RL Circuits

RL Circuits CLAIMED BY VJ SERVERA

The Main Idea

An RL circuit is one that contains resistors and inductors. The simplest type of these circuits involves one resistor and one inductor, and is called a first order RL circuit. The order of the circuit is equal to the number of inductors involved in the circuit. In this section we will study the following circuit:

A Mathematical Model

First we must remember the relationship between current going through an inductor and the voltage drop across it. The relationship is simply

[math]\displaystyle{ {L\frac{d\vec{i}}{dt}} = \vec{v(t)} }[/math]

meaning that the derivative of the current going through the inductor is proportional to the voltage drop across it. The proportionality constant is the inductance [math]\displaystyle{ L }[/math].

Before the switch closes, we know there is no current going through the inductor , and therefore no voltage. The initial conditions are:

[math]\displaystyle{ v(0) = 0 }[/math]

[math]\displaystyle{ i(0) = 0 }[/math]

Now applying Kirchoff's Voltage Law (loop-rule) we obtain the first order differential equation:

[math]\displaystyle{ V = iR + L\frac{di}{dt} }[/math]

Now we solve the equation. Rearranging:

[math]\displaystyle{ \frac{di}{dt} = \frac{V-iR}{L} }[/math]

[math]\displaystyle{ \frac{di}{V-iR} = \frac{dt}{L} }[/math]

Integrating both sides:

[math]\displaystyle{ -\frac{\ln(V-iR)}{R} = \frac{t}{L} + K }[/math]

Since [math]\displaystyle{ i = 0 }[/math] when [math]\displaystyle{ t = 0 }[/math], plugging these values in gives us

[math]\displaystyle{ K = -\frac{\ln(V)}{R} }[/math]

Substituting K back into our equation, getting rid of the denominator, and collecting the logarithms, we have

[math]\displaystyle{ ln(\frac{V-Ri}{V}) = -\frac{Rt}{L} }[/math]

// finish up the proof

A Computational Model

Inductor simulation

This simple online applet shows how the current through the inductor behaves as time passes. Eventually, the inductor acts as a straight wire and the curren through it is constant.

Examples

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