Peter Wheeler: Hi, today, I'd like to talk to you about some concepts related to DC circuits,
DC, direct current, circuits.
There are also AC circuits, which are alternating current.
The difference is, with direct current, the charge moves in one direction through a circuit.
With AC current, the charge is pushed and then pulled, and pushed and pulled.
Today we're talking about DC circuits, and, in particular, we'll talk about simple series
and parallel resistor circuits, and some theorems that help us simplify these circuits, so that
we're only looking at a voltage and a resistance.
What exactly all am I talking about?
Let's get started.
This is a classic series DC circuit.
The direct current source is the battery.
The batter has a polarity.
Charge comes out of the high side, runs around the circuit, if you complete the circuit,
and then goes to the low side.
Typically, the low side is zero volts, and the high side, let's pick a battery voltage,
10 volts.
Now in these DC circuits, we have a law that holds as far as how much current should run
through a circuit, given a certain resistance, and a certain voltage.
That law is Ohm's law.
Ohm said that the push that you see in a circuit is equivalent to the charge that flows, and
the resistance that's pushing back.
Push equals flow times pushback.
Let's pick some numbers.
Let's say we have a resistor value of, I don't know, let's pick 100 ohms, make it simple.
Ohms is the symbol for, this is Greek Omega, for resistance.
The symbol we use for current is amperes, and A, and the symbol we use for voltage is
volts.
We've got 100 ohms, we've got ... excuse my mess.
We've got 10 volt battery.
In this case, here, we'll see current flow in this direction.
The symbol to show current is I. It's from the French, intensité du courant, but regardless,
it's a letter I.
Now, so we have a voltage, we have 10 volts.
We have a resistance.
If we rearrange Ohm's law, okay.
If we take 10 volts and divide it by 100 ohms, we're going to get zero point ... let's just
do that, one, zero amps.
Or, in engineering dimensions, we like to use powers of three.
This would be milli-amps, equal to 100 times 10 to the minus three amps.
Okay?
Okay, that's a very simple circuit.
Now I'm going to add a small complication to this circuit, and I'm going to add another
resistor.
Resistors, when you add them in series, they add together.
Let me just clean the board, just a touch.
Just to make this look a little cleaner to our viewers ... let me take the same circuit,
and add another resistance to it.
Now, in this one here, let's use the 10 volts again.
This is the plus side, this is the high side, and let's use ... we can use the same thing.
We can do 100 ohms, let's leave it there.
Here we'll put 400 ohms.
Let's just do some understanding what's going on in a circuit.
First of all, this is the high side, so the charge is going to come out of the high side
of the battery, and it's going to come through those two resistors, and come back to the
low side of the battery.
In doing so, we're going to develop a potential across these two resistors.
There's going to be a current flowing, and a resistance pushing back, which means there's
a voltage across here.
Same thing.
Current is flowing, resistance is there, I times R, a voltage will be across that resistor.
We call that voltage drop.
The voltage at this point will be 10 volts.
The voltage at this point will be less than 10 volts, and the voltage at this point will
be zero volts.
Now one other symbol that you often see in these circuits is this symbol here.
This symbol is for ground.
This essentially sets a reference.
A battery, by itself, floating in space, is some high side and low side, but with respect
to what?
With respect to what?
What we do is we anchor one side of it to a chassis, and then, say, mother earth.
Ground.
And so this is a ground symbol.
That just says, when you see that, it's zero volts.
It's zero volts.
Now that means we have 10 volts up here, we have some less voltage because of the drop
here, and then we have zero volts, at this point.
What is that?
We could say, well, let's go back to Ohm.
If he said V equals I times R, and in this case, now, we have R is equal to ... our total
R is equal to, in this case R one and R two, and that sum is 500 ohms.
Oops, 500, give me a sec. 500 ohms, that's the total resistance.
Then this current, that is flowing, is equal to 10 volts divided by 500 ohms.
That equals 0.02 amps, which is equal to 20 milli-amps.
I want you to check my math.
That's one divided by 50.
That's 0.1 divided by five.
That's 0.1 divided by five, 0.02, 20 milli-amps.
Okay, 20 milli-amps.
We have 20 milli-amps running through this circuit.
That's the I, flowing through this circuit.
Now if that is the case, there's going to be a voltage drop across here, and a voltage
drop across here, so that we result with zero volts.
The voltage drop across the 400 ohms, and the voltage drop across the 100 ohms, will
add up to 10 volts.
Is that the case?
If we look at R times I equals V, Ohm's law, one more time.
Then we would have, for R1, 100 ohms, time 0.02, is equal to two volts.
And for R2, we would have 400 ohms times 0.02, and that will equal eight volts.
Across here, this point here, we'll have eight volts.
Zero volts at this point, and, as you know, 10 volts at that point.
A two volt drop, across the 100 ohms, an eight volt drop, across the 400 ohms.
I'm going to clean this circuit up a bit, come back to this particular design, here,
and we're going to do that calculation in another fashion.
Coming back to the same circuit, I'd like to calculate the voltage across the 400 ohms
in a slightly different manner as before.
Ohm's law still applies.
Okay?
Ohm's law still applies.
If we look at this circuit, we see that there is one current, which we calculated, it was
20 milli-amps, based on the total resistance of 500 ohms, and 10 volts pushing it.
If we look at that, the current through this resistor ... and again, let's do R1, and R2.
The current through R1 is going to be, basically, is going to be V over R. Okay?
We look at the current through R1, and then we look at the current through R2.
Again, it's V over R. There's some voltage here.
The voltage here, and the voltage across R1, will add up to the source voltage.
If we look at this, we say, okay, this is some percentage of that, but they have to
add up to the total.
In other words, if I have 400 ohms, over a total of 500 ohms, and I multiply that by
the source, I would get the voltage across R2.
Really?
How do you do that?
In other words, what I said was, if I take R2, and divide it by R1, plus R2, and multiply
it by the source voltage, I will get the voltage across R2.
Okay, so if we try to understand that, we can develop this as an equation.
We can say the ratio of the voltages, so the voltage across R2, over the total voltage,
is equal to, in this case, here, it's going to be the current through, in this case R2,
and the total was the current through R1, plus R2, and we see these cancel.
If we take our voltage source to the other side, for example, if I multiply by V source
here, and V source here ... because in the equation, whatever you do to the left side,
you have to do the right side, so then we see these to cancel.
We're left with V across R2 is equal to R2 over R1, plus R2 times the source.
Same thing here.
This formula holds.
This is known as a voltage divider rule.
It's a quick and dirty way to come up with, instead of calculating total resistance, and
therefore voltage divided by total resistance to get the current, then multiplying the current
through the resistor, still holds, fundamental still holds.
There's another way of doing it, saying this is a part of the total pie, and if I take
this as a percentage of the total pie, what is left?
If we do the numbers again, remember, we had eight volts, from the previous discussion,
we would have, in this case here, four fifths time 10 is equal to four times two, is equal
to eight volts.
Same number.