Loading...

Hello, my name's Dr.

George, and this lesson is called, "Linking current, potential difference, and resistance," and it's part of the unit, electric fields and circuit calculations.

The outcome, which I'll help you achieve by the end of the lesson, is "I can describe the relationships between current, potential difference, and resistance." Here are the key words for the lesson.

I'll just explain what those three quantities are that were in the outcome.

So electric current is a measure of the amount of charge passing a given point in a circuit in one second.

And resistance is a measure of how difficult it is for current to flow in a circuit.

Potential difference is a voltage, or you can think of it as a measure of the push on electrons in a circuit.

It's also useful to know that a series circuit has only one loop for electrons to go around, and we'll be meeting the unit of resistance later, the ohm.

The lesson has three parts, current and resistance in a series circuit, current and p.

d.

in a series circuit, and calculating current, resistance, and p.

d.

So let's start.

Here we have a diagram of a series circuit.

It's just got one cell, a 1.

5 volt cell, and one lamp.

If we add more lamps to the circuit, it increases the overall resistance in the circuit and that makes the current decrease.

What it doesn't change is the potential difference of the cell, that stays the same.

So cells have a constant potential difference, but the resistance increases with more lamps and then that makes the current decrease.

Low resistance, we get a high current.

High resistance, we get a low current.

And let's look more closely at that relationship.

Say we have one lamp with the resistance of R ohms, and say that gives us a current of I A.

If we have two lamps, we've doubled the resistance, it's 2R, and that actually halves the current, we get 1/2 I A.

If we double the number of lamps again, we've doubled the resistance again, it's now 4R ohms, and the current is a quarter of what it originally was, it's 1/4 I A.

This is a particular kind of relationship between the two variables.

They're not proportional.

One goes up when the other goes down.

And in a very particular way that we call inverse proportion, we say that current is inversely proportional to resistance, which we're measuring in ohms. So now a question for you, what will the current be if seven more identical lamps are added to this circuit so that there are now eight lamps in series? With short questions like this, I'll wait five seconds, but you may need longer, in which case press pause, and press play when you've chosen your answer.

The correct answer is 1/8 A.

If one lamp with a resistance of R ohms gives us a current of 1 A, then eight lamps will have a resistance of 8 R ohms, eight times as much, giving us a current of 1/8 A, 1/8 of what the current was before.

So well done if you realised that.

So, as we've been saying, current is inversely proportional to resistance.

If the resistance doubles, the current halves.

Another way of saying that is the current changes by one over the change in resistance.

So whatever the resistance multiplies by, the current multiplies by one over that.

We could write that using a mathematical symbol.

Current is proportional to one over the resistance, or we can say current is inversely proportional to the resistance.

So how should the resistance in the series circuit change to increase the current by a factor of four? Should it double, quadruple, halve, or quarter? Press pause if you need time to think, and press play when you've chosen.

The answer is that the resistance needs to quarter for the current to increase by a factor of 4.

I showed you that current is proportional to one over resistance.

It's also true then that resistance is proportional to one over the current.

If current increases by a factor of 4 multiplies by 4, then resistance decreases by a factor of 4, it divides by 4.

When two quantities are inversely proportional, when one quantity is multiplied by a number, the other quantity is divided by that number.

And now a task for you.

Have a look at circuit one.

The lamp is working, and there's an ammeter there measuring the current.

The potential difference, p.

d.

for short, of the battery is 3.

0 V.

Now here's the question.

A resistor with the same resistance as the lamp is added to make circuit two, that you can see on the right, I'd like you to predict and explain what will happen to the current through the ammeter, and what will happen to the current through the lamp.

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

What will happen to the current in the ammeter? Well, it will be half of the original current.

And that's because the resistance of the lamp and resistor together is double that of the lamp alone.

We've doubled the resistance so we've halved the current.

The current through the lamp will also be half of the original current, and that's because the resistance of the whole circuit has doubled and the current is the same everywhere in a series circuit.

So when you have a circuit that's a single loop, the current's the same in all places.

And now for the second part of this lesson, current and p.

d.

in a series circuit.

So we're going to look more at potential difference now.

Think about an electric cell as shown here.

In everyday life we might call this a battery, but strictly speaking it's a cell.

And a cell has a negatively charged end and a positively charged end, and you can often see that written on it.

Let's say that this cell has a potential difference of 1.

5 V between the ends.

We usually abbreviate potential difference to p.

d.

, so that's what I mean if I say p.

d.

during this lesson.

Now let's have two cells in series, that makes what we call a battery.

A battery is two or more cells in series.

There's going to be a higher charge on the ends, and we're going to get double the voltage, 3.

0 V, so we have a greater p.

d.

So if we now put that battery into a series circuit, we could measure the p.

d.

across the battery like this with a voltmeter, and we can measure the p.

d.

across the lamp, and what we find is that they're the same.

And by the way, electrical leads don't affect p.

d.

So even if they're quite long, they're not going to make a difference.

So what will happen to the lamp here? If a cell is removed from the battery, will it get brighter, stay the same, or get dimmer? Press pause if you need time to think, and press play when you've chosen your answer.

The correct answer is that the lamp will get dimmer.

We've reduced the p.

d.

, and the p.

d.

is what gives the electrons a push around the circuit.

And we've said that there's a positive and negative end of a battery, and here it's been drawn on.

Notice that the positive end is the long line in the symbol for cell, and the negative end is a shorter thick line.

When a lamp's connected to the battery, the charge can flow around the circuit, and the lamp now has positive and negative charge at each end as well.

And it's the electric field from the battery that causes charge to form across components.

So the battery has positive and negatively charged ends, there's an electric field around that, and that electric field then extends around the whole circuit causing each component to have a positive end and a negative end.

If we add cells to a series circuit, we get a larger current.

If we double the number of cells, we double the current.

If we treble the number of cells, we treble the current.

And that kind of relationship is direct proportion.

Current is directly proportional to the p.

d.

of the battery in a series circuit.

So another way of saying that is the current changes by the same factor as the p.

d.

If the p.

d.

gets multiplied by a number, the current will be multiplied by the same number.

We could write that using the mathematical symbol for proportion.

This statement here means current is proportional to p.

d.

Now take a look at this circuit.

What would need to happen to the p.

d.

of the battery, perhaps by swapping for another battery, to keep the current the same if the lamp is removed? Does the p.

d.

need to increase, stay the same, or decrease? Press pause while you're thinking, and press play when you've chosen your answer.

The correct answer is the p.

d.

would need to decrease.

So if we remove the lamp, that decreases the resistance in the circuit, which would increase the current.

If we want to bring the current back to what it was before, we're going to need to decrease the p.

d.

And now another question.

What will the current be if two more identical cells are added in series to the circuit shown here, and they're added the same way round as the original cell? Is it 1 A, 2 A, 3 A, or 4 A? Press pause and press play when you've decided.

And if two more identical cells are added here, we're going to get a current of 3 A.

Thinking about why, if one cell causes a current of 1 A to flow, adding two cells makes a battery of three cells.

We've tripled the p.

d.

, so three cells will cause a current of 3 A to flow.

Now let's compare current and p.

d.

in a series circuit.

Current's measured in amps using an ammeter in series in the circuit.

So we connect the ammeter within the loop.

P.

d.

is measured in volts, and we connect a voltmeter in parallel.

So we connect it either side of what we're trying to measure the p.

d.

across.

In a series circuit, the current through each component is the same, and the p.

d.

across the component adds to the p.

d.

across the battery.

So if there's only one component, it will have the same p.

d.

across it as the battery.

If there were two components, their p.

d.

's will add to the p.

d.

across the battery.

Thinking about that, which of the following will happen if a resistor is added in series to a series circuit? Will the p.

d.

of the battery increase or decrease, and will the current increase or decrease? Press pause while you're thinking, press play when you're ready.

There's only one correct answer here and that is that the current in the circuit will decrease.

We've increased the resistance, which makes it more difficult for current to flow.

The p.

d.

of the battery will neither increase nor decrease.

It's not affected by what's happening in the circuit that it's connected to.

So well done if you realised that that was the only correct statement.

Now I'd like you to read the sentences here and fill in the gaps.

The sentences describe potential difference and current in the series circuit, and those are the only words I'd like you to use in those gaps.

So obviously, you'll use them more than once.

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

And here are the correct answers.

Potential difference is measured in volts with a voltmeter.

Current is measured in amps with an ammeter.

In a series circuit, the current is the same everywhere.

In a series circuit, the potential difference of the components adds to the potential difference of the battery.

If another lamp is added, the potential difference of the battery will stay the same.

So well done if you've got most or all of those.

Now onto part three of the lesson, calculating current resistance and p.

d.

Take a look at this circuit.

The current flowing in a circuit can be affected by changing both the p.

d.

and the resistance.

So current is affected by both of those things.

Adding cells would increase the p.

d.

, and that increases the current.

But adding components would increase the resistance, and that decreases the current.

So these two things have opposite effects.

Now have a look at the circuits on the right.

Circuit one is changed into circuit two by adding an identical cell and replacing the motor with a resistor.

And then we find that circuit two has a lower current than circuit one.

So can you work out what is the resistance of the resistor in circuit two compared to the motor in circuit one? Is it larger, the same, or smaller? Press pause while you think, press play when you've decided.

The resistance of the resistor must be larger than the resistance of the motor because we've increased the p.

d.

of the battery and that on its own would increase the current.

But the current has decreased, and the only other change we've made is to change a motor into a resistor so that resistor must have more resistance than the motor, so that even though the p.

d.

's increased, there's a greater effect from increasing the resistance in the circuit.

Well done if you work that out.

So as I've said, both p.

d.

and resistance affect current.

Looking at those relationships again, current is directly proportional to p.

d.

but inversely proportional to resistance.

We could combine those two relationships into one statement.

Current is proportional to p.

d.

over resistance.

You can see how this relationship is coming from the other two.

And in fact, it turns out that current is equal to p.

d.

over resistance as long as we use the right units for these quantities.

So in symbols, we say I equals V over R, I is the symbol for current.

Current is measured in amps, potential difference in volts, and resistance in ohms. So if we use those units, this equation works.

And we can apply this equation to any individual component in a circuit.

We can look at the current through it, the p.

d.

across it, and the resistance of it.

And they will be related using I equals V over R.

So what's the current if the p.

d.

across a resistor of 5 ohms is 10 volts? Press pause, press play when you've worked it out.

The current is 2 A.

And the reason why, is current is p.

d.

divided by resistance.

So 10 divided by 5, which gives us 2.

And we've used the standard units of volts and ohms, so we will get an answer in amps.

We can rearrange the equation to get potential difference if we know the resistance of a component and the current in it.

What we can do is multiply both sides by resistance.

And if you divide by a number but then multiply by the same number, you're back to where you started.

So we could say those two operations cancel out.

So what we have is current times resistance equals potential difference, or we might like to write it the other way around, p.

d.

equals current times resistance.

You don't have to write that many steps if you're more confident with rearranging formula.

In symbols, we can write this as V equals I times R.

Now what is the p.

d.

across a lamp of resistance 8 ohms if the current through it is 0.

2 A? Press pause, and press play when you've done your calculation.

Correct answer is 1.

6 V, because p.

d.

is current times resistance.

0.

2 times 8 is 1.

6 V.

Well done if you're getting these calculations right.

And we can have a third arrangement of the equation.

We might want to find the resistance of a component if we know the current through it in a circuit and the potential difference across it at the same time.

To rearrange, let's multiply by resistance again.

So that cancels out on the right.

Now we have current times resistance equals p.

d.

If we divide both sides by current, on the left, multiplying something by current over current, well that's just the same as multiplying by one, it doesn't do anything.

So we can cross that out and we can say that resistance is p.

d.

over current.

Or in symbols, we can say R equals V divided by I.

Here's a series circuit with an ammeter and a voltmeter included, and their readings are written next to them.

What's the resistance of the resistor in this circuit? Press pause while you're thinking, and press play when you've chosen your answer.

The correct answer is 20 ohms, because resistance is p.

d.

divided by current.

The p.

d.

across the resistor is 3.

0 V.

The current in the resistor is 0.

15 A, because the current in a series circuit is the same everywhere.

So resistance is 3.

0 divided by 0.

15, which is 20 ohms. Now, some longer questions for you.

I'd like you to give your answers to these questions to two significant figures, because the measurements here are written to two significant figures.

Press pause while you're writing your answers, and press play when you're ready and I'll show you the correct answers so you can check yours.

Here are the answers to the first two questions.

First one, a light bulb has resistance, 6.

0 ohms. What is the current through the light bulb when the p.

d.

across it is 1.

5 V? And it's a good idea to start by writing down the quantities that you know, and also writing down the equation that you're going to use, in the arrangement that you're going to use.

If you need to, you can write your rearrangement of the equation first.

So current is p.

d.

divided by resistance, so 1.

5 V divided by 6.

0 ohms and we get 0.

25 A.

And that's written to two significant figures because the zero at the beginning is not a significant figure.

Question two, a motor is connected to a 6.

0 V battery and a current of 0.

75 A flows through it.

What is the resistance of the motor? The resistance of a component, it's p.

d.

across it, divided by the current flowing through it.

So 6.

0 V divided by 0.

75 A, which is 8.

0 ohms. And again, that is written to two significant figures.

Question three, a light bulb has resistance 0.

65 ohms, what p.

d.

is needed to make a current of 0.

20 A flow through it? Well the p.

d.

across the component is the current flowing through it times the component resistance, 0.

20 A times 0.

65 ohms, 0.

13 V to two significant figures.

And question four, a current of 0.

70 A flows through a resistor when there is a p.

d.

of 4.

2 V across it.

What is the resistance of the resistor? The resistance is a p.

d.

divided by the current, 4.

2 V divided by 0.

70 A, which is 6.

0 ohms. Now, what current flows through the resistor when there's a p.

d.

of 6.

0 V across it? Well, we know the resistance from the answer to part A, and we now have a different p.

d.

across the resistor, so there'll be a different current.

And that current is the p.

d.

divided by the resistance, 6.

0 V divided by 6.

0 ohms, which is 1.

0 A.

And that's written to two significant figures.

If you just write 1 A, then you've only written it to one significant figure, and that suggests that you don't know what the number after the decimal point is, and we do know that it's zero.

So well done if you got most or all of those questions right.

And now we've reached the end of the lesson.

So here's a summary.

Current is inversely proportional to resistance and directly proportional to p.

d.

Current equals p.

d.

over resistance, or in symbols I equals V over R.

The equation can be rearranged to give V equals I times R, or R equals V divided by I.

Where the current I is measured in amps, potential difference V is measured in volts, and resistance R is measured in ohms. So well done for working through this lesson.

I hope now that you feel you're able to use the relationship between current, p.

d.

, and resistance for a component in a circuit.

I hope to see you again in a future lesson.

Bye for now.