Lesson video

Hello, my name's Dr.

George.

This lesson is called Interpreting I-V Graphs, and is part of the unit Circuit Components.

Here's the outcome for this lesson.

I can take readings from I-V graphs and use these to describe the changing resistance of a component.

I'll be using these key keywords, and if you need to remind yourself of the meanings anytime, you can come back to this slide.

There are three parts to the lesson.

They're called resistance of a diode, describing I-V graphs, and resistance of a fuse.

Let's start by taking a look at this I-V graph for a diode.

We can calculate resistance from an I-V graph by reading a PD and a current that goes with it, and then using the equation resistance is V divided by I.

Here's an example, let's take this point here.

V is minus.

78 volts, and I is 0.

00 milliamps.

We can't actually read the current to two decimal places from this graph, but let's imagine we have a larger graph where we can read it that precisely.

We calculate the resistance at that value of PD using V value by I, but we find ourselves dividing by zero, which we can't really do.

Dividing a number by zero gives infinity, so this is showing us that the resistance is infinite or at least very high.

Let's go across to a place where the PD is positive.

We've switched the connections to the power source around.

Here we have V is.

39 volts, but the current is still 0.

00 milliamps.

Again, we're dividing by zero, so the resistance is still infinite or extremely large.

Moving across to this point here, we actually have a current, it's very tiny.

We can't see that clearly on this version of the graph, but the current is.

40 milliamps, that's.

40 thousandths of an amp, which is.

0040 amps.

And dividing the PD,.

49 volts, by that current in amps, we get 1,225 ohms. So we have quite a large resistance, but it's not infinite anymore.

Looking at one more point, reading V is.

78 volts, and at the same point on the graph, I is 101 milliamps, we have to convert to amps because the equation R equals V divided by I gives us an answer in ohms only if V is in volts and I is in amps.

Calculating that, we get 7.

7 ohms, which is much lower than what we had before.

At the previous point, the resistance was over 1,000 ohms. So the resistance has decreased to a much smaller value, and this is the kind of behaviour we see for a diode.

We have a resistance that varies depending on the PD across it.

Now I'd like you to try a calculation.

What is the resistance of the diode at the point shown on the graph? And when I ask you a question, I'll wait for five seconds, but you may well need longer, in which case press Pause and press Play when you are ready.

The correct answer is 4.

4 ohms because we take the V and the I, divide V by I, and that's the answer to two significant figures.

So a diode behaves in the following way, at negative PD, this left-hand side of the graph, no current can flow.

The resistance is very high or infinite, and we say that when the diode is connected in this way so that the PD is negative, the diode is reversed biassed.

When we connect it so that the PD is positive, we say that the diode is forward biassed.

But only at PD above positive.

7 volts do we get current significantly flowing and the resistance decreases rapidly.

When the PD is between zero and about.

7 volts, current can't flow as the threshold PD,.

7 volts, hasn't been reached.

So the threshold PD is the positive PD that you need to reach for there to be significant current flowing through the diode.

So at which point on this graph does the diode have the lowest resistance? The answer is c.

We saw that on the left-hand side of the graph where the graph lies on the x-axis, we have no current, we have very high resistance.

Then when the graph starts to curve upwards, the resistance decreases, and at c, it's lower than it is at b.

Now can you sketch a graph? Can you remember how the graph looks for how current varies with PD for a diode? And sketch it for both negative and positive PD on the same graph.

And then add these five labels to the right regions of your graph.

Actually, each label may be referring to a point, or it may be referring to a section of the graph.

Press Pause while you're doing this, and press Play when you're ready to check your answer.

And here's what the answer should look like, this kind of shape on the left, the graph is on the x-axis.

And at the threshold PD, the current starts to rise and then it rises up in a straight line.

It's infinite resistance where the current is zero, rapidly decreasing resistance when the current starts to increase, and then quite low resistance and very low resistance at the far right of the graph.

And that's how a diode behaves.

And that type of behaviour is actually useful in a lot of different types of circuit.

Let's move on to the second part of the lesson, describing I-V graphs.

We can use I-V graphs, current voltage graphs, to describe the resistance of a component.

Here's an I-V graph for a type of component called an ohmic resistor.

And the graph is a straight line, and it passes through the origin, the point 0,0.

For this component, the resistance is the same at every point on the line.

And that's what we mean by an ohmic resistor, a component that has a constant resistance, or at least a resistance that's constant as long as the temperature stays the same.

This line is for a different ohmic resistor.

It has a higher resistance.

If we look at any particular value of the PD, we can see that for the low resistance resistor, the current is higher.

Now, a diode is of course not an ohmic resistor.

Its I-V graphs is not a straight line through the origin, and that's because its resistance changes as the PD across it changes.

At high PD, a high current flows as the resistance is low.

At low PD, a low current flows not only because the PD is low, but also because the resistance of the diode is higher.

If you look at the lower point, you can see that, if this were a straight line through the origin, the point at that PD would be higher up.

It would have a larger current than it actually does here.

And which of the following statements about a diode in forward bias just above the threshold PD is correct? Press Pause while you read these and choose your answer.

And the correct answer is b.

We're at a low PD, and you just above the threshold PD of.

7 volts, and at low PD, the resistance of a diode is high at low current flows.

Well done if you pick that one.

Now back to ohmic resistors.

They have a constant resistance in both forward and reverse directions, so that's different from a diode which behaves differently depending on direction of the PD.

Here's a resistor with a low resistance.

And here's a resistor with higher resistance.

Now take a look at this graph.

You're only seeing the part of it for negative PDs and negative currents.

Which of these lines represents a resistor with the highest resistance? What matters here is the size of the PD and the size of the current, and we're going to have the highest resistance for a, because with the same value of PD, a has the smallest current compared with the other resistors.

The resistance of a filament lamp increases with PD, and so its I-V graph is a curved line.

It's not an ohmic resistor.

And its resistance is high when the PD is high.

And its resistance is low when the PD is low.

Again, if this were an ohmic component, these points would line a straight line through the origin, but they don't because the resistance changes with the PD.

Now, here's part of an I-V graph for a filament lamp.

At which point on this graph does the lamp Have the highest resistance? At its point c.

We've seen this is when the PD is at its largest that we have the highest resistance.

If we read this graph from zero PD towards the left to larger PDs, we find that the current isn't increasing as much as we would expect for an ohmic conductor as the PD increases.

What's happening there is the resistance of the lamp is increasing.

Now, for each of these three graphs, can you identify the component that produced it and describe the relationship between PD, resistance and current for that component? And refer to the graphs when you are answering this question.

Press Pause while you write down your answers.

And when you are ready, I'll show you some example answers.

So here are the answers.

The first graph shows a diode with negative PD, zero current flows, because the resistance is very high.

So we're describing what's on the graph, and we're also explaining it by referring to the resistance.

With positive PD, the resistance is very high until a threshold PD is reached, and then it decreases rapidly.

This means that the current is zero below the threshold PD but increases rapidly above the threshold PD.

The second graph is for a resistor, an ohmic resistor.

The graph shows that current is directly proportional to PD because it is a straight line that passes through the origin, so the resistance remains constant.

This is true when the PD is applied across the resistor in either direction.

And the third graph is for a filament lamp.

The current increases when the PD increases, but not as much as it would if the resistance of the filament lamp remain constant.

As the PD increases, the resistance increases causing a lower-than-expected current.

This is shown by the graph curving away from a straight line towards a lower current.

The graph is symmetrical for positive and negative PDs, which shows that the resistance of the lamp varies in the same way whichever way round the PD is applied.

So compare these answers with your answers.

See if you missed anything, and check that you understand the answers that I've shown you here.

And now we're ready for the third part of this lesson, resistance of a fuse.

An electrical fuse is a component that's designed to limit the flow of current to a device.

This photo shows a fuse inside a plug, and this particular type of fuse is a five amp fuse, which means it will only allow a current of five amps and below.

If a higher current flows, the fuse will blow.

And when I say blow, I do not mean blow up.

The symbol for a fuse is this, so take care not to confuse it with the resistor symbol, which doesn't have a line going through the middle of it.

There's a fine wire inside the fuse, represented here, and that completes a circuit.

But if the current gets too high, the fuse wire heats up and it melts, and then there's a break in the circuit and current doesn't flow anymore.

The fuse wire behaves in a similar way to the filament wire in a lamp, at least up until the point when it melts.

Now here are four values of fuses, 3 amps, 5 amps, 10 amps, 13 amps.

And I'd like you to say which of them is suitable for an electric kettle that normally has a current of 11 amps passing through it.

And the only one here that's suitable is the 13-amp fuse.

If we used a 3, 5 or 10-amp fuse, it would actually break when the kettle is operating normally because they're designed to break at those current shown, and the kettle needs 11 amps.

So when we choose a fuse for an appliance, we should use the fuse that has the lowest value that's above the current the appliance needs.

Let's take a look at what's happening in a fuse wire.

It's made of metal, so it has an ion lattice that's a regular repeating arrangement, a lattice of ions, which are atoms that have lost one or more of their electrons.

Those electrons are free to move from one ion to another, and the movement of those electrons is what gives us a current, a flow of charge.

When a current flows through the fuse wire, it makes the ions vibrate more vigorously, more energetically than they would without a current.

So they vibrate all the time but more energetically if a current flows.

So this picture shows a simple model of that lattice at a low current, and these lines around the ions represent small vibrations, and the small blue dots are representing free electrons, and the lines behind those represent their movement.

If we increase the current, the vibrations of the ions increase, and the electrons move faster.

And at an even higher current, everything is moving even more energetically.

If the PD across the wire is too high and the metal ions vibrate too vigorously, then this means the wire is very hot and it will melt.

And this melting of the wire turns off the circuit, and that's what the fuse is designed to do.

It's used as a safety measure.

So which of the following will cause a fuse wire in a circuit to melt? The only correct answer here is a, the current being too high.

It will heat up the fuse wire and it will melt.

The room temperature is never going to be high enough to melt the fuse wire.

The resistance of the fuse wire will change as the ions vibrate more vigorously because when they vibrate more vigorously, there are more frequent collisions between the ions and the moving electrons.

So at low PD when there's low current, there's low resistance in the fuse wire.

But if the PD increases, that causes a higher current, and that causes the ions to vibrate more energetically, and the resistance is higher.

And the resistance is even higher if the current increases further.

Which of the following describes the resistance of fuse wire just before it melts? The answer is it's at its maximum value.

The resistance increases as the temperature of the fuse wire increases, and the highest temperature it ever gets to is the temperature just before it melts.

And there are a couple of longer questions for you.

Can you sketch the positive part of an I-V graph for a fuse wire to show what happens as the PD is until the fuse wire melts? And label the point at which the wire melts.

Now, I haven't shown you an I-V graph for a fuse wire, but I've told you enough to be able to work out roughly what it should look like.

And then describe the relationship between PD, resistant, current and motion of the metal ions for the fuse wire.

And use your graph to support your description.

Press Pause while you write your answers, and press Play when you're ready to check them.

Here are some example answers.

Your graph should have roughly this shape.

It's the same shape as the graph for a filament lamp because a fuse behaves in a similar way.

And the wire melts at the far right-hand end of your graph.

We can't have a higher PD or a higher current in the fuse wire because it's now broken.

When the PD across the fuse wire increases, the current increases.

This causes the electrons to move faster and collide more frequently and more energetically with the metal ions.

The metal ions vibrate more vigorously causing an increase in the temperature of the wire and even more frequent collisions.

The higher rate of collisions causes the resistance to increase, and so the current is not as high as expected.

And that's why the graph curves.

Eventually the wire gets hot enough to melt.

Well done if you drew a similar graph and made some or all of those key points.

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

So here's a summary.

A diode behaves differently depending if it is in forward bias or reverse bias.

In reverse bias, and when the PD is below the threshold voltage in forward bias, the resistance is so high that a current cannot flow.

At PDs above the threshold voltage, the resistance of a diode is very low and a current can flow.

The resistance of ohmic resistors and filament lamps varies in the same way whichever way round the PD is applied.

Resistors have a constant resistance, and the resistance through filament lamps and fuses increases with greater currents.

I hope you've enjoyed the lesson, and I hope you now feel confident about reading and interpreting I-V graphs for components.

I hope to see you again In a future lesson.

Bye for now.