---
title: "Series RL circuit with 283 sin(300t+90°)"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-rl-283sin300t-4-ohm-0-01h"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rl-283sin300t-4-ohm-0-01h
site: BCT Engineering
---

# Series RL circuit with 283 sin(300t+90°)

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

- Asked in: 2071 Shrawan, 10 marks
- Video: [RLC Series Circuit: Impedance, Power Factor, Active Power, & Reactive Power Analysis with AC Supply](https://www.youtube.com/watch?v=QpRtsF6RlBQ) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rl-283sin300t-4-ohm-0-01h
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A series circuit consists of resistance equal to 4Ω and inductance of 0.01 H. The applied voltage is 283 sin (300t + 90°)V. Calculate the following: i) Power factor ii) Expression for i(t) iii) The power dissipated in the circuit iv) Voltage drop across each elements v) Draw a phasor diagram

## Short answer

Power factor is 0.8 lagging. Current i(t)=56.6 sin(300t+53.13°) A. Power dissipated is 6407 W. RMS voltage drops: V_R=160.1 V, V_L=120.1 V.

## Answer

Given R = 4 Ω, L = 0.01 H, v = 283 sin(300t + 90°) V. So Vm = 283 V, ω = 300 rad/s.

- X_L = ωL = 300 × 0.01 = 3 Ω
- Z = 4 + j3 = 5∠36.87° Ω

#### i) Power factor
cos φ = R/Z = 4/5 = **0.8 lagging**

#### ii) Expression for i(t)
Im = Vm/Z = 283/5 = 56.6 A. The current lags the voltage by 36.87°, so its phase is 90° − 36.87° = 53.13°.

**i(t) = 56.6 sin(300t + 53.13°) A**

#### iii) Power dissipated
I_rms = 56.6/√2 = 40.02 A

P = I_rms² R = 40.02² × 4 = **6407 W** (≈ 6.4 kW)

Check: V_rms I cos φ = 200.1 × 40.02 × 0.8 = 6407 W.

#### iv) Voltage drops (rms)
- V_R = I R = 40.02 × 4 = **160.1 V** (peak 226.4 V)
- V_L = I X_L = 40.02 × 3 = **120.1 V** (peak 169.8 V)

Check: √(160.1² + 120.1²) = 200.1 V = 283/√2.

#### v) Phasor diagram (I drawn as reference)
```text
                         V = 200.1 V
                       ↗ ↑
                     ╱   │
                   ╱     │ V_L = 120.1 V (leads I by 90°)
                 ╱       │
               ╱ 36.87°  │
   ───────────●──────────┴──────► I = 40.02 A
               V_R = 160.1 V (in phase with I)
```
V_R is in phase with I, V_L leads I by 90°, and their sum V leads I by 36.87°. In absolute terms I is at +53.13° and V at +90°, so the whole diagram is simply rotated by 53.13°.

## Check yourself

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

1. With R = 4 Ω and X_L = 3 Ω, the power factor is:
   A. 0.8 leading
   B. 0.75 lagging
   C. 0.8 lagging
   D. 0.6 lagging
   Answer: C. Z = 5 Ω, cos φ = R/Z = 4/5 = 0.8 lagging. (Key point: Impedance and pf)

2. The voltage has phase +90°. What is the current phase?
   A. 90° − 36.87° = 53.13°
   B. 90° + 36.87°
   C. −36.87°
   D. 90°
   Answer: A. Current lags voltage by 36.87° so its phase is 53.13°. (Key point: Current phase)

3. Which expression gives the power dissipated?
   A. I_m^2 R = 56.6^2 × 4
   B. I_rms^2 R = 40.02^2 × 4 = 6407 W
   C. I^2 X_L
   D. V_m I_m
   Answer: B. Real power uses rms current; the inductor consumes none. (Key point: Power dissipated)

4. How do V_R and V_L combine?
   A. V = V_L only
   B. V = √(V_R^2 + V_L^2) = 200.1 V
   C. V = V_R − V_L
   D. V = V_R + V_L
   Answer: B. V_R is in phase with I and V_L leads by 90°, so they add at right angles (160.1 V and 120.1 V). (Key point: Voltage drops)

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