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Hello, my name's Dr.

George, and this lesson is called Analysing Parallel Circuits.

It's part of the unit Electric Fields and Circuit Calculations.

Here's the outcome for the lesson.

I can use the rules for current and potential difference and the equation I equals V divided by R to analyse parallel circuits.

And here are the keywords for the lesson.

You probably know most of them, but you can come back to this slide anytime if you need to remind yourself of the meanings.

The lesson has two parts, current and p.

d.

in parallel circuits and current, p.

d.

, and resistance in parallel circuits.

This circuit has three parallel branches.

That's because you can see three complete loops that each contain the cell, and the current through the cell equals the sum of the current in each of those three branches.

So if you know the current in each branch, you can work out the current through the cell just by adding them together.

Or if you know the current in the cell and in two of the branches, you can work out the current in the other branch.

So here's a question about that.

What is the current through the battery in this circuit? For short questions like this, I'll wait for five seconds, but if you need longer, you can just press pause and press play when you've chosen your answer.

The current in each of those branches will flow through the battery.

So the total current in the battery is.

34 amps, the sum of these two.

Now, it might have looked strange to you that the battery is in the middle of this diagram, it's often shown at the top, but we still have two complete loops, each of which contains the battery and two other components.

So this is a parallel circuit with two branches.

Now, in this circuit, what is the current through the lamp? The correct answer is.

08 amps.

And that's because the currents in the three branches add up to the.

26 amps through the cell.

The value of the current through each component depends on the resistance of the component and the p.

d.

of the cell.

In this circuit, there are three different components, but they all happen to have the same resistance, so they will have the same current through them.

So same resistance, same p.

d.

means you'll get the same current.

Now, which of the following statements is correct? Take a look at the circuit.

The lamps each have a of three ohms. The resistor has two ohms and the motor has resistance one ohm.

Press pause if you need more time to think.

And the answer is that the current is the same in branches one and two because if you add up the resistances of the components in each branch, you'll find that it's six ohms both times.

Now, the p.

d.

across each parallel branch in a circuit is the same, but the p.

d.

in a single branch is shared between the components in that branch.

So, for example, in this circuit, there's 1.

5 volts p.

d.

across the cell, which means there will be 1.

5 volts across these components altogether and 1.

5 volts across these components altogether.

The p.

d.

is shared between the components within a single branch.

Here we have two identical lamps in the branch, so they will share the 1.

5 volts, and so they'll each get.

75 volts, half of the p.

d.

that's across the whole branch.

So which of the following rules are the best to use to find the p.

d.

across the motor in this circuit? Press pause while you choose your answer and press play when you're ready.

So it's useful to know rule a, the p.

d.

across each parallel branch in a circuit is the same.

That tells you that the p.

d.

across the branch with a motor in it is 1.

5 volts, but it's also useful to know that the p.

d.

in a single branch is shared between the components.

That tells you that some of that 1.

5 volts will be across the motor and some will be across the resistor.

In fact, it's possible to work out how much is across each 'cause they have the same resistance, so they'll have half of the 1.

5 volts across each of them,.

75 volts.

C is a correct statement, the current through the cell is equal to the sum of the currents in the branches, but it's not useful here.

We don't actually know any of the current in this circuit, so we can't use that to help us.

And now another question.

What is the p.

d.

across lamp one in this circuit if we know that the lamps each have a resistance of three ohms, the resistor has two ohms, and the motor one ohm? The correct answer is.

75 volts, and let's see why.

Lamp one has resistance three ohms, which is half the resistance in its branch.

So it has half the 1.

5 volts p.

d.

across it.

Well done if you got that.

And now some more for you, so press pause while you're working on these, write down your thinking, and press play when you're ready to check your answers.

Now, let's take a look at the answers.

Circuit one on the left.

The missing current is 1.

8 amps because the current in the two branches add to equal the current in the cell, which is three amps.

In circuit two, we want to know two of the missing currents.

Well, the six ohm resistor will have half the current that the three ohm lamp has, twice the resistance, half the current.

So the six ohm resistor has current.

25 amps through it.

The current through the ammeter will be the same as the current through the cell, and that's the sum of the current in the branches, which makes.

75 amps.

And now for the third question, we need to find the p.

d.

across lamp one and across each of resistors two and three.

And we know that the lamps are identical to each other and the resistors are identical to each other.

Looking at lamp one, it will have the same p.

d.

across it as the identical lamp that's in series with it.

They're in series, so they'll have the same current, they've got the same resistance, so they're going to have the same p.

d.

So the p.

d.

across lamp one is.

8 volts.

Then we can find the p.

d.

across resistor two because the two lamps on that resistor share the two volts of the cell.

So the p.

d.

across resistor two is two volts take away the.

8 volts across each lamp, giving us.

4 volts.

Now, the resistors have half the resistance of the lamps, they must do because they have half of the p.

d.

of the lamps across them.

The p.

d.

across the middle branch is going to be shared in the ratio two to one to one.

Now, just think about that.

The p.

d.

across each component in series is proportional to the resistance of that component.

It's double if the resistance is double, for example.

So here, the ratio of the resistances is two to one to one, lamp to resistor to resistor.

And so the ratio of the p.

d.

is also two to one to one.

That means that that total p.

d.

of two volts is shared into four shares and the lamp gets two of those shares and the resistors each get one.

Dividing two volts into four shares gives us.

5 volts.

And so resistor three has.

5 volts across it.

Well done if you got many or most or even all of these right.

And if you're finding them difficult, you might want to go back and have a look again at how they were solved.

Now let's move to the second part of this lesson, current, p.

d.

, and resistance in parallel circuits.

If you add a resistor in parallel to a circuit, that allows more current to flow as there's an extra route that current can flow down.

And so this actually causes a decrease in the total resistance of the circuit.

We've added a resistor and yet we've decreased the resistance.

We have made it easier for current to flow because we've added a branch that wasn't there before.

We can also see that the current in the cell is greater than it was before.

Now, which of the following will cause the total resistance in the circuit to decrease? Adding a resistor in parallel to the cell, adding another branch with a resistor to the circuit, or adding a resistor in series with the cell? Press pause if you need some time to think.

Well, adding a resistor in parallel to the cell is adding another branch into the circuit, it's giving another route for current to flow through, another loop for current to flow around.

Also, adding another branch with a resistor to the circuit anywhere in the circuit if you add another branch, you've given current another route to flow around.

A resistor can be added in parallel to another component.

So this resistor, instead of being in parallel with both of those lamps, it's just in parallel with one of them.

And the resistance of that section of the circuit decreases because there are now two ways for current to flow through it.

So this also causes a decrease in the total resistance of the circuit.

Again, we've added a resistor, but because we've added it in parallel, we've decreased the overall resistance.

Now, what happens to the resistor combination when another identical resistor is added to a first resistor in parallel in a circuit that is running? So you can see in the diagrams what's happened.

Does the resistance double and current double? Does the resistance halve and the current double? Does the resistance double and the current halve, or does the resistance halve and the current halve? Press pause while you're thinking about that, and press play when you've chosen your answer.

The correct answer is the resistance halves and the current doubles.

We now have added an identical root because these resistors are identical for current to flow through.

So current will continue to flow through the first resistor, the same size current, and the same size current again will flow through the second parallel resistor.

So the current has doubled and that means the resistance overall must have halved.

If the p.

d.

and current in a branch are known, the resistance can be calculated using the equation I equals V divided by R, that's current is potential difference divided by resistance.

We can rearrange to find the resistance to R equals V divided by I.

To rearrange, you use the same skills that you would use to rearrange your formula in maths.

And substituting in the values that we know here, we know that the potential difference across the resistor is 1.

5 volts, the current through it is.

6 amps, and we get 2.

5 ohms. Now, you may think there seems to be a parallel branch here containing a volt metre, but actually a volt metre doesn't allow current to flow through it.

So you can think of the voltmeter as looking into the circuit from the side without actually becoming part of the circuit.

The ammeter also doesn't affect the circuit, it has no significant resistance.

So this circuit is simpler than it looks.

It is just a cell in series with a resistor and with two measuring instruments looking on.

Now, I'm going to show you an example question how to solve it and then I'll ask you a similar question.

So in this circuit, what is the resistance of the resistor? Well, let's look at what we know about the resistor.

We know that we could use the equation, I equals V divided by R rearranged to find R if we know V and I.

Well, we do know the current through the resistor is shown next to the resistor and the voltage across the resistor is the same as across the cell because the resistor is in parallel with the cell and it's not sharing that p.

d.

with anything else in its branch.

So we write down the quantities we know and then we write down the equation in the arrangement that we need to find the missing quantity, R equals V to row by I, substitute in the values and calculate, and we get 2.

3 ohms. So now here's a question for you to try.

What is the resistance of the motor in this circuit? Press pause while you're writing your solution and press play when you're finished.

So, first of all, write down what you know.

We do know the potential difference across the motor.

It's the same as the potential difference across the cell.

It's not sharing that p.

d.

with anything else in its branch.

The current is shown, so we need to find R.

We'll use R equals V divided by I, substitute in the values, and we get 30 ohms. So well done if you got that.

Now, what about if we know the p.

d.

across a component and we know its resistance, but this time we don't know the current in it, but we have a diagram of the circuit that it's in? We can use I equals V divided by R.

This time we don't need to rearrange it.

And so we substitute here the p.

d.

across this resistor, the resistance of the resistor, and we find the currents through the resistor.

Again, I'll show you an example step by step and then you can try one.

So what's the current through the lamp in this circuit? Write down the quantities we know, the potential difference across the lamp, same as the cell, 1.

6 volts.

We know its resistance too.

So write down the equation in the form you need for finding I, current, substitute in the known values and we calculate a current of.

08 amps.

Now one for you to try.

So again, write down the quantities we know and the quantity we're looking for.

Use the equation in the right arrangement to find I, the current.

Do the calculation,.

3 amps.

And of course, there's one other thing we can use this equation for.

We might know the current through a component when it's in a particular circuit and the resistance of the component, and we want to know the p.

d.

across the component when it's in that circuit.

This time we rearrange the equation to V equals I times R.

And for this motor the current is shown as.

4 amps.

The resistance is 6.

0 ohms. So we calculate a p.

d.

across that motor of 2.

4 volts.

And another pair of examples.

I'll show you one and you try the next one.

In this circuit, what is the p.

d.

across the resistor? Well, we can find V if we know I and R, and we do.

The current in the resistor is the current in the ammeter,.

4 amps, and the resistance of the resistor is shown as 10 ohms. So we use the equation in this arrangement to find the p.

d.

across the resistor, which is 4.

0 volts.

Now you try this one.

So again, we write down the quantities that we know and the one we're trying to find.

We write down the equation we're going to use, substituting the values that we found by looking at the diagram, and we calculate 15 volts.

Well done if you've got most or all of these examples right.

And if you didn't, you might want to go back and look at the solutions again.

And now a short question, which of the following equations is correct? The only correct one here is V equals I times R.

It can be arranged in two other ways to make I subject and R the subject, but these arrangements in a and c are not the right ones.

And now here are some longer questions for you.

I'd like you to calculate the missing current p.

d.

or resistance values in each of these circuits.

Press pause while you're working on this and press play when you're finished.

So let's take a look at the solutions.

Circuit one.

The resistance of the resistor can be worked out like this.

The potential difference across it is the same as the p.

d.

across the cell, 4.

0 volts, because it's not sharing that p.

d.

with anything else in its branch.

The current is shown as.

2 amps, and we use the equation R equals V divided by I, and we get 20 ohms. The current through the lamp.

Again, we're going to use the equation.

We can work out the current through the lamp if we know V and R for the lamp, which we do.

The lamp, again, doesn't share the p.

d.

of the cell with anything else in its branch, so it has four volts across it, and its resistance is shown.

So I equals V divided by R and we get.

4 amps.

Now let's look at circuit two.

We have three currents to find.

Let's look at the current through the lamp.

The p.

d.

across the lamp is three volts.

It's the same as the p.

d.

across the cell, and the resistance is shown as five ohms, and current is V divided by R.

We get.

6 amps.

Current through the resistor.

We can do exactly the same kind of calculation, but with a different resistance and we get.

4 amps.

Now we can work out the current through the cell, which is the sum of the currents in the two branches.

So we get 1.

0 amps in the cell.

And circuit three, we have two missing currents and a missing potential difference.

First of all, the p.

d.

across the cell is the same as the p.

d.

across each branch.

And it's useful to know that because we can find the p.

d.

across the middle branch.

We know enough about that lamp to find the p.

d.

across it.

We know the current in it, we know its resistance.

So the p.

d.

across it is I times R and we find that it's 1.

6 volts, so there must also be 1.

6 volts across the cell.

What about the current through the resistor? Well, we do know the p.

d.

across the resistor, it's the only component in its branch, so it has 1.

6 volts across it, the same as the cell.

Its resistance is shown.

So we can use I equals V divided by R to find the current through it, and we get.

08 amps.

And finally, the current through the cell is the sum of the currents in the two branches.

So we get.

28 amps.

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

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

The current through a cell or battery in a parallel circuit is equal to the sum of the current in the branches.

The p.

d.

across the components in a branch of a parallel circuit is shared, depending on the resistance of the components and adds up to the p.

d.

across the cell or battery.

Values of current, p.

d.

, and resistance can be found using the equation I equals V divided by R and the rules for current, p.

d.

, and resistance.

Well done for working through this lesson.

There's some quite challenging ideas in there and you have to use problem solving to puzzle out the answers to some of these circuits questions.

You may find that you need a bit more practise to really get confident at these.

I hope to see you again in a future lesson.

So bye for now.