---
title: "RL series circuit current, pf, power"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:rl-series-230v-numerical"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/rl-series-230v-numerical
site: BCT Engineering
---

# RL series circuit current, pf, power

Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits

- Asked in: 2081 Kartik (new course), 3 marks
- Video: [Calculating Series RL Circuit Amps, Ohms, and Volts](https://www.youtube.com/watch?v=uEBqAFSWGaU) — Dave Gordon
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/rl-series-230v-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

A coil having a resistance of 5 Ω and inductance of 30 mH in series are connected across a 230 V, 50 Hz supply. Calculate current, power factor, and power consumed.

## Short answer

XL = 2π×50×0.03 = 9.425 Ω, Z = √(5²+9.425²) = 10.669 Ω. Current I = 230/10.669 = 21.56 A. Power factor cosφ = 5/10.669 = 0.469 lagging, power consumed P = I²×5 = 2.32 kW.

## Answer

**Given:** $R=5\ \Omega$, $L=30$ mH, $V=230$ V, $f=50$ Hz.

**Inductive reactance:**

$$
X_L=2\pi fL=2\pi\times50\times0.03=9.425\ \Omega
$$

**Impedance:**

$$
Z=\sqrt{R^2+X_L^2}=\sqrt{5^2+9.425^2}=10.669\ \Omega
$$

**Current:**

$$
I=\frac{V}{Z}=\frac{230}{10.669}=21.56\ \text{A}
$$

**Power factor:**

$$
\cos\phi=\frac{R}{Z}=\frac{5}{10.669}=0.469\ \text{(lagging)}
$$

**Power consumed:**

$$
P=I^2R=(21.56)^2\times5=2323.7\ \text{W}
$$

(Check: $VI\cos\phi=230\times21.56\times0.4686=2323.7$ W.)

**Answer: $I=21.56$ A, pf $=0.469$ lagging, $P=2.32$ kW.**

## Check yourself

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

1. What is the inductive reactance of the 30 mH coil at 50 Hz?
   A. 1.5 Ω
   B. 9.425 Ω
   C. 0.0094 Ω
   D. 94.25 Ω
   Answer: B. XL = 2 pi f L = 2 x pi x 50 x 0.03 = 9.425 Ω. (Key point: Reactance)

2. How is the current found?
   A. I = V/(R + XL) = 16 A
   B. I = V/R = 46 A
   C. I = V/XL = 24.4 A
   D. Z = sqrt(R^2 + XL^2) = 10.669 Ω, so I = 230/10.669 = 21.56 A
   Answer: D. R and XL add at 90°, so Z is their phasor sum, sqrt(25 + 88.8) = 10.669 Ω. (Key point: Impedance and current)

3. Which gives the power factor and power consumed?
   A. pf = R/Z = 0.469 lagging; P = I^2 R = 2.32 kW
   B. pf = XL/Z = 0.88; P = I^2 XL
   C. pf = R/Z = 0.469 leading; P = V I = 4.96 kW
   D. pf = 1; P = V^2/R
   Answer: A. pf = R/Z for a series R-L circuit (lagging), and real power is dissipated only in R: I^2 R. (Key point: Power factor and power)

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