---
title: "RL circuit current as a function of time"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:app-rl-circuit-current"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/app-rl-circuit-current
site: BCT Engineering
---

# RL circuit current as a function of time

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2074 Asoj; 2073 Shrawan; 2070 Chaitra; 2069 Chaitra (asked 4 times)
- Video: [Find i(t) in RL circuit. | First Order Circuit | Electrical Engineering](https://www.youtube.com/watch?v=bq_kXw4dF8M) — 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/mathematics_i/app-rl-circuit-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

A resistance of 100 ohms and an inductance of 0.5 Henry are connected in series with a battery of 20 volts. Find the current in the circuit as a function of time.

## Short answer

Solve L di/dt + Ri = E with i(0)=0. The current is i(t) = 0.2(1 − e^(−200t)) A.

## Answer

A series RL circuit with a constant EMF E obeys Kirchhoff's voltage law:

L di/dt + R i = E

Given R = 100 Ω, L = 0.5 H, E = 20 V, and (as usual) the current is zero at the instant of closing the switch, i(0) = 0.

#### Set up
0.5 di/dt + 100 i = 20 → di/dt + 200 i = 40

This is a linear equation with integrating factor e^(200t):

d/dt(i e^(200t)) = 40 e^(200t)

i e^(200t) = (40/200) e^(200t) + K = 0.2 e^(200t) + K

i = 0.2 + K e^(−200t)

#### Apply i(0) = 0
0 = 0.2 + K → K = −0.2

**i(t) = 0.2(1 − e^(−200t)) amperes**

#### Interpretation
- Steady-state current E/R = 20/100 = 0.2 A, reached as t → ∞.
- Time constant L/R = 0.5/100 = 0.005 s (5 ms); after about 5 time constants (25 ms) the current is within 1% of 0.2 A.

## Check yourself

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

1. Which equation does Kirchhoff's voltage law give for the series RL circuit?
   A. L di/dt + R i = E
   B. L di/dt - R i = E
   C. L i + R di/dt = E
   D. R di/dt + L i = E
   Answer: A. The voltage drops across L (L di/dt) and R (Ri) sum to the EMF E. (Key point: Governing equation)

2. For di/dt + 200 i = 40, what is the integrating factor?
   A. e^(200t)
   B. e^(-200t)
   C. e^(0.5t)
   D. e^(40t)
   Answer: A. The integrating factor is e^(integral of 200 dt) = e^(200t). (Key point: Integrating factor)

3. Using i(0) = 0 in i = 0.2 + K e^(-200t), what is K?
   A. -0.2
   B. 200
   C. 0.2
   D. 0
   Answer: A. 0 = 0.2 + K gives K = -0.2. (Key point: Initial condition)

4. What are the steady-state current and the time constant?
   A. 0.2 A and 0.05 s
   B. 5 A and 0.005 s
   C. 0.2 A and 200 s
   D. 0.2 A and 0.005 s
   Answer: D. Steady state is E/R = 0.2 A and the time constant is L/R = 0.5/100 = 0.005 s. (Key point: Steady state and time constant)

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