# Video: Understanding Appropriate Methods for Accurately Measuring the Resistance of Circuit Components

A student wants to investigate how the current through a fixed resistor changes as the potential difference across it increases. Use the circuit symbol shown in the table to draw a diagram for a circuit that he could use. Describe how the student could use this circuit to investigate how the current through a fixed resistor varies with the potential difference across it. The student’s results are shown in Figure 1. A line of best fit has been drawn on the graph. Describe how the current through the fixed resistor changes as the potential difference across it increases. Use Figure 1 to estimate the current through the fixed resistor when the potential difference across it is 11 V. Use the axes below to sketch the current–potential difference characteristics for a diode. Label the axes appropriately.

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### Video Transcript

A student wants to investigate how the current through a fixed resistor changes as the potential difference across it increases. Use the circuit symbol shown in the table below to draw a diagram for a circuit that he could use. The circuit symbols given are those of a battery, a variable resistor, a fixed resistor, a voltmeter, and an ammeter.

Now the student wants to investigate how the current through a fixed resistor changes as the potential difference across that fixed resistor increases. So let’s start our circuit diagram with a fixed resistor, because this is the main focus of our experiment. In other words, we can’t measure the current through a fixed resistor if we don’t have a fixed resistor in the circuit.

But anyway, secondly, if we want to measure the current through that fixed resistor, then we need to use an ammeter. Now whenever we want to use an ammeter to measure the current through a component in a circuit, the ammeter needs to be connected in series with that specific component. So we’re trying to measure the current through our fixed resistor here, and we’ll connect our ammeter in series with our fixed resistor.

Now the next thing that we need to consider is that there needs to be a potential difference across our fixed resistor, because this is our independent variable. We’re changing the potential difference across our fixed resistor and seeing how the current changes. So we need to have something in the circuit that provides a potential difference across our fixed resistor. And looking at the five circuit symbols we’ve been given to use in our circuit, the only thing that can do this is the battery. And so we’ll connect our battery into the circuit as well.

Now as we already said earlier, we’ll be using the ammeter to measure the current going through our fixed resistor. But then we should also have something that measures the potential difference across our fixed resistor, because there’s no point in us measuring the current through this fixed resistor and seeing how that current specifically depends on the potential difference without having a device that helps us measure the potential difference.

You know, if we don’t know the potential difference across our fixed resistor, then how will we know how the current varies as a result of the potential difference? So we need to include a voltmeter in our circuit to measure the potential difference across our fixed resistor. And it’s important to know that voltmeters need to be connected in parallel with the component that we’re trying to measure the potential difference across. So we’ll start out a parallel branch across the fixed resistor, and we’ll stick a voltmeter on this parallel branch.

Now it’s at this point that we can realize that the whole point of this experiment is to change the potential difference across the fixed resistor and see how the current varies. But then if we were to complete the circuit as it is now, then there will be no way to change the potential difference across the fixed resistor because the battery only provides a set specific potential difference. And this cannot be changed.

Lucky for us though, this is where the variable resistor comes into play. What we can do is to connect the variable resistor in series with our fixed resistor and then we can complete the circuit. Now because our fixed resistor and variable resistor are in series, the potential difference provided by the battery in this circuit will have to be shared across the fixed resistor and the variable resistor. So, for example, let’s say that our battery produces 12 volts of potential difference. Well, these 12 volts will have to be shared amongst the two resistors.

So let’s say that, initially, we set the variable resistor to have the same resistance as the fixed resistor. In this situation, the potential difference across the fixed resistor will be six volts and the potential difference across the variable resistor will also be six volts. But then the fact that this is a variable resistor comes into play.

We could, for example, increase the resistance on the variable resistor. We can use Ohm’s law to tell us what will happen because of this. Now if we’ve increased the resistance on the variable resistor, the potential difference across that variable resistor will increase as well, because these two resistors are in series and so the current through both of them must be the same. So in other words, increasing the resistance of the variable resistor increases the potential difference across it as well. And so the potential difference across the variable resistor might now become something like seven volts, for example.

But then if seven of the 12 volts have been used up on the variable resistor, then there are only five volts to go across the fixed resistor. And this way, we have successfully varied the potential difference across the fixed resistor, which will show up on this voltmeter because it will measure five volts now instead of six volts. And we can see how the reading on the ammeter will change as a result of this.

Therefore, using the five components given to us in table one, we can create this circuit to measure the current through a fixed resistor and see how it changes as the potential difference across that same resistor is varied. Now we’ve just drawn a diagram to show how we could conduct this experiment. But let’s look at how we could explain this in words.

Describe how the student could use this circuit to investigate how the current through a fixed resistor varies with the potential difference across it.

Now we’ve just discussed the answer to this question in great detail. So we just need to put it into words.

Firstly, we can say that the battery provides a constant potential difference across both resistors together. So like we said earlier, for example, the battery could provide something like 12 volts of potential difference across both resistors together. It doesn’t have to be 12 volts, but that’s just an example we’re using here.

Next, we can say that Ohm’s law — 𝑉 is equal to 𝐼𝑅 — tells us that as we increase the resistance of the variable resistor, which we’ll call 𝑅 sub var, the potential difference across the variable resistor increases as well, 𝑉 sub var. And hence, the potential difference across the fixed resistor, 𝑉 sub fix, decreases. So the point of this is that we can vary 𝑉 sub variable, the variable resistor’s potential difference, to vary 𝑉 sub fix, the potential difference across the fixed resistor. And we can measure the potential difference across the fixed resistor with the voltmeter. And finally, we can measure the current, the dependent variable in this experiment, with the ammeter.

And so that is how we can use our circuit to investigate how the current through a fixed resistor varies with the potential difference across it. Now in this case, the student does go on to conduct this experiment. Let’s take a look at their results.

The student’s results are shown in figure one. A line of best fit has been drawn on the graph. Describe how the current through the fixed resistor changes as the potential difference across it increases.

Okay, so in figure one, we can see that, on the vertical axis, we have the current in amperes, and this is the current through the fixed resistor. And on the horizontal axis, we’ve got the potential difference in volts across the fixed resistor. We’re being asked to describe how the current through the fixed resistor changes as the potential difference across it increases. In other words, we’re being asked to give the relationship between the current and potential difference.

Now we can see that there’s been a line of best fit drawn on the graph as well. This fits the data quite nicely, and it’s a straight line going through the origin or zero, zero. Now the fact that it’s a straight line also going through the origin takes off both of the requirements to show that the two quantities on the graph are directly proportional to each other. In other words, our description of how the current through the fixed resistor changes as the potential difference across it increases is that the current increases proportionally to the potential difference. This is the relationship between the current and the potential difference.

Now let’s use the data that the student has collected to work out what we’d expect to see for a potential difference that the student did not measure.

Use figure one to estimate the current through the fixed resistor when the potential difference across it is 11 volts.

So we can see on a horizontal axis to see that there is no data point for when the potential difference is 11 volts. At this point, all we have is the line of best fit, since the data point for which the highest potential difference was measured is this data point here, which corresponds to nine volts.

However, luckily, our line of best fit is a good fit for the data, so we can use it to estimate where the current for the fixed resistor should be if the potential difference across it was 11 volts. And the way to do this is to simply go across from the point of the line of best fit corresponding to 11 volts of potential difference, and so we go across until we meet the vertical axis. And by the way, this is best done with a ruler to ensure maximum accuracy.

Now at this point, we just need to read off what this value of the current is. The way to do that is to realize that this point on the vertical axis represents 0.5 amperes of current and this point represents 0.6 amperes of current.

Now as well as this, we’ve got one, two, three little divisions between 0.5 and 0.6. And those three little divisions divide the distance between 0.5 and 0.6 into four equal segments. Therefore, each little division in gray represents a quarter of the distance between 0.5 and 0.6. So this line represents 0.5 amperes of current. The next line up is going to be 0.525. The next one is going to be 0.550. The third one is going to be 0.575. And then this one is 0.6 as we’ve seen already.

However, the place that we have to take our reading is somewhere between 0.550 and 0.575. But it’s much closer to 0.575, and hence that’s what we’ll say our current is when the potential difference across the fixed resistor is 11 volts.

However, realistically, for this graph, it’s okay to give any reading between 0.550 and 0.575. Anywhere in that range is a valid answer, but we will say that, for a potential difference of 11 volts, the current through the resistor is 0.575 amps. Okay, now we’ve considered many different types of circuit components, including fixed resistors, variable resistors, batteries, voltmeters, and ammeters. Let’s now take a look at diodes.

Use the axes below to sketch the current–potential difference characteristics for a diode. Label the axes appropriately.

Okay, so a diode is an electrical component which has a circuit diagram that looks something like this. And the reason that a diode is drawn like this is because it only allows current to pass through it in one direction. In other words, no current can pass in the opposite direction. And this triangle bit shows the direction of the current that’s allowed to pass if we’re considering conventional current.

In other words then, if we were to set up a circuit with a battery, an ammeter, and a diode and we were considering conventional current which flows from the positive terminal of a battery to the negative terminal, then we would see that this current is allowed to flow because it’s flowing in the correct direction as labeled by the diode circuit diagram. And hence, the current would be able to complete the circuit and the ammeter would show some nonzero reading. Let’s say, for example, it shows something like 0.5 amps.

If however we were to flip the direction of the diode so that it’s now pointing this way, then there would be no current in the circuit because the potential difference from the battery is trying to push the current this way, but the diode does not allow any current to flow in the opposite direction to the arrow in its circuit diagram. And hence, there’ll be no current in the circuit. The ammeter will measure zero amps.

And of course, instead of a battery, we could connect some sort of variable potential difference source or even set up the diode in series with a variable resistor like we did in the first part of this question. Regardless though, any current trying to be pushed in this direction, that is, against the direction of the diode’s arrow, will not be allowed regardless of how large the potential difference is.

So let’s start by labeling our axes as potential difference on the horizontal axis and current on the vertical axis, because in this case the potential difference is the independent variable which we change and the current is the dependent variable which change is based on the potential difference.

So anyway, the situation that we’ve drawn in this diagram here is one for which we can say that the potential difference across the diode is negative because the terminals of the battery are oppositely aligned to the direction that the diode allows current to pass. And as we’ve already said, the current is going to be zero for any value of this potential difference. So if we say that the origin is here, any negative potential difference is gonna be to the left; any positive potential difference is to the right.

Similarly, any negative current is going to be down on the vertical axis and positive current is going to be up, at which point we can plot a flat line along the entirety of the negative potential difference axis, because as we said, for any value of potential difference, regardless of what it is, the current is always going to be zero if the potential difference is negative.

Now what happens to the current when the potential difference is positive, that is, when we’ve lined up the diode and the battery the correct way round? Well, as we said earlier, the battery will be able to push some current through the diode.

However, interestingly, this doesn’t happen for any positive voltage. There is still some sort of threshold voltage that’s different for every diode. And the battery must have a potential difference larger than this in order to be able to push a current through the diode. In other words, our blue line over here continues on until we get to some value of the potential difference, which we can call the threshold voltage. And once we cross the threshold voltage, the line does something like this, because once we get to this voltage, current is allowed to pass through the diode. And then as the potential difference increases, the current also increases. Therefore, this is our graph showing the current–potential difference characteristics for a diode.