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Hello, my name's Dr.

George, and this lesson is called Analysing Series Circuits, Including Complex Calculations, as part of the unit: Electric Fields And Circuit Calculations.

The outcome for the lesson, which I'll be helping you achieve by the end, is, "I can use circuit rules and the equation I equals V divided by R to analyse series circuits." Here are the key words for the lesson.

I'll introduce them as we go along, but this slide is here in case you want to come back and check the meanings anytime for yourself.

There are three parts to the lesson, called Current and P.

in Series Circuits; Analysing Series Circuits; and Two Step Analysis of Series Circuits.

Let's get started.

Here's a representation of current in a series circuit.

Remember, a series circuit is a circuit that's a single loop and current is the flow of electrons around the wires and components.

And we can measure current as the amount of charge flowing past a point per second.

And as the animation shows, current is the same everywhere in a series circuit.

So it's the same in the cell as it is in the wires or in the lamp.

You could choose any point in the circuit, and the amount of charge flowing past per second would be the same.

Zooming in on what's happening in a metal when current flows, a metal consists of a lattice of ions.

That's a regular arrangement of atoms that have lost one or more of their electrons, and those electrons are able to move freely because they're only loosely held.

They're not bound to a single atom.

So here's an electron.

Here's a metal ion in this diagram.

And the electrons can drift through the metal lattice.

By the way, the model here doesn't show the charges, but the ions would be positively charged and the electrons would be negatively charged.

An electrical cell causes positive charge to build up on one side, and negative charge on the other side.

And when there's a charged object or charged part of an object, there's an electric field around that, an area of influence on other charges.

It's that field that pushes electrons around a circuit, because when you connect the battery into the circuit, there's a field then around the whole of the circuit pushing on the electrons.

In fact, the field exerts a force on anything that has charge, so it exerts a force on all the charges in the metal.

All the free electrons that are everywhere in the wire all around the circuit will move at the same time, in the same direction, due to the field.

Meanwhile, the metal ions, which are positively charged, do experience a force, but they're fixed in place so they don't move.

Here's a circuit diagram, and the current, if we measure it at the first ammeter, will be the same as the second ammeter position, and the same here, and here, and here.

It's the same everywhere.

So now, a question for you.

Look at the circuit here.

"What is the current through bulb X?" With these short questions, I'll wait five seconds, but if you need longer, press pause and press play when you've chosen your answer.

And the correct answer is 0.

15 amps, because the current through the bulb is the same as the current through the ammeter.

The current in this series circuit is the same everywhere.

Now the potential difference or P.

of a cell determines the size of the electric field around the cell.

And if you put the cell into a circuit, there's an electric field around the whole circuit, and it's that electric field that pushes charge around.

It pushes electrons around the circuit.

Here's a cell with a P.

of 1.

5 volts, and if you put two cells like that in series with each other, we'll have double the P.

, 3.

0 volts.

The electric field that pushes the charge, the electrons, is like the slope of a ramp that causes a force on a ball.

If the ramp gets steeper, the force on a ball increases, and if the P.

in a circuit is greater, the force on the electrons is greater.

And the push on the electrons due to a 1.

5 volt cell is causing a large current in this circuit.

And as we've seen, the current is the same everywhere in a series circuit, but if we increase the resistance by adding another resistor, so there's now two resistors instead of one, the field can't push any harder.

The cell hasn't changed, it's still the 1.

5 volt cell, and so the current will decrease.

So the same potential difference, the same push from the cell, but greater resistance means lower current.

And that decreased current is the same everywhere in the circuit.

So, "What happens if a resistor is added in series to the circuit shown?" Press pause if you need more than five seconds to decide.

Press play when you're ready.

And the answer is the current will decrease.

If you add anything to a series circuit that has resistance, then you'll increase the total resistance and you'll decrease the current.

But what if we add another cell? If we do that, we increase the P.

It's gone from 1.

53 volts here, and we increase the size of the electric field around the circuit.

That increases how hard the electrons are pushed, and so the current increases.

Let's check you're paying attention.

What will happen if another cell is added in series to the circuit shown? Press pause if you need more time to think.

There are two correct answers here.

If we add another cell, the push from the battery will increase.

We can call it a battery now.

It's made of two cells.

And because of that, the current will increase.

Well done if you spotted both of those.

Take a look at this voltmeter.

It seems to be measuring the P.

across the cell, but we could represent that like this, or we could draw it like this.

Perhaps this looks as though it's measuring the P.

across the lamp.

And in fact, in all three of these positions, it is measuring the P.

across the cell, and at the same time, it's measuring the P.

across the lamp.

And those two will be the same.

So we'll get the same voltmeter reading if we connect it any of those ways.

Here's something a bit more like what it would look like in real life.

We take the voltmeter, and we can move the connections anywhere in between the cell and the lamp, and notice that the reading on the voltmeter doesn't change.

So here again we have a voltmeter.

It's measuring the P.

across the cell, and at the same time the P.

across the lamp.

What if we add more components? Let's add a resistor and a motor like this.

Turns out the voltmeter reading stays the same.

It's still measuring the P.

across the cell, which hasn't changed.

And at the same time, it's measuring the P.

across all three of these components.

So in a series circuit, the total P.

across the components is the same as the P.

across the cell, but they share that P.

in a sense.

Each of them is going to have a smaller P.

than the total P.

And how that's shared is that the P.

across a component is determined by its resistance, how hard it is to push current through it.

So a higher resistance requires a higher P.

to push the same current through.

So, say this resistor had a higher resistance than the other two components, it would have the largest P.

across it.

Now, which component here has the highest P.

across it? Press pause if you need more time to think.

The correct answer is this one: the motor.

Because it has the largest resistance, it needs the highest P.

to make the same current flow through it.

Remember, the current will be the same through all these components, because they're in a series circuit.

I now have some more questions for you to think about.

In each of these series circuits, there is a missing P.

I'd like you to try to work out what that P.

should be.

So press pause for as long as you need, and when you've written down your answers, press play, and then I'll show you what the correct answers are.

So here are the answers.

In question one, there's just one component and one cell.

So the P.

across that component, a lamp, is the same as across the cell, 1.

5 volts.

In question two, the P.

s across the two components, the resistor and the motor, should add to the P.

across the cell.

So to make 1.

5 volts, we have 0.

6 volts plus 0.

9 volts.

In three, the P.

across the cell will be the sum of the P.

s across the three components, and those three add up to 3.

0 volts.

And in four we have two cells, so that makes a battery with a total P.

of three volts, and the P.

across the three components should then add up to three volts.

So the P.

across the motor is 0.

Well done if you got most or all of those.

Let's move on to the second part of this lesson, Analysing Series Circuits.

You may already know this equation.

It relates the current P.

and resistance for a component in a circuit.

So, for any component that's in a circuit, the current through it equals the P.

across it divided by its resistance.

We can write that in symbols like this, where I is the current measured in amps, potential difference is measured in volts, and resistance is measured in ohms. Let's try doing some calculations using this equation.

So in each pair of calculations, I'll show you the first, and then you try the second.

The first question: "Calculate the current through a bulb if the voltage across it is 12 volts, and it has a resistance of 40 ohms." Well, it's a good idea first to write down the quantities that we've been given, summarise those, and notice what is the quantity we're trying to find.

And then we can think what arrangement of the equation we need, and we can use the arrangement we've just seen, I equals V divided by R, substitute in the values, and write down the answer.

You could write 0.

3 amps.

30 is written here, because the calculation was made using numbers with two significant figures, so the answer here has been written with two significant figures, but don't worry about that too much.

So here's a question for you to try.

Calculate the current through a motor if the voltage across it is 10 volts and it has a resistance of 25 ohms. Press pause while you're writing out your answer.

You first of all write down the quantities, write down the equation, substitute the values in, and we do 10 volts divided by 25 ohms and we get 0.

40 amps.

But what if we know the resistance of a component and we know the current through it, but we want to know what P.

there will be across it? We'll need to rearrange the equation.

To do that, we could multiply both sides by R.

And then on the right, we have divide by R times by R.

If you divide by a number and then multiply by the same number, that doesn't do anything, so we can cross those out, and we get I times R equals V, or you might prefer to write it the other way around: V equals I times R.

When you're rearranging an equation, you don't have to write as many steps as this.

Just write as many steps as you're comfortable with.

And remember, you're just using the same skills here that you would use in a maths lesson to rearrange your formula.

So now, here's a question.

I'll show you how to solve this one.

"What is the P.

across a resistor if it has a resistance of 38.

0 ohms, and a current of 0.

40 amps passes through it?" Again, write down the quantities, write down the arrangement of the equation that we need, that's V equals I times R.

Substitute the values in, do the calculation, and we get 15.

2 volts.

And now here's one for you.

"What's the P.

across a thin wire if it has a resistance of 26 ohms and a current of 0.

50 amps passes through it?" Press pause while you're writing out your solution.

So we write down the quantities, we write down the equation, substitute the right quantity in the right place, do the calculation, and we get 13 volts.

And what if we want to measure the resistance? We might actually not know the resistance of a component.

We put in a circuit, we measure the current through it, we measure the P.

across it, and we can use that to find its resistance, but we'll need to rearrange the equation again.

Let's multiply by R again, and that cancels out.

And now, let's divide by I, because if we do that, then on the left we are both multiplying by I and dividing by it, and that has no overall effect.

We can say those cancel out, and we're left with R equals V divided by I.

So now a question: "What is the resistance of a motor if the voltage across it is 24.

0 volts, and the current through it is 0.

70 amps?" R is V divided by I, that's the arrangement of the equation we need.

Substitute in 34.

3 ohms. And one for you.

"What's the resistance of a thermistor if the voltage across it is 8.

0 volts, and the current through it is 0.

20 amps?" Press pause while you're writing your answer.

So we start by writing the quantities down.

The arrangement of the equation that we need, R equals V divided by I.

Substitute in the known quantities, and calculate the resistance.

We get 40 ohms. Well done if you've been getting these right.

And now, one more question about the equation.

Which equation is correctly arranged to find the missing quantity in the circuit shown? Press pause if you need more than five seconds.

And all of these equations are correct arrangements of the equation, but the one that's useful here is I equals V divided by R, because the current in the circuit is the one thing that we don't know.

And now, here's a set of questions for you.

So I'd like you to try these.

Write out your answers using the techniques I've shown you.

Write the quantities, write the equation in the arrangement that you need, then substitute in the numbers and calculate.

Take as long as you need to do that.

Press pause while you do, and press play when you're finished.

So let's check the answers to these questions.

Light bulb has resistance 2.

5 ohms. What's the current if the P.

across it is 7.

5 volts? We need the I equals V divided by R arrangement of the equation.

We substitute in the values, we get 3.

0 amps.

If you just put three amps, that's okay.

Question two, "The P.

across a motor is 12 volts.

If its resistance is 24 ohms, what's the current through it?" Again, we need I equals V divided by R.

Substitute in and we get 0.

5 or 0.

50 amps.

Question three.

This time we're asked for resistance.

We know the current is 0.

50 amps, the voltage is 4.

5 volts.

We substitute those into the equation, we get 9.

0 ohms. We're not actually told the potential difference across the light bulb here, but if the light bulb on the battery are the only things in the circuit, then the P.

across the light bulb is the same as the 4.

5 volts across the battery.

Question four, we have a buzzer.

It's just another type of component.

Has a resistance 7.

5 ohms. We want a current of 0.

20 amps to flow through it.

What P.

is needed? So now we need the V equals I times R arrangement of the equation.

And when we substitute in, we get the answer 1.

5 volts.

And finally, we have two 1.

5 volt cells connected into series.

So we need to find the total P.

across the cells.

We can just do that little calculation first.

And then, we know the current through a motor.

The resistance of the motor, it's going to have the P.

across it, the same as across the cells.

0 volts, divide that by the current, 0.

30 amps, and we get 10 ohms resistance.

Well done if you're getting these right.

And now, the third part of this lesson: two step analysis of series circuits.

So we're going to be answering some longer questions where we have to do a couple of steps to find out something about a circuit.

So, as we've seen, the current is the same everywhere in a series circuit.

If it's 0.

14 amps here, then it also is here, and here, and here, and here, and here.

And that's a useful thing to keep in mind when we're solving problems to do with circuits.

So the position and resistance of each of these components is different, but the current is always the same.

Let's see how we can use that.

The lamps in this circuit are identical.

Which of the following statements about the current through the lamps is correct? Press pause if you need more than five seconds, and press play when you've chosen your answer.

The correct answer is C: lamp three has a current of 0.

3 amps.

It's true, because the current is the same everywhere in a series circuit.

The other two are not true.

You don't divide the current by three to get the current in each lamp.

Each lamp has 0.

3 amps.

It's the same everywhere.

The sum of the P.

across components in a series circuit equals the P.

across the battery.

Here's an example.

So we have one volt plus three volts plus two volts across those components, adds up to the six volts across the battery.

So look at this circuit.

What is the P.

across the resistor? And the answer is 1.

1 volts, because the P.

s across the three components should add up to the 1.

5 volts across that cell.

Here's a way you could calculate that.

So in a circuits question, values of current, P.

and resistance may not be given for a component like the lamp below, but can be found using the values for other components.

So, let's say we're asked for the resistance of that lamp.

At the moment, we don't seem to know anything about it.

If we want to know the resistance, we can calculate that from the P.

and the current.

But those aren't shown.

Or are they? Well, the current in the ammeter is 0.

15 amps.

So the current in the lamp will also be 0.

15 amps.

What about the P.

across the lamp? Well, the P.

s across the two components should add up to the total P.

of the cells.

And that gives us a P.

across the lamp of 1.

2 volts.

Remember the ammeter is just a measuring instrument, and it doesn't have a P.

across it.

And now, if we want the resistance of the lamp, we can find it.

It's V divided by I, which gives us 8.

0 ohms. So you can see this is a multi-step problem if we're asked to find the resistance of the lamp, but we can find it.

As we've seen, the P.

across each component depends on the resistance of that component, and we can use that to work out missing values.

If the lamp here has a resistance of two ohms, the resistor must have double that resistance, four ohms, because it has double the P.

across it.

And remember, the greater the resistance, the greater the P.

needed to make the same current flow through these components in series.

So what's the resistance of the motor here, the symbol with the M inside it? Well, the motor has three times the P.

across it of the lamp.

So, that must be because it has three times the resistance.

So the motor has resistance six ohms. Well done if you spotted that.

"And what's the total resistance of these three components?" The answer is 12 ohms, and here's why.

The resistance of the lamp is two ohms with a P.

across it of one volt.

The total P.

here is six volts, so the total resistance must be six times the resistance of the lamp, to have six times the P.

across it.

So six times two: 12 ohms. Well done if you've got that.

You can find out information about resistors in a circuit from the rest of the components sometimes.

So here, we have three identical resistors, and there's also a lamp.

The current through the resistors must be 0.

2 amps, because that's the current through the ammeter.

The P.

across the resistors must be the P.

across the cell minus the P.

across the lamp, 0.

9 volts.

So we can work out the resistance of the three resistors altogether from the current through them and the P.

across the three of them.

So R is V divided by I, and we get to their total resistance is 4.

5 ohms, but they're identical.

So each one of them must have a resistance of 4.

5 divided by three: 1.

5 ohms. Again, a multi-step problem.

Now, "The lamps in this circuit are identical.

What is the resistance of one of the lamps?" Press pause while you work this out, and press play when you're ready.

The correct answer is five ohms. I'll show you why.

The resistance of all the lamps altogether is the P.

across all of them divided by the current, and the P.

across them must be 2.

0 volts.

It's the same as the P.

across the cell.

There's no other components in the circuit.

So we get a total resistance of 20 ohms. They're identical lamps, so each one has the same resistance.

So resistance of each one must be 20 divided by four, which is five ohms. And here are some multi-step circuit problems for you to try.

Write down your answers step by step, press pause while you're working these out, and press play when you're ready to check your answers.

Now I'll show you step by step how to answer each of these.

First we need to find the resistance of this lamp, and to do that, we want to know V and I, so we can use the equation.

The P.

across the lamp is the P.

across the battery minus the P.

across the motor.

That gives us 2.

4 volts.

The current through the lamp is the same as in the ammeter, 0.

60 amps.

And now, we can use R equals V divided by I to work out the resistance of the lamp, which is 4.

0 ohms. Question two: we have two buzzers with the same resistance.

What's the resistance of each? We can work out their resistance from V divided by I, and the voltage across the two buzzers together must be the same as across the battery.

It's 3.

0 volts.

Divide that by the current.

It's the same everywhere in the circuit, 0.

40 amps.

And the resistance of the buzzers together is 7.

5 ohms. If they're identical, they must have the same resistance, so half each of that, which is 3.

8 ohms if we round it to two significant figures.

In question three, we need to find the resistance of the motor.

We can first find the P.

across the motor.

It's the P.

across the battery minus the P.

s across the bulb and the buzzer.

We get 1.

8 volts.

The current is 0.

09 amps, the same as in the ammeter, and resistance is V divided by I.

And using those values, we get 20 ohms. The P.

across the resistor here, and we don't actually know the potential difference across the power supply, but we do know that the resistor has twice the lamp resistance.

We don't know the P.

across the lamp either, but we could work that out.

It's V equals I times R.

The current in the lamp is the same as the current in the ammeter, and we know the resistance of the lamp, so we get 3.

0 volts across the lamp.

Now, if we know the resistor has twice the resistance of the lamp, it will have twice the P.

that's across the lamp.

So it'll be twice 3.

0 volts.

There's six volts across the resistor.

That was a bit of a harder question.

You may find that you need more practise with this type of question.

It's not always easy at first.

It's a kind of puzzle to solve.

"How do I find out what I need to be able to work out what the question is asking me for?" But if you practise, you should find you get better at these.

And now, we've reached the end of the lesson, so I'll give you a summary.

"The current in a series circuit is the same everywhere in the circuit.

The P.

of the battery equals that across the components in the circuit.

The current in a series circuit depends on the P.

and the resistance in the circuit, according to the equation: current equals P.

divided by resistance, or I equals V divided by R.

This equation can be rearranged so that either V or R is a subject to the equation.

If two of the three quantities, I, V or R are known, the remaining one can be algebraically found." So, well done for working through the whole lesson.

There was a lot there to think about, and I hope I'll see you again in a future lesson.

Bye for now.

I've finished the video