Hello, my name is Dr.

George, and this lesson is called Potential Divider Circuits.

It's part of the unit Circuit components.

The outcome for this lesson is I can calculate p.

d.

across one of two components in a series circuit with a battery.

I'll be using these key words during the lesson.

If you need to check the meanings anytime, you can come back to this slide.

The lesson has three parts.

They're called a potential divider circuit, variable resistors and variable resistors in potential divider circuits.

So let's start by looking at potential dividers.

Here's a series circuit, a single loop, and the p.

d.

across the cell is the same as the total p.

d.

across all the components.

That's always true for a series circuit.

So these two readings will be the same.

But the way that the two components share that p.

d.

depends on their resistances.

Let's say this resistor has a higher resistance and the lamp has a lower resistance, in that case the p.

d.

across the resistor will be higher than the p.

d.

across the lamp.

One way of thinking about that is that in a series circuit, the current is the same everywhere, and if the current is the same in these two components, a larger p.

d.

is needed to push that current through the component with higher resistance.

Now what's the reading on the volt metre in this circuit? I'm not looking for a calculation, but just one of these options.

And when I ask a question, I'll wait for you for five seconds, but you may need longer.

In which case, press Pause and press Play when you have your answer ready.

Out of these two components, the nine ohm resistor has most of the resistance, and so it has most of the p.

d.

, most of the six vaults across it.

In this circuit, the total p.

d.

across the resistors is the p.

d.

across the cell, six volts.

And the total resistance with resistances in series, you just add them together, so it's 15 ohms. Let's look at the proportion of the total resistance that each of these resistors has.

It's 2/3 for the 10 ohm resistor and the five ohm resistor has 1/3 of the total resistance.

The p.

d.

across these resistors is in the same proportion.

So the 10 ohm resistor has 2/3 of the six volts, which is four volts, and the five ohm resistor has a third of the six volts, which is two volts.

If you're comfortable with using ratios, you can use those here instead.

The ratio of the resistances is two to one, and that's the same as the ratio of the p.

d.

's So if you know how to share six volts in the ratio two to one, you can work it out that way.

What's the p.

d.

across the six ohm resistor in this circuit? And it's three volts because the six ohm resistor has 3/4 of the total resistance, eight ohms, and so it has three quarters of the total p.

d.

3/4 of four volts is three volts.

Well done if you got that.

Now you probably know the current in a series circuit is the same everywhere in the circuit.

Wherever we put this ammeter, we'll get the same reading.

So looking at this circuit again, the current's the same everywhere.

The total resistance is 15 ohms, so we can find the current using I = V divided by R.

And if we do that, we get 0.

4 amps.

And then we can use a different method for finding the p.

d.

across each resistor.

We can use the resistance of that resistor and the current, which is now known.

So using V = IR, the left hand resistor, we get four volts.

And for the right-hand resistor, which has the same current of 0.

4 amps, we get two volts.

An answer that agrees with the way we calculated it previously.

Now which of the following statements are correct? Press Pause while you read these and think about your answer.

It's true that the current through each component in a series circuit is the same.

The p.

d.

isn't, unless the component happened to have identical resistances.

It's also true that if a component in a series circuit has a higher resistance than another component, it will have a higher p.

d.

across it.

Well done if you picked out both of those.

Now we can think of a circuit like this as something called a potential divider, because the potential is divided between these two resistors and it's divided in a way that's proportional to their resistances.

The p.

d.

across a component with a higher resistance is a greater proportion of the cell p.

d.

As we've already seen, the p.

d.

's here are four volts and two volts.

Now which of the following statements about a potential divider circuit with two resistors, one that has a higher resistance than the other is correct? The correct answer is a, the p.

d.

across the higher value resistor is higher.

Make sure you remember that.

It's very useful to know.

And now here are four questions for you.

I'd like you to work out the missing p.

d.

for each circuit.

So everywhere there's a question mark, there's a p.

d.

for you to calculate.

And in fact, in question one, there are two missing p.

d.

s.

So press pause while you work out your answers and press play when you're ready to check them.

I'll show you how to work out each one of these.

So in question one, there were two missing p.

d.

's across the cell and across the right hand resistor.

The p.

d.

across the second 11 ohm resistor will be three volts since it has the same resistance and so the same p.

d.

as the first resistor.

The p.

d.

across the cell will be the sum of the p.

d.

's across the resistor, so 3 plus 3, which is 6 volts.

In question two, we needed to find the p.

d.

across the 10 ohm resistor.

Well, the total resistance is 50 plus 10, which is 60 ohms, and the 10 ohm resistor has a fraction 1/6 of that p.

d.

10 divided by 60 is 1/6.

And 1/6 of the cell p.

d.

, six volts, is simply one volt.

So there'll be a p.

d.

of one volt across the 10 ohm resistor.

There is another way you could calculate each of these, and that's to work out the total resistance in the circuit and use that and the cell p.

d.

to find the current.

And then for a resistor, you can find its p.

d.

using the current and its resistance, but that's a slightly longer method.

Question three, we needed the p.

d.

across the 25 ohm resistor.

Well, the total resistance is 100 ohms here, and the 25 ohm resistor has 25/100, or a quarter of the resistance in the circuit.

And a quarter of the cell p.

d.

is 1.

5 volts.

You can work that out on paper or you can use a calculator to find 25/100 times 6 or six divided by four.

Question four, we need the p.

d.

across the three ohm resistor.

Again, we can find the total resistance, that's 18 ohms. And the three ohm resistor has 3/18 of the total resistance in the circuit.

So it must have 3/18 of the cell p.

d.

and 3/18 of 9 we can work out is 1.

5 volts.

Well done if you're getting these right.

Now in the second part of the lesson, we're going to look at variable resistors in preparation for the third part when we put a variable resistor into a potential divider circuit.

Variable resistor, of course, is simply resistor who's resistance you can change by adjusting a control on the resistor itself.

And you could use a variable resistor to control the brightness of a lamp as in this circuit.

Changing the resistance of the variable resistor will change the p.

d.

across the lamp, because it will change the fraction of total resistance that the lamp has.

This circuit can be used to measure and compare the p.

d.

across the lamp and across the variable resistor, and you're going to investigate that shortly.

You would keep the p.

d.

across the power supply constant so that you're only changing one thing and change the resistance of the variable resistor.

Using two volt metres allows the p.

d.

across the variable resistor to be measured at the same time as the p.

d.

across the lamp.

So what will happen if the resistance of the variable resistor is increased here? What happens is that the lamp gets dimmer, because the variable resistor has a greater fraction of the total resistance, so it gets a greater fraction of the supply p.

d.

And with less p.

d.

across the lamp, it's dimmer.

Now for the investigation, I'd like you to investigate how the p.

d.

across a variable resistor and the p.

d.

across a lamp vary as the resistance of the variable resistor is changed.

So keep the power supply p.

d.

constant and draw out a table like this one before you start taking measurements and write your measurements into it as you go along.

Press Pause while you're doing this.

And when you're finished, press Play.

I hope that went well.

I'll show you some sample results.

If you take a look at these, there's something a little strange here.

This reading here, it doesn't fit with the other points.

The total p.

d.

is the same or almost the same in all of the other rows.

So this is likely to be an anomalous point, a measurement that we can't trust.

We can see that the p.

d.

across the lamp in this row seems normal.

It fits in with these very even steps that the p.

d.

is going up in, so it's likely that the p.

d.

across the resistor is what was measured incorrectly.

And this measurement should be done again.

If we were to sum up what the results seem to show, ignoring the anomaly, the total p.

d.

across the whole circuit didn't change, but the share of the p.

d.

between the two components did.

Now let's move on to the third part of the lesson, variable resistors in potential divider circuits.

Looking again at these results and ignoring the anomaly, the p.

d.

across both the variable resistor and the lamp always adds up to the same amount.

There's a slight difference in row three, but that could be some small experimental error and it's not enough to disprove the overall conclusion.

And this total p.

d.

is the same as the p.

d.

across the power supply.

Now what happens in the circuit shown when the resistance of the variable resistor is increased? The correct answer here is c.

The p.

d.

across the variable resistor increases.

The larger its resistance, the greater its share of the supply p.

d.

So the p.

d.

across the power supply, as we've seen, is the same as the sum of the p.

d.

across the components.

So it's the same as the sum of the p.

d.

across the variable resistor and the lamp in this circuit.

Changing the resistance of the variable resistor doesn't affect the p.

d.

across the branch.

And this is a series circuit, but you could think of it as having a single branch with these two components in it.

And that p.

d.

across the branch is constant, whatever the resistance of the variable resistor.

Here's a potential divider circuit that uses a variable resistor.

So you could think of it in a way as a variable potential divider.

The p.

d.

across the power supply is divided across the lamp and the resistor.

And increasing the resistance of the variable resistor increases the share of the p.

d.

across it and decreases the p.

d.

across the lamp.

Decreasing the resistance of the variable resistor increases the p.

d.

across the lamp.

Now a question for you.

What happens to the p.

d.

across the lamp when the resistance of the variable resistor is increased? The answer is a, the p.

d.

across the lamp decreases because it has a smaller fraction now of the total resistance.

We can also think about the current in the circuit.

The current is affected by the resistance of the variable resistor.

When its resistance increases, that increases the total resistance in the circuit, and so the current decreases.

So which of the following will cause the current through the lamp to increase? The only correct answer here is decreasing the resistance of the variable resistor.

That decreases the total resistance, so the current increases.

And now some longer written questions for you.

Take a look at this circuit.

The resistance of the variable resistor is increased from a low value.

I'd like you to explain what happens to the p.

d.

across the variable resistor and the current through it.

So not just describe, well, then you start with that, but explain why.

And then explain what happens to the p.

d.

across the lamp and the current through it.

Press Pause while you write down your answers and press play when you're ready, I'll show you some example answers.

Check that you've made the same points even if you may well have used different wording.

Question one, when the resistance of the variable resistor is increased, it has a larger proportion of the total resistance of the circuit, or you could say larger fraction, or even larger percentage.

It will have a larger share of the power supply p.

d.

across it, and so the p.

d.

across it increases.

As the total resistance of the circuit is increased and the p.

d.

across the power supply does not change, the current through the variable resistor will decrease.

And now for the lamp.

When the resistance of the variable resistor is increased, the lamp has a smaller proportion of the total resistance of the circuit.

It will have a smaller share of the p.

d.

across the power supply across it, and so the p.

d.

across it decreases.

Because the total resistance of the circuit is increased and the p.

d.

across the power supply does not change, the current through the lamp will decrease.

Well done if your answers contain the same point.

And now we've reached the end of the lesson, so here's a summary.

The sum of the p.

d.

across each component in a series circuit adds up to the p.

d.

across the battery.

The resistance of a component compared to the total resistance of the circuit is equal to the p.

d.

across the component compared to the p.

d.

across all the components.

Increasing the resistance of a variable resistor in a series circuit increases the resistance of the whole circuit and reduces the current flowing through it.

Well done for working through this lesson.

I hope you now feel that you understand what a potential divider is and how it works, and I hope to see you again in a future lesson.

Bye for now.