RL Circuits: Difference between revisions
| Line 28: | Line 28: | ||
<math>\frac{di}{dt} = \frac{V-iR}{L}</math> | <math>\frac{di}{dt} = \frac{V-iR}{L}</math> | ||
<math> | <math>\frac{di}{V-iR} = \frac{dt}{L}</math> | ||
Integrating both sides: | Integrating both sides: | ||
<math>- | <math>-\frac{ln(V-iR)}{R} = \frac{t}{L} + K</math> | ||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 00:44, 12 April 2016
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]
A Computational Model
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