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Okay, now we need to talk about real voltage and current sources.

And that's going to bring us to the point where we can then start talking about

impedance matching. So now, up to now, we've been talking

about only ideal voltage sources. So, an ideal voltage source.

The definition of that is that, it has a fixed voltage no matter how much current

is drawn from the source. And we just represented that with, with

this sort of symbol. Now, a real voltage source can only

provide a limited amount of current. There is a you know, if you take, for

example, a 1 1 half volt battery, it's 1 1 half volts.

but if I, completely short-circuit that battery, it cannot provide an infinite

amount of current. A very large current, but it's limited.

And the way to represent that is to add a small internal resistance inside the

battery. There actually is a small amount of

resistance inside the battery. And that limits the maximum amount of

current available from that source. So then, if we take this model of a real

voltage source here, and then I connect a short circuit across it.

I'm going to have some current flowing through this short.

And the amount of current that flows through that short is going to be limited

by the internal resistance of the battery.

So, let's say this is a 1 and 1 half volt battery.

And it has an internal resistance of 0.1 ohms, for example.

Then, the total amount of current flowing through this short will be limited at 15

amps. So, that's a lot of current, but it's not

infinite. Now the, in a practical sense then we're

getting closer to the idea of impedance matching.

So, in a practical circuit. I want to let's say I'm trying to build a

heater circuit. And I want to get the maximum of power

transfer from the battery into this load resistor to make the maximum amount of

heat. And now, in practice, the load resistance

and the internal resistance form a voltage divider.

So remember, this is inside the battery and this is the external load.

So, the voltage developed across RL, the load, is just going to be familiar

voltage divider equation. So, it's the total voltage times this

factor, which is the ratio RL over the total series resistance RL plus R

internal. Now, we're ready to talk about the

subject of impedance matching. So, this is the same circuit that we just

looked at, you have a volt, voltage source, a real voltage source with some

internal impedance. And we attached some lower resistor and

there's a current flowing around this circuit.

So the question is, what is the value of the load resistance rl that will give us

maximum power transfer from this source to the load?

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Dissipated in this resistor is just the current going through the resistor times

the voltage across the resistor, VL. Then, the current going through the

resistor is just the voltage divided by the series combination of the two

resistances. And the voltage across this resistor is

just the current going through it times RL.

So, I can use this expression for I and plug it in here, and so this becomes V

over R internal plus RL times RL. So, now I can take both of these

expressions. The one for VL and plug it in here.

The one for I and plug it in there. And so, you get the power dissipated in

the low resistor. It goes as V squared.

So, there's a V times a V. And then it, it's proportional to RL

divided by this factor R internal plus R L squared.

So, that's the power dissipated in that resistor, which transferred to that

resistor. And so now what we want to do is find the

value of RL that makes this a maximum, as big as it can be.

Now, we can do this using a little bit of calculus.

And we'll do that in a minute. But first, let's just take a look at a

graph of this. So I just use Microsoft Excel and plotted

this factor as a function of RL over R internal.

And so you, you see that there's a peak here when RL equals R internal.

This curve reaches its maximum, and the value of that maximum is .25.

And so, here's the answer to, the question that RL equals R internal gives

us the maximum of this curve. And at that point, this factor has a

value of one fourth. Now we, as I said, we could also use a

little bit of calculus to find this maximum.

And so, what we have to do is take the factor that we're trying to maximize with

respect to RL. So, you take the derivative with respect

to RL, set the whole thing equal to zero. And then, solve for RL.

If you do that, you find that the solution to that is going to be RL equals

R internal. That's where the peak is going to be.

And then, if I plug that back in, I make RL equal R internal.

The maximum power transfer then is just v squared over 4R internal.

So at that value of rl I get maximum power transfer to the load resistor