so only their difference adds at 90° to I_R. (Key point: Supply current)

4. Why is the power factor unity although I_L and I_C are about 20 A each?
   A. Supply is DC
   B. X_L ≈ X_C so reactive currents cancel (parallel resonance)
   C. Resistor is removed
   D. Inductor current is zero
   Answer: B. X_L ≈ X_C ≈ 10 Ω, so I_L ≈ I_C and the supply provides only I_R. (Key point: Resonance)

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