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Hello, my name's Dr.

George.

This lesson is called Calculating with Electric Charge and it's part of the unit Main's Electricity.

The objective for the lesson is I can explain the equations I equals Q divided by t, V equals E divided by Q, P equals I times V, and V equals I times R, and use them in calculations.

I'll be using these keywords during the lesson, and you can come back to this slide anytime if you need to check their definitions.

The lesson has three parts.

They're called electric charge and current, electric charge and potential difference, and calculations with electric charge.

This lesson is all about electric charge, and by the time we've done some thinking about electric charge, you'll be able to understand why power is current times potential difference in the equation P equals I times V.

Remember the units here.

Power is measured in watts, current in amps, and potential difference in volts.

So we'll come back to that equation later in the lesson.

Now, what do we know about charge? It's a property of materials, it comes in two types, negative and positive, and it can be measured, it's a quantity.

It's measured in units of coulombs.

When an electric current flows through a metal wire, negatively charged electrons move through a lattice, a repeating regular arrangement, of positively charged ions.

Metal ions are atoms that have lost one or more of their electrons.

They're arranged in this regular pattern and they're vibrating all the time, but held in fixed positions.

They can't swap places with each other.

The free electrons, the ones that come from outer shells of metal atoms, are called free electrons because they can move from ion to ion, and they're the reason why current through a metal is possible.

That flow of charge happens because electrons can flow through the metal.

So which of the following is the unit of electric charge? 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've chosen your answer.

And the unit of electric charge is the coulomb, symbol capital C.

The charge on a single electron is really tiny.

In fact, it takes the charge on this many electrons to add up to a total charge of just one coulomb.

When we measure a current of one amp, it means that one coulomb of charge is flowing past a point each second.

So this ammeter in that situation reads one amp.

If two coulombs of charge flow past in two seconds, the current is still one amp because two coulombs in two seconds is equivalent to one coulomb per second.

If a charge of 10 coulombs flows past in four seconds, well, how many coulombs per second is that? It's 2.

5 amps.

5 coulombs per second would come to 10 coulombs in four seconds.

Now a question for you.

What is the current when 60 coulombs of charge flows past in 12 seconds? And the correct answer is 5.

0 amps.

Because the current is the amount of charge per second, 60 coulombs in 12 seconds means there must be five coulombs every one second.

And so to calculate current from charge and time, we divide the charge by the time.

That gives us the number of coulombs per second.

We can write the equation I equals Q over T where as usual current has a symbol I, is measured in amps, and we use a symbol Q for charge, and that's measured in coulombs, and time, T, is measured in seconds.

If we want to work out the amount of charge passing a point in eight seconds and there's a current of 1.

5 amps, we could rearrange the equation I equals Q divided by t and then put in the numbers, or we could just use our understanding of what current means.

A current of 1.

5 amps means 1.

coulombs of charge is flowing past each second, so that's eight lots of 1.

5 coulombs in eight seconds, and eight times 1.

5 coulombs is 12 coulombs.

A similar question for you.

How much charge flows through a washing machine in 30 seconds if the current it uses is 3.

0 amps? And the correct answer is 90 coulombs because three amps is three coulombs per second and there are 30 seconds, so three coulombs every second 30 times, three times 30 is 90 coulombs.

Well done if you got that.

Now, some pupils are discussing what they think current is.

Aisha says, "It is the amount of energy transferred in one second." Sofia says, "Current equals charge times time.

I equals Q times t." And Sam says, "Current is the rate at which charge flows in a circuit." So I'd like you to think about which pupil is correct and how you would improve the incorrect statements to help all of the pupils better understand current.

Press pause while you're thinking about this and press play when you have your answers ready.

Let's take a look at the answers.

Sam is correct.

Current is the rate at which charge flows in the circuit because it's charge per unit time, coulombs per second.

How can we correct the statements made by the others? Well, Aisha's could be changed to current is the amount of charge that flows past each second.

That's another way of saying the rate of flow of charge.

So current is equal to the total charge divided by the number of seconds.

So Sofia's equation is wrong.

Current should be charge divided by time.

Let's move on to the second part of the lesson.

Electric charge and potential difference.

In these two circuits, the current is the same.

If we go from the left hand circuit to the right hand circuit, the number of cells is doubled, so that doubles the p.

, and the number of bulbs is doubled, so it doubles the resistance, and those two effects cancel out to give us the same current.

So in these circuits, the same amount of charge flows round each second.

But as we've said, the p.

across the battery is twice the p.

across the cell in the first circuit.

So in the second circuit, each coulomb of charge transfers twice the amount of energy.

The p.

across this lamp is three volts.

And what that means is each coulomb of charge passing through the lamp will transfer three joules of energy to it.

So you may previously have heard of p.

as being something to do with the push by the cell or battery on the electrons, and that is a way that you can think about it.

But more precisely, the p.

d across a component is the amount of energy transferred to it by each coulomb of charge that passes through it.

Here the p.

across the kettle is 230 volts.

This is a mains circuit, and that means each coulomb of charge passing through the kettle will transfer 230 joules of energy to it.

Now a question.

What does the energy transferred by each coulomb of charge depend on? The correct answer is b, the potential difference across the appliance.

The current is the number of coulombs of charge passing through per second.

That doesn't tell us anything about how much energy each coulomb transfers.

The power rating is how much energy is transferred per second.

But again, that doesn't tell us how much energy is transferred per coulomb.

When we measure a p.

across a component, if we find it's one volt, we know that one joule of energy is transferred to the component by each coulomb of charge passing through it.

If we measure 3.

50 volts, it means that 3.

5 joules of energy is transferred to the component by each coulomb passing through.

If four joules of energy is transferred to an appliance when two coulombs of charge flows through it, the p.

across it is two volts because four joules transferred by two coulombs means two joules transferred by each coulomb, so two volts.

If 30 joules of energy is transferred when five coulombs of charge passes through, the p.

across it, well, it's how many joules are transferred by each coulomb, which must be six.

Five coulombs each transferring six joules, total energy transferred five times six is 30 joules.

So to calculate potential difference from energy and charge, we use this equation.

Potential difference is energy over charge, which we can write as V equals E over Q.

So potential difference is how many joules transferred per coulomb.

So in the equation, we need p.

to be measured in volts, energy in joules, and charge in coulombs and then this relationship works.

And a question for you.

What is the p.

across a lamp if 48 joules of energy is transferred to it by 12 coulombs of charge? Press pause if you need more time to work out your answer.

The correct answer is 4.

0 volts.

And the reason is that the p.

is the number of joules transferred per coulomb, so we can work out the p.

by the total energy transferred divided by the amount of charge that passed through.

48 joules divided by twelves coulombs is four volts, and we can give two significant figures because there were two significant figures in the quantities we used to do the calculation.

Now here's a longer question for you.

Bulb A is from a lamp that operates on mains electricity, while bulb B is from a small torch that runs on a three volt battery.

And the pictures also show the p.

and current for each of these bulbs.

Now compare the amount of charge that passes through each bulb in a minute and compare the amount of energy transferred to each bulb in a minute.

Press pause while you write your answers to these questions and press play when you're ready to check them.

Now let's take a look at the answers.

The same amount of charge passes through each bulb in a minute because the current through each one is the same.

So no calculation was needed there.

They both have a current of 0.

2 amps, that's 0.

2 coulombs flowing through per second.

Much more energy is transferred to bulb A in a minute because the p.

across it is greater, meaning more energy is transferred by each coulomb of charge that passes through it.

So bulb A, 230 joules is transferred to it by each coulomb, but bulb B, three joules transferred per coulomb.

So the same number of coulombs passes through per second, but each of those coulombs transfers more energy to bulb A.

In fact, 73 times more energy because 230 joules per coulomb is 73 times three joules per coulomb.

Well done if you've got those questions right.

Now let's take a look at some calculations with electric charge.

There are several different equations that apply to electric circuits.

The ones that include electric charge can be worked out if you know definitions.

So current by definition is the amount of charge flowing past a point per second, and so the equation is current is charge over time.

The definition of potential difference is that it's the energy transferred per coulomb of charge, so potential difference is energy transferred divided by the amount of charge that passes through.

Now I'm going to show you how to answer a question using one of those equations and then I'll ask you to try a question.

2 coulombs of charge flows round a circuit to ring an electric bell.

If the bell rings for 0.

40 seconds, what is the average current? Now, don't worry about the fact that it's average, that's to allow for the fact that the charge might not flow round at a steady rate, but we can find the average current using the equation I equals Q over T.

The total charge that flows divided by the time it takes.

And substituting in the quantities, we get 3.

0 amps.

And it's appropriate to write that to two significant figures because the quantities we used to work it out were known to two significant figures.

And now here's a question for you.

27 joules of energy is transferred to an electric motor in a circuit.

If 6.

0 coulombs of charge flows through the motor, what is the p.

across it? Press pause while you write out your working, and press play when you're ready to check your answer.

Well, this is the appropriate equation to use, V equals E over Q.

We know the energy transferred is 27 joules, the charge transferred is six coulombs, and we get a potential difference across the motor of 4.

5 volts.

So we've been able to work out the p.

across the motor by knowing the energy transferred to it and the amount of charge that flowed through while that energy was transferred.

Now, power is the rate at which energy is transferred, or in other words, the amount of energy transferred each second.

So we can work it out by dividing the energy transferred by the time taken.

So in this equation, P equals E over t, power is P measured in watts, energy, E, is measured in joules, and time, t, is measured in seconds.

And those are the units you need to use if you're using this equation.

We can combine equations to calculate power in electric circuits.

So if we start with I equals Q over t, current is charge over time, and V is E over Q, potential difference is energy over charge, we could take I equals Q over t and multiply both sides by t.

And if we do that, the ts cancel out on the right hand side and we get Q equals It.

Now let's substitute that into the equation V equals E over Q.

So we get V is E over I times t.

If we multiply both sides by I now, we get IV is E times I over I times t.

We can cancel the I on top and the I on the bottom, so we get E over t is I times V.

But E over t is power, rate of energy transferred, so we can now say that P is I times V.

So that equation we mentioned at the beginning for power, here it is, and we've shown why it's true by thinking about relationships to do with charge.

Now, which of the following equations are correct? P equals IV, is it correct? We've just seen that.

And V equals E divided by Q.

And we've also been using I equals Q divided by t.

All of these are correct equations.

And they're not simply rearrangements of the same equation.

They are three different equations relating different sets of quantities.

Now, some of the equations used with electric circuits aren't obvious from the definitions of the quantities.

So we know power is amount of energy per second transferred, so the equation P equals E over t could be seen as just another way of stating that fact, that definition.

But P equals I times V isn't obvious from the definition of power.

We also know current is the amount of charge that flows past a point per second.

And so I equals Q over t is really a restatement of that definition in a mathematical statement.

But I equals V over R isn't obvious, even if you know the definition of current.

Now I'm going to show you how to answer a question using one of these equations and then give you a question to try.

The p.

across an electric motor is 3.

0 volts and the current through it is 1.

6 amps.

How much energy does the motor transfer every second? First we need to realise that what we're being asked for is the power of the motor, the energy transferred per second.

And we're given the p.

, V, and the current, I.

So a suitable equation is P equals I times V.

Multiply the current by the potential difference and we get 4.

8 watts.

And so the energy transferred per second is 4.

8 joules each second.

Now here's a question for you.

The p.

across a lamp is 4.

5 volts and its resistance is 2.

0 ohms. How many coulombs of charge flow through it each second? Press pause while you write out your answer, and press play when you're finished.

And it helps here to realise that if we're being asked for the number of coulombs per second, then that's the current.

And we're given V and R, and we can calculate current from V divided by R.

Substituting in the numbers we get 2.

3 amps, so 2.

3 coulombs flowing through the lamp each second.

I'll show you another example question.

What is the power of a lamp that has a resistance of 300 ohms when a current of 0.

60 amps flows through it? Well, we're given R and we're given I, and we don't have an equation at the moment to work out P directly from R and I, but we do know that P is I times V, we don't have V, but we do have the relationship between V, I, and R.

And we have I and R, and we can calculate V by multiplying them together.

So the potential difference is 0.

60 amps times 300 ohms, and the power is the current times that, as shown here, and we get 108 watts.

We really should round that to two significant figures because we only have the current to two significant figures, 110 watts.

Now here's a question for you.

What is the power of a doorbell that has a resistance of 60 ohms when a current of 0.

2 amps flows through it? Press pause while you write out your answer to this.

Now, this is similar to the example.

We're given the resistance and the current and we want the power, and we can use these two relationships, P equals IV and I is V over R.

So from the second equation, V is I times R, and from the first equation, P is I times that, so we have the current times the current times resistance, and we get 2.

4 watts.

I hope you're starting to see how to do these.

And I have some more practise for you.

I'd let you to try answering these questions and show all of your working out.

So press pause while you're doing this and press play when you're ready to check your answers.

Now let's have a look at how to answer these.

You were told that in a minute, 108 coulombs of charge flows through an electric motor and asked for the size of the current.

Well, we have to remember that when we're using t in any of these equations, we need the time in seconds.

So one minute is 60 seconds, and I is Q divided by t, so we can work out the current by 108 coulombs divided by 60 seconds.

We get 1.

8 amps.

In question two, we're given the energy transferred to a lamp when an amount of charge flows through it, and we want the p.

And there is an equation that relates to those three quantities, V equals E divided by Q.

We don't need to rearrange it.

We can substitute in E and Q from the question, and we get 4.

5 volts.

Now we're given the p.

across an electric heater and the current through it, and we want the energy transferred every second.

Notice that energy transferred per second is power, so the question is asking us for the power of this heater.

And an equation that relates power to current and p.

is P equals I times V.

Substituting in, we get 1,104 watts, which is the same as 1,104 joules per second.

Although really we should write that to two significant figures because we were only given the current to two significant figures.

Question four.

The p.

across a lamp and its resistance are given, and we're asked for the current.

And if we remember, there is an equation that relates those three quantities, I equals V divided by R.

So we can substitute in, and we get I as 0.

46 amps.

And finally, what is the power of a lamp when we're given the resistance and we're given the current through it? And we don't have an equation that directly relates P, R, and I, but we have these two equations, P equals IV and I equals V over R.

We can rearrange the second equation to find that V is current times resistance and the power is current times V, so current times current times resistance, and we get 11 watts.

So well done if you've got some or all of these right.

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

Current is equal to the amount of charge passing a point per second, and it can be calculated using I equals Q over t or I equals V divided by R.

Potential difference is equal to the energy transferred per coulomb of charge that passes through, and it can be calculated using V equals E divided by Q.

Power is equal to the amount of energy transferred per second, P equals E divided by t, or in an electric circuit, P also equals I times V.

And it's important that you remember the symbols for the quantities and the units to use in these equations.

Time, t, is in seconds.

Energy, E, in joules.

Current, I, in amps.

, V, in volts.

Power, P, in watts.

Charge, Q, in coulombs.

And resistance, R, in ohms. So well done for working through this lesson, and I hope to see you again in a future lesson.

Bye for now.

I've finished the video