- [Narrator] Welcome to an example of an application
of first order differential equations
involving electrical circuits.
In this video we'll model an RL circuit
using a first order differential equation
and then solve an RL circuit problem.
The differential equation L times d i d t plus R times i
equals E of t can be used to model the amount
of current, i of t in amperes, in a circuit containing
a resistance R in ohms,
inductance L in henries,
and the electromotive force E of t in volts.
Another way to say this is:
The sum of the voltage drop across the inductor,
L times d i d t, and the voltage drop
across the resistor, R times i,
is equal to e of t,
often called the impressed voltage,
where, again, L is the inductance, R is the resistance,
where both L and R are constants,
and the current i of t is often called the response
of the system.
Here's a diagram of a basic RL circuit.
Let's take a look at our example.
Two nine-volt batteries are connected to a series
in which the inductance is 1/4 henry
and the resistance is eight ohms.
We want to determine the current i of t
if the initial current is zero.
So from this problem we need to recognize
that L is 1/4 henry, so L is equal to 1/4.
The resistance is eight ohms, so R is equal to eight.
And because we have two nine-volt batteries,
and two times nine volts is 18 volts,
E of t is equal to 18.
So this gives us the differential equation
1/4 times d i d t
plus eight i must equal 18.
Now this is a linear first order differential equation.
So let's go ahead and put this in standard form,
so we'll multiply everything by four.
This'll give us d i d t
plus, eight times four is 32,
so we'll have 32 i, must equal four times 18,
is going to be 72.
Now because this is a linear first order
differential equation, we're going to go ahead and solve this,
using an integrating factor,
where mu of t is going to be equal to e raised to the power
of the integral of 32dt.
Again, this 32 is the function being multiplied
by i, or in this case just the constant, 32.
So our integrating factor is going to be
e raised to the power of 32t.
Now we'll multiply both sides of this differential equation
here by e raised to the power of 32t.
So we'll have e raised to the power of 32t
times d i d t,
plus e to the 32t times 32i,
must equal e to the 32t times 72.
Remember, when solving a differential equation
using this technique, the left side of this equation,
or this side here, is always going to equal the derivative
of the product of the integrating factor
and our function i, or i of t.
So if the left side of this equation is equal
to this derivative here,
then this derivative must also be equal to the right side,
or 72 times e raised to a power of 32t.
Now we can solve this for i
by integrating both sides of the equation
with respect to t.
Let's do that on the next slide.
So now we integrate the left side with respect to t,
and the right side with respect to t.
On the left side the integral and derivative
undo each other, leaving us
with e raised to the power of 32t times i.
On the right side we do have to perform
a u substitution here, where u is going to be equal to 32t,
so differential u is going to be equal to 32dt.
So in performing this substitution notice
how we'll have an extra factor of one over 32.
So the integral here is going to be 72 divided by 32
times e raised to the power of 32t,
plus our constant of integration.
And now to solve this for i,
we'll divide both sides by e to the power of 32t.
So we have i of t is equal to,
this simplifies to one,
so I have 72 over 32, which we'll simplify
in the next step, plus c times e raised
to the power of negative 32t.
And this simplifies to 9/4, so we have i of t
equals 9/4 plus c times e
raised to the power of negative 32t.
Now we'll find the value of our constant.
Remember they told us the initial current was equal
to zero, which means i of zero is equal to zero,
so we'd have 9/4 plus c raised to the power
of e to the zero, which is just one.
This must equal zero, so we have c equals
negative 9/4, so our function i of t
that tells us the current is equal to 9/4
minus 9/4 times e raised
to the power of negative 32t.
So this is the function that we're looking for
that will tell us the current at any time t.
But there are a couple of more things I would like
to mention.
If this is our function i of t,
notice as t approaches infinity, i of t is going
to approach this constant 9/4,
because as t approaches infinity, the second term,
negative 9/4 e raised to the power of negative 32t
approaches zero.
If you can't see why that's true,
we could write this as negative nine divided by four
times e raised to the power of 32t.
In this form you can probably see,
as t approaches infinity,
the denominator increases without bound,
and therefore this is going to approach zero.
So this term that approaches zero
is called the transient term,
and the remaining term, 9/4 in this case,
is called the steady term.
So 9/4 amperes is the steady rate current.
OK, I hope you found this example helpful.