Solving RL Circuit Differential Equation via Laplace Transform

PhysicsCircuit Analysis using Laplace TransformsMediumSTEM

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b) The differential equation for an RL circuit with unit ramp input is given by:

$$iR + L \frac{di}{dt} = t, \ t > 0, \ i(0) = 0$$

Use Laplace transform to solve the equation. (7 marks)

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Step by Step Written Solution

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Step 1

In this problem, we are asked to solve a first-order differential equation representing an R L circuit with a unit ramp input using the Laplace transform method.

Solving RL Circuit Equation

$$iR + L \frac{di}{dt} = t, \quad t > 0, \quad i(0) = 0$$
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Step 2

Let's begin by taking the Laplace transform of both sides of the equation. We will denote the Laplace transform of small i of t as capital I of s.

$$\mathcal{L}\{iR + L \frac{di}{dt}\} = \mathcal{L}\{t\}$$
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Step 3

Using linearity and the derivative property of the Laplace transform, the left side becomes R times I of s, plus L times the quantity s times I of s minus i of zero.

$$R I(s) + L(s I(s) - i(0)) = \mathcal{L}\{t\}$$
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Step 4

The Laplace transform of the ramp function t is one over s squared. We also substitute the initial condition, i of zero equals zero.

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Step 5

Now, we factor out I of s on the left side to begin isolating it.

$$I(s)(R + Ls) = \frac{1}{s^2}$$
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Step 6

Dividing both sides by the quantity L s plus R, we get our expression for I of s. To make calculation easier, let's factor L out of the denominator.

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Step 7

To find the inverse Laplace transform, we need to use partial fraction decomposition on this expression.

Partial Fraction Decomposition

$$\frac{1}{s^2(s + R/L)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + R/L}$$
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Step 8

Multiplying through by the common denominator, we get one is equal to A times s times the quantity s plus R over L, plus B times the quantity s plus R over L, plus C times s squared.

$$1 = As(s + R/L) + B(s + R/L) + Cs^2$$

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About This Question

Subject
Physics
Topic
Circuit Analysis using Laplace Transforms
Difficulty
Medium
Exam
STEM
Question Type
Open Ended

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