Video: Eg17S1-Physics-Q45

A sensitive galvanometer can measure current up to 𝐼_g. A number of multiplier resistances are connected to its coil one at a time to convert it into a voltmeter. The table below records the maximum potential difference 𝑉 measured by the voltmeter, in volts, and the total resistance 𝑅 of the voltmeter, in ohms. Plot a graphical relationship between 𝑉, on the 𝑦-axis, and 𝑅, on the 𝑥-axis.

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Video Transcript

A sensitive galvanometer can measure current up to 𝐼 g. A number of multiplier resistances are connected to its coil one at a time to convert it into a voltmeter. The table below records the maximum potential difference 𝑉 measured by the voltmeter, in volts, and the total resistance 𝑅 of the voltmeter, in ohms. Plot a graphical relationship between 𝑉, on the 𝑦-axis, and 𝑅, on the 𝑥-axis.

Okay, so in this question, we first of all have been told that we’ve got a galvanometer. And this galvanometer can measure current up to 𝐼 g. In other words, there’s a limit to how much current it can measure. Any current above 𝐼 g and the galvanometer won’t register this. Now, we’ve also been told that a number of multiplier resistances are connected to the coil of the galvanometer one at a time to convert it into a voltmeter. Now, multiplier resistances are basically resistors placed in series. So we connect a resistor to the coil of the galvanometer in series with it. And that’s the first multiplier resistance that we connect to it. Then, we can connect another one, also in series. And we do this one at a time, in order to convert the galvanometer into a voltmeter. In other words then, this whole setup is equivalent to just a voltmeter.

Now, in the table, we’ve been given the maximum potential difference 𝑉, measured by the voltmeter, and the total resistance 𝑅 of the voltmeter as well. And these values are given for every time we add a multiplier resistor to the circuit. Now, what we’ve been asked to do is to plot a graphical relationship between 𝑉 on the 𝑦-axis and 𝑅 on the 𝑥-axis. So let’s go about doing that. First of all, let’s clear some space on the screen though. Okay, so what we’ve done here is to take the table from the question and move it to the top left corner of the screen. And we’ve drawn two axes here, ready for our graph to be drawn on them.

Now, we were told in the question to plot 𝑉 on the 𝑦-axis or vertical axis and 𝑅 on the 𝑥 or horizontal axis. So if we wanna plot our graph, we first need to realise that, on the vertical axis, our range of voltages is between 100 and 300. Therefore, we need to be able to comfortably accommodate that on our vertical axis. So if we start by labelling zero on our vertical axis, then we can go up in steps of 100. And so we can see that our range between 100 and 300 will be easily accommodated on the graph.

Similarly, we need to see that the values for 𝑅 range between 500 and 1500. So if we start with zero on the horizontal axis, we can this time go up in increments of 200. And this time, we can see that we can go from zero all the way to 1400. So we may not have allowed ourselves enough space on the horizontal axis. This is why it’s important to be careful when drawing axes. And doing them with pencil first is often a good idea, either that or actually measuring out the amount of space that we’ll need in order to be able to contain our entire range. For now though, we can just extend our axis and add the 1600 label on there. At which point, we can see that all of our values in this table should now fit on our graph.

So let’s start with this pair of values here. We can see that the horizontal axis or 𝑅-value is going to be 500. And the vertical axis or 𝑉-value is going to be 100. So we start by going across 500 on the horizontal axis. So that’s zero, 200, 400, 500, which will be between 400 and 600. And then, once we’ve got to that point, we can draw an imaginary dotted line upwards. And this doesn’t have to be necessarily drawn on the graph. We can just imagine it in our heads. And then, we do the same thing with the vertical axis value. We see that the value of 𝑉 needs to be 100. So we go up along the vertical axis 100. And from that point on, we draw an imaginary or real dotted line horizontally. And the point at which our two dotted lines intersect is going to be the point at which we draw our cross. And so at this point, we have our first point on our graph. And it has a horizontal value of 500 and a vertical value of 100.

Now, a slightly quicker way of doing this, which we can do with our next point on the graph, is to just go across 750. So that’s zero, 200, 400, 600. 700 is going to be exactly in the middle. And 750 will be between that and 800. So here’s 750. And then, from that point on, we simply go 150 up. So we start at zero, go up 100 and 150, which is halfway between 100 and 200. And then, at that point, we draw our cross. So that’s our second point on the graph. Moving on to the next one then, the 𝑅-value is 1000 and the 𝑉-value is 200. So we go across 1000 and go up to 200 and plot our point there. And we can repeat this process for the final two points in our table.

At this point, we’ve plotted all the values of 𝑉 and 𝑅 on our table. So we can even choose to draw a line of best fit if we want to. Oh look! There’s a happy accident. All of the values in our table happen to lie on the line of best fit. And the line of best fit goes to the origin. This will prove useful in the next part of the question. But before we move on to that, let’s realise that we’ve plotted the graph of 𝑉 against 𝑅, just as we were required to do in the question. Let’s look at the next part of the question then.

From the graph, find the measuring range of the galvanometer, 𝐼 g.

Okay, so to answer this, let’s first recall what we said earlier. We said that we had a galvanometer. And to it we were adding lots of multiplier resistances, to form a voltmeter. Now, adding lots of multiplier resistances would not have changed the behaviour of the galvanometer itself in that it could always only measure the maximum current 𝐼 g. In other words, if the current through the galvanometer was higher than 𝐼 g, the galvanometer would not be able to measure this and would still just read 𝐼 g. And so when we added all these multiplier resistances, we allowed the potential difference across the entire part of the circuit to be shared between the galvanometer and the increasing number of multiplier resistances. Now, this allowed us to measure larger and larger values of the voltage as we can see in our table, but not because the galvanometer itself can measure the voltage. Because, remember, the galvanometer was limited to the amount of current that could pass through it. That current was always 𝐼 g.

However, by adding more and more resistors and sharing the potential difference across the resistors, we could calculate the potential difference across the galvanometer, which we’ll call 𝑉 subscript G, by using Ohm’s law, which tells us that this is equal to the current through the galvanometer, which when the current was maximum was 𝐼 g, multiplied by the resistance of the galvanometer 𝑅 G. Then, we could use our understanding of resistors in series, to work out the total potential difference across the entire voltmeter as we’ve called it now. In other words then, the galvanometer would always only be able to measure a potential difference of 𝑉 G. However, the more resistors we added, the more potential difference we can measure, because each resistor would take on some extra potential difference. And then, we could simply calculate what the total potential difference was across the entire voltmeter.

In other words then, what we’ve plotted on this graph is not 𝑉 subscript G. It’s actually the voltage across the entire voltmeter. And this 𝑅 is not just the resistance of the galvanometer. It’s the resistances of the galvanometer and the multiplier resistances combined. However, because the multiplier resistances are in series with the galvanometer, this means that the total current through the entire voltmeter is equal to the total current through the galvanometer. And whenever the voltmeter measures the maximum value of the potential difference, this means that the current through the galvanometer is at its maximum value.

Hence, if we use this graph to find the maximum current through the voltmeter, then this has to be the same as the current 𝐼 g. So how do we go about doing that? Well, we’ve already mentioned Ohm’s law. And we’ve applied it to the galvanometer. However, this time, we’re gonna apply it to the entire voltmeter. We’re going to say that the maximum voltage in each case, which is 𝑉, is equal to the maximum current through the galvanometer and therefore through the entire voltmeter, 𝐼 g, multiplied by the resistance of the voltmeter in each case.

Interestingly, though, because this voltmeter is an ohmic conductor or, in other words, we’ve got a straight line or the gradient of this line is constant, we can simply work out the gradient of the line to give us 𝐼 g. Because if we rearrange our Ohm’s law equation, to say that 𝑉 divided by 𝑅 is equal to 𝐼 g, then because the gradient is constant, this is the same as saying that the change in 𝑉 divided by the change in 𝑅 is equal to 𝐼 g. And of course, the gradient of our straight line is given by the change in 𝑉 divided by the change in 𝑅. So let’s pick two points on our line of best fit to use in our gradient calculation.

Now, luckily for us, we can see that our line of best fit goes to the origin. So let’s use that as one of our points. And for our second point, because, once again luckily, all of our points go through our line of best fit, let’s use the data from the table. Let’s say that we’ll use this point here. In other words, the point that we’re using is this one. And it has an 𝑅-value of 750 and a 𝑉-value of 150. And of course, the origin has point zero, zero.

So first of all, calculating Δ𝑉, that’s the change in the vertical axis value, we’ve got 150 minus zero. So we can say that Δ𝑉 is equal to 150 volts minus zero volts. And similarly, for Δ𝑅, we can say that this is equal to 750 minus zero. So we say that Δ𝑅 is equal to 750 ohms minus zero ohms. And both of these can be simplified down simply to 150 volts and 750 ohms. And by the way, we could also have calculated the value of 𝐼 g by simply dividing one of these values by its corresponding 𝑅-value. And this is only because our line goes through the origin. And it’s a straight line with a constant gradient.

But anyway, coming back to our calculation then, the gradient, which is equal to 𝐼 g, that’s the maximum current through the galvanometer, is equal to Δ𝑉, that’s 150 volts, divided by Δ𝑅, that’s 750 ohms. And we can also see that because we’ve got 𝑉 in its standard unit and 𝑅 in its standard unit as well, our value of 𝐼 g is also going to be in its own standard unit, which is the amp. So calculating the quantity on the right-hand side of the equation, we get 0.2 amps. And therefore, our final answer is that the measuring range of the galvanometer 𝐼 g is 0.2 amps.

Now, just as a quick aside by the way, when we’ve plotted a graph of 𝑉 against 𝑅 for a different component, that may not necessarily display straight line behaviour. If we want to find the current through that component for a particular resistance value or a particular voltage value, we’re much better off simply dividing the voltage at that point by the resistance, in order to give us the current flowing through that component at that time, rather than finding the gradient, because, as we said earlier, the gradient method only works when we’ve got a straight line going through the origin. So it’s important to keep that in mind.

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