---
title: "RL coil on sinusoidal source"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:rl-coil-200sin314t-20ohm-0-1h"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/rl-coil-200sin314t-20ohm-0-1h
site: BCT Engineering
---

# RL coil on sinusoidal source

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

- Asked in: 2074 Chaitra, 6 marks
- Video: [Phasor Diagram of Series RL Circuit](https://www.youtube.com/watch?v=5jI35F6kLa8) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/rl-coil-200sin314t-20ohm-0-1h
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

An alternating source of emf v =200sin (314t) volts is applied to a practical coil with resistance 20Ω and inductance 0.1 H respectively. Determine (i) expression for instantaneous current and power factor (ii) active, reactive and apparent power of circuit (iii) voltage drop on resistor and inductor and (iv) construct phasor diagram for above circuit.

## Short answer

Reactance X_L=31.4 Ω, impedance 37.23 Ω. Instantaneous current is 5.372 sin(314t−57.5°) A, power factor 0.537 lagging. Active power 288.6 W, reactive 453 VAR, apparent 537 VA. Voltage drops: resistor 76 V, inductor 119.3 V.

## Answer

Given v = 200 sin 314t, so Vm = 200 V, ω = 314 rad/s. R = 20 Ω, L = 0.1 H.

- X_L = ωL = 314 × 0.1 = 31.4 Ω
- Z = R + jX_L = 20 + j31.4 = 37.23∠57.5° Ω

#### (i) Instantaneous current and power factor
- Im = Vm/Z = 200/37.23 = 5.372 A
- φ = tan⁻¹(31.4/20) = 57.5° (current lags)

**i = 5.372 sin(314t − 57.5°) A**

**Power factor = cos 57.5° = R/Z = 20/37.23 = 0.537 lagging**

#### (ii) Powers
V_rms = 200/√2 = 141.4 V, I_rms = 5.372/√2 = 3.80 A

- Apparent: S = VI = 141.4 × 3.80 = **537 VA**
- Active: P = I²R = 3.80² × 20 = **288.6 W** (= VI cos φ)
- Reactive: Q = I²X_L = 3.80² × 31.4 = **453 VAR** (lagging)

#### (iii) Voltage drops (rms)
- V_R = I R = 3.80 × 20 = **76.0 V** (peak 107.5 V)
- V_L = I X_L = 3.80 × 31.4 = **119.3 V** (peak 168.7 V)

Check: √(76.0² + 119.3²) = 141.4 V.

#### (iv) Phasor diagram
```text
  V_L = 119.3 V ▲          ↗ V = 141.4 V
                │        ↗
                │      ↗
                │    ↗
                │  ↗  φ = 57.5°
                O──────────────────► I (reference)
                └── V_R = 76 V ──┘
```
Take I as reference along the horizontal axis. V_R is in phase with I, V_L is 90° ahead of I, and the supply voltage V is their phasor sum, leading I by 57.5°.

## Check yourself

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

1. For R = 20 Ω, L = 0.1 H, ω = 314, the impedance is:
   A. 31.4 Ω
   B. 20 + j31.4 = 37.23 Ω
   C. 20 + 31.4 = 51.4 Ω
   D. 20 − j31.4 Ω
   Answer: B. X_L = 314 × 0.1 = 31.4 Ω and Z = √(20^2 + 31.4^2) = 37.23 Ω. (Key point: Impedance of coil)

2. What is the instantaneous current?
   A. 10 sin(314t)
   B. 5.372 sin(314t − 57.5°)
   C. 3.80 sin(314t − 57.5°)
   D. 5.372 sin(314t + 57.5°)
   Answer: B. Im = 200/37.23 = 5.372 A and current lags voltage by 57.5° in an RL circuit. (Key point: Current expression)

3. What is the power factor?
   A. 0.843 lagging
   B. 0.843 leading
   C. 0.537 lagging
   D. 0.537 leading
   Answer: C. cos φ = R/Z = 20/37.23 = 0.537 lagging. (Key point: Power factor)

4. Why does V_R + V_L (76 + 119.3) not equal the supply 141.4 V?
   A. The drops are in phase
   B. Measurement error
   C. Resistor drops are negligible
   D. V_L leads V_R by 90°, so they add as phasors: √(V_R^2+V_L^2)
   Answer: D. V_R and V_L are 90° apart so supply = √(76^2 + 119.3^2) = 141.4 V. (Key point: Voltage drops)

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