Lesson Video: Design of the Voltmeter Physics

In this video, we will learn how to describe the combining of a galvanometer with a multiplier resistor to design a DC voltmeter.

14:43

Video Transcript

In this video, we’re going to be looking at the design of the voltmeter, a device that we can use to measure the voltage or potential difference across a component in a circuit. In a circuit diagram, we can represent a voltmeter with an uppercase 𝑉 inside a circle. And in this circuit, the voltmeter is being used to measure the voltage drop across this resistor. In this video, we’ll see how we can construct a voltmeter by using a galvanometer and a resistor connected in series. We’ll also see how we can calculate the required resistance of this resistor in order to build a voltmeter capable of measuring a given maximum voltage.

So to start off with, let’s consider a cell. And let’s say that this cell has a certain voltage 𝑉 that we want to measure. A simple way that we can try to do this is to connect a galvanometer in series with the cell. Let’s quickly recall that a galvanometer is a device which can measure the magnitude and direction of a current using a needle on a dial. So in this circuit, because the cell is applying a voltage to the galvanometer, this produces a current which we can call 𝐼. This causes the needle on the galvanometer to deflect. And as long as the current isn’t too big, the deflection will be proportional to the current.

Now, Ohm’s law tells us that the voltage applied to a conductor is equal to the current in that conductor multiplied by the resistance of that conductor. In other words, the voltage across our galvanometer, which is the same as the voltage supplied by the cell, is equal to the current in our galvanometer multiplied by the resistance of the galvanometer, which we can call our 𝑅 G. So if we know the resistance of the galvanometer and the galvanometer tells us the current in the circuit, then we can work out the voltage of the cell simply by multiplying these two numbers together. So in this simple case, it looks like a galvanometer can function as a voltmeter.

The deflection of the needle is proportional to the current in the circuit. And Ohm’s law tells us that the current in the circuit is proportional to the voltage. Therefore, the deflection of the needle is proportional to the voltage. However, there is a problem with just using a galvanometer as a voltmeter. This is down to the fact that galvanometers are really sensitive, and typically they can only measure up to a maximum current in the microamp or milliamp region. So, for example, we might find that the needle on our galvanometer reaches maximum deflection for a current of 100 microamps in either direction. And this means that any current over 100 microamps will also just cause maximum deflection of the needle.

What this means is that we can use a galvanometer as a voltmeter, but it would only be capable of measuring voltages within a very limited range. If the galvanometer has a maximum deflection current of 𝐼 G, then it means it will reach maximum deflection for a voltage equal to 𝐼 G times 𝑅 G. So this expression basically tells us the voltage range of our galvanometer. If we want to increase the range of voltages that we can measure, then we need some way of limiting the current in this circuit to prevent the galvanometer’s needle from reaching maximum deflection.

Fortunately, there’s a pretty simple solution. All we need to do is connect a resistor in series with the galvanometer. The function of this resistor is that it increases the overall resistance in the circuit, thus decreasing the current in the galvanometer. This means that, together, these two components could be connected to a larger potential difference without the needle on the galvanometer of reaching maximum deflection. And this is actually all we need to build a voltmeter, just a galvanometer and a resistor connected in series.

In the context of voltmeter design, the extra resistor which we have attached here is known as a multiplier resistor. And we can say it has a resistance 𝑅 M. The reason it’s called a multiplier resistor is that it effectively multiplies the maximum voltage that the galvanometer could measure on its own. We can see how this works by applying Ohm’s law to our voltmeter as a whole. Ohm’s law tells us that the voltage across the voltmeter, which once again is the same as the voltage supplied by the cell, is equal to the current in the voltmeter multiplied by the total resistance of the voltmeter.

This means that the full deflection voltage of our voltmeter, in other words, the range of our voltmeter, is given by the full deflection current of the galvanometer multiplied by the resistance of the voltmeter. Here, it’s useful to remember that for resistors connected in series, the total resistance is given by the sum of the individual resistances. This means that the total resistance of our voltmeter is equal to the resistance of the multiplier resistor plus the resistance of the galvanometer. In other words, 𝑅 V equals 𝑅 M plus 𝑅 G. So overall we can write 𝑉 equals 𝐼 G times 𝑅 M plus 𝑅 G.

This is a really useful formula that tells us the voltage range that our voltmeter can measure based on the full deflection current of the galvanometer, the resistance of the multiplier resistor, and the resistance of the galvanometer. We can get another useful formula if we rearrange this expression to make 𝑅 𝑀 the subject. To do this, we start by multiplying out the brackets on the right-hand side of the expression to give us 𝑉 equals 𝐼 G 𝑅 M plus 𝐼 G 𝑅 G. We can then subtract 𝐼 G times 𝑅 G from both sides of the equation and then finally dividing both sides of the equation by 𝐼 G.

Finally, we’ll just swap around the left and right sides of this expression to give us 𝑅 M equals 𝑉 over 𝐼 G minus 𝑅 G. This expression tells us the sides of multiplier resistor that we need to use in order to build a voltmeter with a range of 𝑉, using a galvanometer with a resistance 𝑅 G and a full deflection current 𝐼 G. Now, one other important thing to mention is that when we make a voltmeter by connecting together a resistor and a galvanometer, we need to make a couple of modifications to the galvanometer. The first issue we need to address is that galvanometers can measure current in either direction. This means that generally they have a zero in the middle of the dial and the needle will deflect either to the right, with the current flowing one way, or to the left when the current is reversed.

Now, if we’re building a direct current or DC voltmeter, that means we only want it to measure a potential difference in one direction. This means that we can get rid of half of the scale as we’re only interested in this bit, which indicates a current with a certain direction. Now, the other issue we need to address with our galvanometer is that at the moment it measures current. However, we’ve shown in this equation that if our galvanometer has a maximum deflection current 𝐼 G, then the voltmeter that we build using this galvanometer will have a maximum deflection voltage 𝑉. We can use this expression to calculate the value of 𝑉 that we would write in place of 𝐼 G on the galvanometer’s dial.

For example, if we’re using a galvanometer that have a resistance 𝑅 G of 100 ohms and a full deflection current 𝐼 G of 100 microamps and we’re using a multiplier resistor with a resistance equal to five kiloohms, then the range of our voltmeter 𝑉 would be given by 100 times 10 to the negative six amps, that’s 𝐼 G, multiplied by 5000 ohms, that’s 𝑅 M, plus 100 ohms. That’s 𝑅 G, which works out at 0.51 volts, which we could then write at the maximum deflection position on the voltmeter’s dial. So once we’ve chosen the value of our multiplier resistor, connected it in series with the galvanometer, and calibrated the scale, our voltmeter’s ready to be used.

Of course, measuring the voltage of a cell isn’t the only application for a voltmeter. More commonly, we might use a voltmeter to measure the potential drop of the individual components in a circuit like this. Here, we have a cell and two resistors wired in series. And let’s say that these resistors have resistances of 𝑅 one and 𝑅 two, respectively. Now, in this circuit, the cell provides a voltage, which we’ll call 𝑉, and this creates a current, which we’ll call 𝐼. Now, if we were analyzing a circuit that looked like this, it might be useful to measure the voltage that’s dropped across each of these resistors.

And to measure the voltage drop across 𝑅 one, for example, we would attach a voltmeter in parallel with 𝑅 one. And of course, we now know that a voltmeter is essentially comprised of a multiplier resistor with a resistance 𝑅 M and a galvanometer with a resistance 𝑅 G. Now when we look at a voltmeter in an application like this, we can see that the multiplier resistor actually fulfills another useful function. At this point in the circuit, the incoming current is split into two smaller currents.

Let’s say that the current flowing through the voltmeter is called 𝐼 V and the current going through resistor 𝑅 one is called 𝐼 R. The splitting of the current could potentially cause a problem as it threatens to decrease the size of the current that flows through the resistor 𝑅 one. And once again, Ohm’s law shows us that if the current decreases, then the voltage will decrease, meaning that connecting a voltmeter here actually threatens to decrease the voltage we’re trying to measure, which obviously isn’t what we want from an accurate measuring device. Very fortunately, this problem is actually solved by the presence of the multiplier resistor. This resistor insures that the overall resistance of the voltmeter is relatively high.

This means that only a very small amount of current flows through the voltmeter. So since the current through a voltmeter 𝐼 V is very small, this means that the current through the resistor, 𝐼 R, is approximately equal to the current in the rest of the circuit, 𝐼. The upshot of this is that connecting a voltmeter in parallel with 𝑅 one makes only a negligible change to the current in 𝑅 one, therefore making a negligible change to the voltage across 𝑅 one. So now that we’ve seen how a voltmeter is designed and used, let’s try answering a practice question.

The circuit diagram represents a galvanometer connected with a multiplier resistor. The multiplier resistor has a resistance 50 times that of the galvanometer. What is the ratio of the current in the galvanometer, 𝐼 G, to the current in the multiplier resistor, 𝐼 M?

So in this question, we’ve been given a circuit diagram that shows a galvanometer and a resistor called a multiplier resistor connected in series with a cell. Let’s start by recalling that the term multiplier resistor describes a resistor used in the construction of a voltmeter. Specifically, it’s the name given to a resistor, which is connected in series with a galvanometer, as is the case in this circuit. This combination of a multiplayer resistor and a galvanometer creates a voltmeter. So effectively, this circuit diagram shows a voltmeter being used to measure the voltage of a cell. Now, it is possible to use a galvanometer on its own to measure a voltage. However, galvanometers are so sensitive that they’re only able to measure voltages within a very small range.

The function of the multiplier resistor in a voltmeter is that it greatly increases or multiplies the maximum voltage that the galvanometer can measure. In a voltmeter, we typically find that the resistance of the multiplayer resistor, which we can call 𝑅 M, is much greater than the resistance of the galvanometer, which we can call 𝑅 G. As we can see, the same is true in this question. We’re told that the multiplier resistor has a resistance 50 times that of the galvanometer.

We’re then asked to calculate the ratio of the current in the galvanometer 𝐼 G to the current in a multiplier resistant 𝐼 M. So let’s start by writing down an expression for each of these currents in terms of their resistances, which we’ve been given some information about. We can do this using Ohm’s law, which tells us that the current in a conductor is equal to the voltage across that conductor divided by the resistance of that conductor. So we can say that the current in the galvanometer 𝐼 G is equal to the voltage across the galvanometer, which we could call 𝑉 G, divided by the resistance of the galvanometer, which is 𝑅 G.

Similarly, we can say that the current in the multiplier resistor 𝐼 M is equal to the voltage across the multiplier resistor, which we’ll call 𝑉 M, divided by the resistance of the multiplier resistor 𝑅 M. Now it’s really important to note that 𝑉 G and 𝑉 M are not necessarily the same. It’s tempting to assume that each of these voltages is simply the same as the voltage supplied by the cell, which we could call 𝑉 C. However, this isn’t the case. When we have resistors in series connected to a cell, as we do in this question, then the voltage drops across each component will add up to the total voltage supplied by the cell.

Now, the question is asking us to find the ratio of the current in the galvanometer 𝐼 G to the current in the multiplier resistor 𝐼 M. And one way of expressing the ratio of 𝐼 G to 𝐼 M is to calculate 𝐼 G over 𝐼 M, which must be equal to 𝑉 G over 𝑅 G divided by 𝑉 M over 𝑅 M. Dividing this fraction by this fraction is the same as multiplying this fraction by the reciprocal of this fraction, which gives us 𝑉 G over 𝑅 G times 𝑅 M over 𝑉 M, which is equivalent to 𝑉 G 𝑅 M over 𝑉 M 𝑅 G.

Now, the question tells us that the multiplier resistor has a resistance 50 times that of the galvanometer. In other words, 𝑅 M equals 50 𝑅 G. This means that we can substitute 50 𝑅 G in place of 𝑅 M in this expression, which then enables us to cancel the common factor of 𝑅 G in the numerator and the denominator, leaving us with 50 𝑉 G over 𝑉 M. Okay, so this simplifies our expression. However, we still don’t have a numerical value for this ratio. And we’re not able to calculate the actual values of 𝑉 G and 𝑉 M without first knowing the voltage supplied by the cell.

However, to help us, we can recall that when we have resistors connected in series with a cell, the size of the voltage drop over each resistor is proportional to its resistance. In other words, a bigger resistor will use a bigger proportion of the total available voltage supplied by the cell. Now, because we’re told that the resistance of the multiplier resistor is 50 times that of the galvanometer, that means that the voltage drop over the multiplier resistor, 𝑉 M, is 50 times the voltage drop over the galvanometer, 𝑉 G. Substituting 50 𝑉 G in place of 𝑉 M in our expression tells us that the ratio of 𝐼 G to 𝐼 M is 50 𝑉 G over 50 𝑉 G, which is equal to one. And this is the answer to our question.

However, there is a simpler way of answering this question that doesn’t require us to use any algebra. In fact, we don’t even need to know how a voltmeter is designed. Nor do we need to know anything about the resistances of the galvanometer and the resistor. In fact, it’s enough just to see that these two components are connected together in a single series circuit. In a series circuit, the rate of flow of charge, in other words, the current, is the same at every point, which means the current in the galvanometer 𝐼 G must be the same as the current in the multiplier resistor 𝐼 M. And if 𝐼 G equals 𝐼 M, then 𝐼 G over 𝐼 M is one. If we have a multiplier resistor connected in series with a galvanometer, then the ratio of the current in the galvanometer to the current in the multiplier resistor is one.

Let’s finish by recapping the key points that we’ve learned in this video. Firstly, we saw that a voltmeter could be made by connecting a galvanometer in series with a resistor known as a multiplier resistor. The multiplier resistor increases the voltage range of the galvanometer and prevents it from greatly affecting the voltage that’s being measured. We’ve also seen that to build a voltmeter with a voltage range 𝑉, using a galvanometer with a resistance 𝑅 G and a full deflection current 𝐼 G, the required resistance 𝑅 M of the multiplier resistor is given by this expression. And we can rearrange this expression like this in order to calculate the voltage range of a voltmeter. This is a summary of the design of the voltmeter.

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