Peter Wheeler: We've talked about Ohm, V equals IR.
We've talked about Kirchoff's voltage law, sum of the voltage drops around a closed
loop equals 0.
We've talked about, in this case here, a silicon diode, which has a polarity, but it has a
0.7 voltage drop to it.
We've also talked about a voltage divider circuits in a two resistor circuit, how to
calculate the voltage in between the two resistors.
Let's look at what Thevenin was all about.
When you look inside, look at your PC, and you see a port, you plug into the port, there's
a voltage there and a current that will come out, but there's also a bit of an internal
resistance.
You can plug anything into those ports.
For example, if it's USB, which has power in ground, it doesn't know if it's a mouse.
It doesn't know if it's a scanner.
It doesn't know what you're plugging in.
Effectively when you look at a circuit, and let's take this circuit here and here like
this ... Let's put another resistor here.
All right.
Let me use some numbers.
Why don't we use like 10k ohms?
10k ohms sounds about right.
What?
15k ohms?
Yeah.
Why not?
I need a voltage.
5 volts?
All right.
What we do is in our circuits we tend to apply a load.
A load is just something that requires current.
Okay?
A load.
These loads can be different.
I mean, if it's some kind of a heating device and you keep turning the thermostat, you're
going to draw more or less current.
The thing is what we'd like to know is really if this is a black box on this side here,
regardless of the load that's supplied, at least let me understand what's coming out
of my black box.
What is the voltage, and what is the internal resistance, which then would drop the voltage
as current is pulled?
If we look at this circuit here, the interesting part is the current is only flowing in this
loop here.
I'm going to just disconnect the load, so we're not confused.
I'm just going to disconnect this load.
Okay?
It's external.
The current is only flowing in this loop here, which means that we have a voltage at this
point here.
Now, do we have a voltage at this point here?
The answer is yes.
The voltage at this point here, let's call it VX, is going to be the same as the voltage
at this point here.
No current will flow, but the presence of charge, the presence of the pressure pushing
the charge, I can see it from this side of this resistor.
It's like cars driving on a road.
Here, unfortunately, there's a huge chasm and nobody's going anywhere, otherwise we'd
have a problem.
The electrons, or charge as it be, would just park themselves here, which means that there's
no voltage across here.
These two are the same.
No current flows there, but you do see a voltage here.
This concept of an open source voltage sometimes is a bit problematic with new students.
Regardless, we know now that we have a current flowing over here, so we have basically ...
Let's use our voltage divider formula here, so V across the 15k ohms is equal to VS.
Let's use this formula here.
It'll be 15k divided by the sum of those two, 25k, and VS in this case here was 5 volts,
so we have 5 times 15, times 10 to the 3, divided by 25, times 10 to the 3.
Okay?
That's 15, 30, 45, 60, 75 divided by that is equal to 3 volts.
Okay?
At this point here VX is equal to 3 volts, as is equal to here, 3 volts.
Now, we're talking about Thevenin.
He says, "It's one thing to know what is the voltage presented outside of this black box.
What is the voltage presented to the load outside of this black box, but internal to
the box there is an internal resistance.
As current is drawn there will be a voltage drop across the internal resistance."
What he tried to do was create a model of not only the voltage source presented to the
load, but also the internal resistance.
How do you figure out what the internal resistance is?
Well, essentially you've got to put your eye over here.
You put your eyeball here.
All right.
That's you.
You look into this device, into this circuit.
The first rule we had with Thevenin is disconnect the load.
Okay?
We disconnect the load.
That's the first thing.
The second thing we do is we short out voltage sources.
Let's just do that.
If that is the case, then we're going to remove this voltage source and replace it with a
wire.
As you know, the resistance of this wire, at least the wires we use in the labs, is
negligible.
Actually, let's use, for conversational purposes, 0.
We have to short this particular voltage source out, which means were bringing this resistor,
we're bringing it down to this point.
We've actually got two resistors in parallel, I'm looking in from this side, and one resistor
in series.
Again, R1, R2, R3.
Let me put it here, R1 in parallel with R2, and that is added to R3. 10k in parallel with
15k.
The first rule is that resistors in parallel, the final value of resistance looking at parallel
resistors, is less than the smallest resistor there.
I'm going to use a quick formula.
I'm going to use R1 times R2, divided by R1 plus R1, which is a calculating resistors
in parallel, this here.
That's going to be ... Here we have 150 times 10 to the 3, times 10 to the 3.
Correct?
We're going to divide that by 25 times 10 to the 3.
That's going to leave us with 6 times 10 to the 3, 6k ohms for the two parallel resistors.
This becomes 6k ohms.
We're adding to that another 10k ohms.
The total resistance that we're looking at from this side of the circuit into this black
box is 16k ohms.
What Thevenin says is instead of working your ohms in Kirchoff's every time you put a load
on there, why don't we just simplify the circuit by putting in a battery, in this case here
of source 3 volts, and a series resistor of 16k ohms for this particular circuit?
Then go ahead and attach your load and do your calculations.
We've simplified three resistors down to one in this case here, and we've also simplified
the voltage that's present on the circuit that is contributing to then whatever load
you add.
Again, disconnect the load first in order to determine what is the voltage that you
see?
If you leave the load connected, you will have a different number here all together.
Secondly, once you've disconnected the load and determined the voltage, determine the
resistance looking into this Thevenin ... We call that Thevenin resistance.
Determine the Thevenin resistance from shorting the voltage source there.