Video: GCSE Physics Higher Tier Pack 1 • Paper 1 • Question 14

GCSE Physics Higher Tier Pack 1 • Paper 1 • Question 14

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Video Transcript

A student measures the resistances of different pieces of copper wire. Each piece of wire is cylindrical and has the same length of 40 centimeters, but a different cross-sectional area. Figure one shows the results.

Okay, so what the student is doing here is to take different pieces of copper wire and measure their resistance. However, importantly, the student is keeping a few things constant between the different copper wires. Firstly, the fact that each piece of wire is made from copper is being kept constant. Secondly, each piece of wire that the student uses is cylindrical. And thirdly, each piece of wire is 40 centimeters long.

This is important because the only thing that the student is changing then between different pieces of copper wire is the cross-sectional area of each piece of wire. So this is the independent variable of the experiment. All the things that are kept constant — such as the pieces of wire being made from copper, being cylindrical, and being 40 centimeters long — are control variables. This way the student can get a good understanding in their experiment of how the resistance of a wire depends solely on the cross-sectional area of the wire. And of course, the resistance, therefore, is the dependent variable of the experiment. And figure one shows the results of the experiment.

So from figure one, we can clearly see that as the cross-sectional area of the wire increases, the resistance of the wire decreases. In other words, this piece of wire, which is 40 centimeters long, is made from copper and is cylindrical, but has a small cross-sectional area, has a large resistance. However, this piece of wire, which is also 40 centimeters long, is made from copper and is cylindrical, but has a large cross-sectional area, has a small resistance.

Now, we’re told that the student needs a piece of wire that is 40 centimeters long and has a resistance of less than two milliohms. We’re told to determine the minimum wire diameter that the student must use. The diameter, 𝑑, of the wire can be calculated using the following equation: 𝑑 is equal to the square root of four 𝐴 over 𝜋, where 𝐴 is the cross-sectional area of the wire.

So first things first, we’re told that the student needs a piece of wire that is 40 centimeters long. Well, thank goodness for that because the student only has pieces of wire that are 40 centimeters long. That’s all they’ve used in the experiment. Remember, this was one of their control variables. So this piece of information, we don’t need to worry about. However, we’re told that the piece of wire needs to have a resistance of less than two milliohms. And we’ve been asked to find the minimum wire diameter that the student must use.

Now, this is slightly problematic because the student conducted an experiment to see the effect of cross-sectional area on resistance, not diameter. However, luckily for us, we’ve been given a relationship between the diameter, 𝑑, of the piece of wire and the cross-sectional area, 𝐴, of the piece of wire. And that’s this relationship here. So first things first, let’s use the graph, figure one, to find the cross-sectional area of a piece of wire that has a resistance of two milliohms.

To do this, we can start moving towards the right from two milliohms, on the resistance axis of the graph, until we meet the curve. At this point then, we can start going downwards until we intersect the horizontal axis. Now, this point intersects at roughly 3.4 millimeters squared. This means that any piece of wire that has a resistance of two milliohms — also, of course, assuming that it’s made from copper and it’s 40 centimeters long and is cylindrical and yada yada yada — this piece of wire will have a cross-sectional area of 3.4 millimeters squared.

Now, bear in mind, a student needs to find a piece of wire that has a resistance of less than two milliohms. On the graph, we can see that as any piece of wire having a resistance of less than two milliohms will, therefore, have a cross-sectional area greater than 3.4 millimeters squared. This is because as the resistance of the piece of wire decreases, its cross-sectional area increases. We can see this from the inverse relationship between the cross-sectional area and the resistance in figure one.

In other words then, what we’ve just found here is the minimum cross-sectional area of the piece of wire. Let’s, therefore, call this term 𝐴 sub min, minimum cross-sectional area. However, because, once again, we need to find the minimum wire diameter, we need to work out the minimum diameter of the piece of wire using this equation, using the value of the minimum cross-sectional area that we’ve just found.

So let’s say that the minimum wire diameter that we’re trying to find is called 𝑑 sub min, minimum diameter. And we know from our equation that this is equal to the square root of four multiplied by 𝐴 sub min, the minimum cross-sectional area, divided by 𝜋. Essentially, what we’ve done here is substituted a general diameter for the minimum diameter that we’re trying to find out, 𝑑 sub min. And we substituted a general area for the piece of wire with a minimum cross-sectional area that we’ve just found.

So all that needs doing here then is to substitute in the value of 𝐴 sub min. When we do this, we find that 𝑑 sub min is equal to four times 3.4 millimeters squared, that’s 𝐴 sub min, divided by 𝜋. Now, because we’ve substituted in our cross-sectional area in millimeters squared, our final answer for 𝑑 sub min is going to be in millimeters. And that’s exactly what we’ve been asked to give our answer in. So when we evaluate the right-hand side of this equation, we find that 𝑑 sub min is equal to 2.080628 dot dot dot millimeters. However, this is not our final answer.

Now, because we’ve not specifically been told what accuracy to give our answer to, we’ll round this answer to three significant figures. So, significant figure number one, number two, and number three. We’re going to be rounding off over here. But, we need to look at the next significant figure to see whether we round that third significant figure up or if it stays the same. Now, this value is a zero. Zero is less than five. So the third significant figure stays the same; it doesn’t round up. And so, to three significant figures, the minimum wire diameter is 2.08 millimeters. This is the minimum diameter of the wire that the student must use so that the wire has a resistance of less than two milliohms.

The piece of copper wire with a resistance of two milliohms is connected to a circuit. While in the circuit, 2400 coulombs of charge flows along the wire in 160 seconds. At what rate does the wire dissipate energy to its surroundings? Write down any equation do you use.

Okay, so what we actually have to find in this part of the question is the rate at which energy is dissipated by the wire to its surroundings. Now, we can say that the rate of energy dissipated by the piece of wire is equal to the total energy dissipated by the wire divided by the time taken for that energy dissipation to occur. Because, this is what the rate of anything means. It’s how quickly something happens. Or, that same something — in this case, the total energy dissipated — divided by the time taken for that something to happen.

Now, at this point, we can recall a different equation. We can recall the power dissipated, which we’ll call 𝑝 here, by a component in a circuit is equal to the total energy dissipated by that component divided by the time taken for the energy dissipation to occur. Now, this is interesting because the right-hand side of this equation looks identical to the right-hand side of this equation, apart from, of course, we’ve written this in words and this one in symbols. But they look mathematically identical because 𝐸 represents the total energy dissipated and 𝑡 is the time taken for the energy dissipation to occur.

This means that the rate of energy dissipated by the piece of wire in the circuit is the same thing as the power dissipated by the piece of wire in the circuit. And this is backed up by the fact that right at the end of the question, we have to find the power of the piece of wire, in this case, in watts. And of course, watts are the standard unit of power. So we’re on the right track here, with our power dissipated. Let’s then erase the long and wordy equation on the left-hand side of the screen. And note that we’re actually trying to find the power as the final answer to our question.

Now, logically then, we would assume that in order to find the power dissipated by the wire in the circuit, we’d first have to work out the energy dissipated by the wire in the circuit and the time taken for this to happen. Now, we have been given a time in our question. However, instead of working with energy and time, there’s a much much more convenient way of working this out. To understand what we’re talking about here, we first need to realize that we’ve actually placed this piece of copper wire in a circuit. And we can recall an equation that gives us the power dissipated by a component in an electrical circuit.

We can recall then that the power dissipated by a component in an electrical circuit is equal to the current flow through that component squared multiplied by the resistance of that component. Now, in both cases, the left-hand side of the equation here and the equation here is referring to the power dissipated by a component in a circuit. In this case, the component is the piece of copper wire. Therefore, if we’re actually trying to find this power, we don’t need to bother with this. We can just find the value of 𝐼 squared multiplied by 𝑅.

And this becomes much more convenient because, firstly, we’ve been given the resistance of the piece of copper wire. We’re told that the resistance 𝑅 is equal to two milliohms. And secondly, we’ve been given enough information to quickly work out the current through the piece of copper wire. Because, we’ve been told the amount of charge flowing through the piece of copper wire and the amount of time in which that charge flows through the wire. Why is this relevant? Well, we can recall that the current in a component in a circuit is equal to the total charge that flows through that component divided by the time taken for that total charge to flow.

Now, because we know the charge flowing through the component and the time taken for this charge to flow, we can work out the current in the component. So let’s say then that the final power that we’re trying to work out, the power dissipated by the piece of wire in the circuit, is equal to the current through that copper wire squared. But remember, the current is equal to the charge flowing through that component divided by the time taken for the charge to flow. And we need to square the current. And in order to find the power, we need to multiply the current squared by the resistance of the component. Now essentially, what we’ve done here is we’ve rewritten 𝑃 is equal to 𝐼 squared 𝑅. But we’ve substituted in 𝐼 is equal to 𝑄 over 𝑡.

At this point, all that’s left for us to do is to substitute in the numbers. However, before we do that, let’s remember that we need to find our final answer in watts, the standard unit of power. Well, if we want to find our power in standard units, then everything else in this equation needs to be in its own standard units as well. So the charge, firstly, that we’ve been given is 2400 coulombs. Now, the coulomb is indeed the standard unit of charge. So we are okay there. Secondly, the time for which the charge flows to the circuit, 160 seconds. Well, the second is the standard unit of time. So we’re okay there as well.

Finally, the resistance of the piece of copper wire, we’ve been told that this is two milliohms. Ah! We’ve reached the problem. The standard unit of resistance is ohms not milliohms. Whatever are we going to do? Of course, we’re gonna convert to the standard unit which is ohms. To do this, we recall that one milliohm is equivalent to one thousandth of an ohm. We can then multiply this equation by two on both sides. On the left-hand side, we’re left with two times one milliohm, which is two milliohms. And on the right-hand side, we’ve got two times one thousandth of an ohm.

Simplifying both sides of the equation, we find that two milliohms is equal to 0.002 ohms. At which point, we also have the resistance in its standard unit, which is, of course, ohms. Oh, happy days! We can plug in our values now. I think that’s how it goes. Anyway, so we say that the power dissipated by the piece of copper wire in the circuit is equal to, firstly, the charge, which is 2400 coulombs, divided by the time taken for that charge to flow, which is 160 seconds. And we square this value. And we finally multiply this by the resistance, which is 0.002 ohms. Then, we evaluate the right-hand side of this equation. And we find that the power dissipated is equal to 0.45 watts. And so, we’ve just worked out that the rate at which the wire dissipates energy to its surroundings is 0.45 watts.

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