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Hello, my name's Dr.

George, and this lesson is called Adding Components to a Parallel Circuit.

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

The outcome for this lesson is I can use the rules for current and potential difference in a parallel circuit and explain the effects of adding a component.

Here are the key words, and you can come back to this slide anytime if you want to remind yourself of the meanings.

The lesson has two parts called Adding Components to Parallel Circuits and Rules for Parallel Circuits.

So let's start.

When a lamp is connected to a battery, extra electric charge builds up on either side of the lamp, one side of it becomes positive and one side becomes negative as with the battery itself.

If you add another lamp in parallel, you get the same amount of charge building up on either side of the new lamp as shown here.

Each branch of the circuit behaves like a separate series circuit and it gets the full potential difference of the battery.

So one series circuit would be this larger loop going through one of the bulbs, and the other series circuit would be this smaller loop going through the other bulb.

And the electric field from the battery is the same for the series circuits and for the parallel branches.

So which statement is true about the brightness of the bulbs in these circuits? With these short questions, I'll wait five seconds, but if you need longer, press pause while you're thinking and press play when you've chosen your answer.

And the correct answer is that all three bulbs are equally bright, and that's because each of these bulbs has the full PD of the cell across it, and if the bulbs are identical, they have the same resistance, then the same PD is gonna push the same current through each of these cells.

Each branch of a series circuit allows electric current to flow through it.

So the current through the cell is the sum of the current in the branches.

So the sum of the current in these two branches is the same as the current in the cell.

Adding branches to a battery in parallel causes the current to increase.

So when we add this extra branch, the current through the battery is larger.

And here we add another branch and we get a bigger current again in the battery.

However, this does mean the battery will run out of energy more quickly.

In this circuit, what will happen to the reading on the ammeter when the switch is closed? Press pause if you need more time to think.

And the correct answer is that the current through the cell will increase and so the reading on the ammeter will increase.

We'll now have current not only flowing through one branch with a resistor but also through the other branch, and the current in the cell will be the sum of those two currents in the branches.

In this series circuit, the same current flows through the cell and the lamp because current is the same everywhere in a series circuit.

But if we add this extra branch, a resistor in parallel, additional current flows because now current is flowing through the lamp, but current is also flowing through the branch containing the resistor.

The current through the cell increases because the current through the cell is the current through the lamp plus the current through the resistor.

The current through the lamp hasn't changed by the way, but we've added current through the resistor.

And so, the total current in the cell is greater.

But if the current in the cell is greater, that must mean that the resistance of the whole circuit has decreased, and it has, by adding a second branch, we've given two routes for current to flow and that decreases the effective resistance of the circuit.

What's the total resistance when two different resistors are connected in parallel as shown? And you don't need to know an exact answer in ohms, but just is it going to be more or less or the same as either of these two resistances? Press pause if you need more than five seconds to decide.

And the answer is that the total resistance will be less than three ohms. Imagine that we just had a single branch with a 3 ohm resistor in it.

If we add another branch with a five ohm resistor, we now have current through both of those branches.

And so, the overall resistance has decreased, we've got a higher current altogether.

What if instead of adding a branch, we add a component to a branch that's already there? If we do, the resistance of that branch increases and so the current in that branch decreases.

That makes the current in the battery decrease because the current in the battery is the sum of the currents in the two branches.

Meanwhile, the current in the lower branch hasn't changed because the resistance of that branch hasn't changed.

So which of these lamps will be brightest if they are identical lamps? The answer is lamp 3, because it's in a branch that has less resistance altogether than the other branch.

You could see that lamp 3 is in the larger loop and there's just the cell and the one lamp in that loop.

In the smaller loop, there is lamp 1 and 2 and the cell, so there's a larger resistance there and a smaller current.

And so, lamps 1 and 2 won't be as bright as lamp 3.

Well done if you realised tat.

And now a task for you, I'd like you to try building a circuit, build the circuit on the left and measure the currents in the branches, you probably won't have three ammeters, but you can just move the same ammeter into those three positions one at a time.

And then you're going to make a change to the circuit as shown in the second diagram.

You'll be adding a lamp into the lower branch, measure the current again, and then you'll change the circuit again and you'll have three branches each with one lamp and measure those currents as well.

So press pause while you're doing this investigation and press play when you're finished.

I'll show you some example results, your currents won't be exactly the same as this, your PD of your cell might have been different and your resistances of your lamps might have been different.

But check whether you saw this sort of thing, that the current in each branch in one and three is the same, and the current through the cell is the sum of those currents in the branches.

In two, the current in the cell is still the sum of the currents in the branches, but the current in the two branches are different from each other this time, there's more resistance in the branch with two lamps and so there's a lower current in that branch.

Now we can summarise what we've learned about how parallel circuits work with a set of rules that we'll look at in this second part of the lesson.

The electric field around a cell pushes charge around the circuit, and you can think of the electric field as being a bit like the slope of a ramp, which causes a force on a ball.

Two cells in series have a stronger electric field and a larger total potential difference and they push the charges harder around the circuit so that causes a greater current.

So here we have a kind of model, a way of looking at it that the steeper the slope, the faster the ball rolls down and the greater the PD and the field around the cells, the faster the electrons flow around the circuit.

Now what about connecting cells in parallel? They will have the same potential height as one cell, here's a model for trying to imagine what that's like.

With this ramp model, the force on the ball is the same, it's going to roll down at the same speed.

And it turns out, there's a similarity here with cells in parallel, the push from their electric field will be the same.

And a combined PD of the two cells in parallel is the same as the PD of a single cell.

So take a look at these circuits, which lamp is dimmer? Press pause if you need more time to think.

The correct answer is they both have the same brightness.

Remember, the PD that we get from these two cells in parallel is the same as from the single cell.

So the PD across the lamp is the same in both circuits, and that's going to push the same current through the lamp both times.

Now what is the combined PD of these two cells? The combined PD is 1.

5 volts.

For two identical cells in parallel, the combined PD is the same as for one of them.

But two cells in parallel contain a bigger store of chemicals, and it's a chemical reaction inside a cell that provides the energy to push charge around a circuit.

So two cells in parallel will last longer than one cell in the same circuit.

Now, which of the following statements about electric cells connected in parallel compared to an individual cell is correct? Do parallel cells push electrons with a stronger field? Do they have a larger store of chemicals? Or do they allow a greater current to flow? Press pause if you need time to think.

The only correct statement here is B, parallel cells have a larger store of chemicals.

They don't push electrons with a stronger field, it's the same field strength and they don't allow greater current, it's the same current.

And now I'm going to summarise the rules for current and potential difference in parallel circuits.

But if you like, you could pause the video here and try and see which of these you could fill in yourself.

So let's take a look at what belongs in this table.

The rule for parallel circuits.

For current, current through each branch in a parallel circuit adds up to the current through the battery.

You could say that the current through the battery splits when it reaches a junction, and some of it goes into each branch.

If you add branches with components in them, the current through the battery increases, you would've seen that I hope in your investigation.

And if you add more components to a branch that's already there, the resistance in that branch increases and the current in that branch decreases.

So those are the things to know about current in parallel circuits, and those will be useful when you're analysing circuits later.

Now for PD, the PD across the battery adds up to the PD across the components in each separate branch.

So for example, let's say the PD across the battery is two volts.

The PD across all the components in each of the branches will be two volts.

The PD across all branches stays the same when we add more branches with components in them.

So adding another branch doesn't affect the PD across each branch.

Now if you add components to a branch, the PD across the whole branch stays the same as before.

But you could say that PD is shared between the components in the branch, so each component will now have a lower PD than before.

When you're answering questions about parallel circuits, these rules will help you.

Now what happens to potential difference across the lamp if the resistor is removed? Does it increase, decrease, or stay the same? Correct answer is it stays the same.

This is following the rules about PD in parallel circuits, it's the same across each branch, doesn't matter how many branches there are.

So if we remove the resistor, we've effectively removed that branch, that doesn't affect the PD across the lamp.

Now in this question, I'd like you to take a look at the three circuits and comment on the brightness of the lamps in the circuits below.

Say whatever you can about which ones have the same brightness or more or less bright, the lamps themselves are identical.

And explain your reasoning, explain how you know.

And then, I'd like you to try to explain which lamps will stay lit for the longest, if we just leave these circuits running, which ones last longest? So press pause while you're working on that and press play when you've finished and I'll show you some example answers.

So here are the answers.

In circuit 1, the two lamps are equally bright.

This is because they both get the full cell PD.

In circuit 2, the lamps are equally dim, this is because they share the cell PD.

In circuit 3, the single lamp is as bright as in circuit 1, and the two lamps in series are as dim as the lamps in circuit 2, 'cause they share the cell PD.

So the single lamp in circuit 3 is the only thing in its branch, it gets the whole cell PD, the other two lamps only get half each.

So well done if you picked up on most of those points.

Now question two, the lamps in circuit 2 will last the longest since they draw the least current.

The lamps in circuit 3 will last longer than the lamps in circuit 1, as the total resistance of the circuit is greater than that of circuit 1.

So circuit 3 will draw a bit less current than circuit 1, but the lamps in circuit 3 will go out before the lamps in circuit 2 because the total resistance is lower than that of circuit 2.

So circuit 3 will draw more current than 2.

The lamps in circuit 1 will go out first since they draw the most current.

So well done if you got the correct order for that, it's all about how much current the cell is providing.

And when it's providing more current, the cell's going to transfer all of its energy more quickly and run out sooner.

And now we're at the end of the lesson, so here's a summary.

In a parallel circuit, the PD of each branch is the same as the PD of the battery.

Adding more branches to the battery does not change the PD across each branch.

In a parallel circuit, the current in a battery is equal to the sum of all the currents in the branches.

Adding branches of components to a circuit reduces the total resistance, and therefore increases the current through the battery.

Adding components to an individual branch increases the resistance of that branch.

So well done for working through this lesson and for doing the investigation, and I hope to see you again in a future lesson.

Bye for now.