This is a short video on potential dividers, but before we talk about potential dividers I
just need to introduce two new symbols for two new parts of a circuit. This first symbol on the left
means a variable resistor, which really just means a resistor with a resistance that can be changed
to a number that we want. So basically we can control the resistance in that resistor and we
can change it. The symbol on the right is for a potentiometer or a rheostat. That's another name
for a potentiometer. A potentiometer is a resistor that we can attach a wire to at different parts,
splitting it into 2 resistors with one part in series with the attached wire and the other in
parallel with the attached wire. I'm going to show you some animations on what that looks like now.
So if we have a potentiometer set up like this you can see that there's a wire connected to the
potentiometer itself, to the resistor itself, and that arrow just shows you where it's connected.
Specifically, this circuit here would behave like this circuit here, with 2 non-potentiometer
resistors. Basically what's happening is that wire is splitting the resistance in half. So that half
of the resistance is experienced by a current in series and the other half is in parallel,
with that other resistor so you can see that that's what's happening here. It's kind of
just splitting the resistor in 2. And because this is at the exact center and I'm assuming
that there's a uniform resistance throughout the potentiometer, the resistance of each individual
part of the potentiometer would just be the total divided by 2. If we split that exactly in half,
if we moved it up here to three quarters of the way up, would look something like this
where the top part has a resistance of one-fourth of the total and the bottom
part has a resistance of three-fourths of the total. If we move the wire all the way to the top
it would look like this. You can see that there's no longer any current going through
the resistive part in series, it's all in parallel now. And if I brought it down here
all of it would just be in series. So now we’re ready to talk about potential dividers themselves.
A potential divider is any circuit arrangement that is designed to reduce the voltage by a
certain fraction. In IB potential divider problems they just mean problems where you
have to choose resistor values to give certain parts of the circuit specific voltages.
So I'm going to do an example problem with that variable resistor. So our problem is that the
light bulb in the circuit is designed to work at 9 volts. What resistance should we give the
variable resistor to give the battery 9 volts? So it seems pretty simple. We've got a total
voltage in the circuit of 21 volts and we want that light bulb to have a voltage of 9 so we
need to figure out exactly how much resistance in that parallel resistor will give it a voltage
of 9. So as with most physics problems I'm not really going to try to visualize this. Instead,
what I'm going to do is to just start writing down everything that I know about this problem.
So first of all I can come up with an equation for the equivalent resistance of the parallel
resistors and I’m going to be calling the variable resistor just “r.” That's the thing I'm trying to
solve for in this problem. And when I combine these I get the equivalent resistance is equal
to this equation here, and that means that the equivalent resistance of the total circuit,
because that parallel arrangement is in series with a 2 ohm resistor. This is what the equivalent
resistance of the full circuit would be like. Just the sum of those two resistances. I'm now going
to bring this over here and continue to solve the problem. I know that the light bulb needs 9 volts,
and because it's in parallel with the variable resistor
that means that the variable resistor is also going to receive 9 volts.
And because this 9 volt drop is in series with the 2 ohm resistor that means that those
two voltages have to add together 9 plus the voltage in the 2 ohm resistor has to equal 21.
So that means that the voltage in the 2 ohm resistor is going to be 12 volts,
and using V = IR I can find that the total current in the 2 ohm resistor is 6 amps.
So again I’m just starting to find information based on what I know.
So that means that the total current in the circuit is 6 amps because it
hasn't been split up yet when it gets to the 2 ohm resistor.
And I can now make a connection between my equivalent resistance, my current,
and my voltage, because I know the total voltage, I know the total current, and I have an equation
for the equivalent resistance. So this is what that looks like when I plug these in.
And now I have an equation with just one variable R, so I’m going to be able to
solve for r which is the goal of this problem. I want to know what resistance
that variable resistor should have. So just plugging this in using algebra
I get 2.4 ohms for the resistance of the variable resistor. So if we give the variable
resistor that exact resistance, the light bulb is going to receive 9 volts of voltage.
So that is how you do a potential divider problem. Again a potential divider problem
really just means trying to find an arrangement or a specific resistance that will cause another
part of the circuit to experience the voltage that you would like it to have.
The second example is using a potentiometer, and the questions about potentiometers don't get too
complicated. So here this is just asking where can we set the potentiometer so that the light bulb
receives the greatest voltage? And where can we set it so that it will receive the least voltage?
So the place where the light bulb would experience the greatest voltage is where
it would be in parallel with the potentiometer, because if it's in parallel that would mean that
both the potentiometer and the light bulb are receiving the full 18 volts from the battery,
whereas if they're in some form of series with each other that means that they're
going to be splitting up the voltage. So to make the light bulb in parallel
with the potentiometer we would need to put the line up here specifically.
So in this case the current can either go through the full potentiometer or through the light
bulb like this. So this is the light bulb being perfectly in parallel with the full potentiometer,
and the place where it would experience the least voltage is where we would put the potentiometer
entirely in series with the light bulb, and what's happening here is that the light bulb is now in
parallel with a part of the wire that doesn't have any resistance. So this part of the wire
down here, the only part that's in parallel with the light bulb, has zero resistance,
and because resistance is voltage drop divided by current that means that in this area the voltage
drop is also going to be zero. So there's no voltage drop across this wire which means that
there's also going to be no voltage in the light bulb either which actually also means that there's
no current in the light bulb, because the light bulb does have a resistance, so if there is a
resistance but there's no voltage, no current can occur either so the light bulb would experience
the most voltage when it's in parallel with the full potentiometer, and it would experience
the least voltage when it's just in parallel with that empty wire below the potentiometer.