Lesson Explainer: Design of the Ammeter Physics

In this explainer, we will learn how to describe the combination of a galvanometer with a shunt resistor to design a DC ammeter.

An ammeter is a device that can be used to measure the current in a circuit. As we shall see, we can make such a device using a galvanometer along with a resistor.

Since the ammeter’s design makes use of a galvanometer, let’s begin by reminding ourselves how a galvanometer behaves.

A galvanometer is a device that responds to the direction and magnitude of electric current. The following figure shows a galvanometer.

As shown in the sketch, the galvanometer has a needle that can deflect to either side of zero. This needle deflects whenever there is a current through the galvanometer.

So, if we apply a potential difference across the galvanometer, resulting in a current through it, the needle will deflect to one side of the zero. This is shown in the sketch below.

As shown in the sketch, if we reverse the polarity of the potential difference such that the current goes the opposite way through the galvanometer, the needle will deflect to the opposite side of the zero.

So, we can see that a galvanometer provides a means of measuring the amount of current through it. In fact, it turns out that the deflection of the galvanometer’s needle away from the central zero is proportional to the magnitude of the current, up until the point at which the needle reaches the end of the scale.

This limit, at which the needle is pointed fully to one end of the scale, is illustrated below.

In this situation, we say that the needle has a maximum deflection. The current that causes the needle to just reach this maximum deflection is the maximum value of current that can be measured using this galvanometer. The exact maximum current that can be measured using a galvanometer varies depending on the device but is typically on the order of microamperes or milliamperes.

For example, if we have a galvanometer with a full-scale deflection of 500 μA, then this particular device would be able to measure the intensity and direction of a current as long as this current is smaller than 500 μA.

If we, however, want to use a galvanometer as an ammeter, we will run into two problems.

The first problem is that there is clearly a limit on the intensity of the currents we can measure with a galvanometer. Specifically, we are limited to currents below the full-scale deflection current of the galvanometer. This represents an issue if we want to measure currents outside this range.

We might think that if we could somehow extend the range of a galvanometer, then this would provide an accurate way of measuring larger currents too. However, there is a second problem: the galvanometer has its own internal resistance. Let’s see why this represents a problem by considering a simple circuit.

In this circuit, the cell provides a potential difference 𝑉 across a resistor of resistance 𝑅, and so there is a current 𝐼.

We can recall that Ohm’s law tells us that, for such a circuit, we have 𝑉=𝐼𝑅.

We can rearrange this by dividing both sides by the resistance 𝑅 to get an expression for the current in the circuit in terms of the potential difference and resistance: 𝐼=𝑉𝑅.

Let’s now see what happens when we try to use a galvanometer in order to measure the value of 𝐼. Imagine that the range of the galvanometer and the values of 𝑉 and 𝑅 are such that 𝐼 is less than the full-scale deflection current of the galvanometer.

Adding in a galvanometer in series with the other components, we get the following circuit.

However, we said earlier that the galvanometer has some resistance of its own. Let’s make that clear by explicitly drawing in this resistance in our circuit.

We have labeled the resistance of the galvanometer 𝑅. We can see that we now have two resistors connected in series.

We can recall that when we have two resistors connected in series, the total resistance is given by the sum of the individual resistances. So, in our case, if we label the total resistance 𝑅, then we have that 𝑅=𝑅+𝑅.

If we now apply Ohm’s law to the circuit as a whole, we can see exactly why this is a problem. Let’s use the form of the equation in which the current is the subject. We will label this current 𝐼. The total resistance of our circuit is now 𝑅, not just 𝑅. Substituting this into Ohm’s law, we have 𝐼=𝑉𝑅.

Then, we can substitute in 𝑅=𝑅+𝑅 to get 𝐼=𝑉𝑅+𝑅.

Let’s now compare this equation to our original equation without the galvanometer, in which 𝐼=𝑉𝑅. The fact that the total resistance of the circuit has changed means that the intensity of the current has also changed.

This means that the galvanometer, the device we were trying to use in order to measure the current, has actually changed the value of the current we wanted to measure. This is a bit like having a weighing scale that changes the mass of the object you place on it, or a ruler that changes the length of the thing you were trying to measure with it.

Fortunately, it turns out that there is a way we can deal with this problem of the galvanometer changing the current in the circuit. We can do this by adding a resistor in parallel with the galvanometer.

This resistor is referred to as a shunt resistor, and we have labeled its resistance 𝑅.

We can recall that when we add a parallel branch to a circuit, the potential difference across each branch will be the same. Meanwhile, the current gets split such that there is a current in each of the two parallel branches. So, instead of all of the current flowing through the galvanometer, some of the current now follows the other path through the shunt resistor instead.

Let’s look at a quick example.

Example 1: Finding the Potential Difference across Two Parallel Branches in a Circuit with a Galvanometer

The circuit diagram represents a galvanometer combined with a shunt resistor. The emf of the source connected to the galvanometer and the shunt is 3.0 V. The circuit does not represent a circuit in which the galvanometer and shunt correctly function as an ammeter.

  1. What is the potential difference across the shunt? Answer to one decimal place.
  2. What is the potential difference across the galvanometer? Answer to one decimal place.

Answer

Part 1

In the diagram for this question, we have a circuit containing a galvanometer in parallel with a shunt resistor. In this case, the resistance of the galvanometer has not been explicitly included as a resistor in the circuit diagram, but we know that it does have some resistance.

This first part of the question is asking us to find the potential difference across the shunt resistor in this diagram.

We can recall that the emf of the voltage source in a circuit is equal to the total potential difference across the components that the current travels through as it goes around the circuit.

In this case, we have two parallel branches, meaning that there are two possible routes for the current to take. The total potential difference across each of these loops must be equal to the emf of the voltage source.

The loop passing through the shunt resistor is marked in pink in the diagram below.

We know that the total potential difference across this loop must be equal to the emf of the source, which is 3.0 V.

We will assume that the wires have no resistance so that all of the resistance along the loop marked is in the shunt resistor.

This means that all of the 3.0 V potential difference over the pink loop is across the shunt resistor. So, we know that the potential difference across the shunt resistor is equal to 3.0 V.

Part 2

This second part of the question is asking us for the potential difference across the galvanometer.

The second of the two complete loops in the diagram, which is the loop passing through the galvanometer, is marked in orange in the diagram below.

As with the first part of the question, we know that the total potential difference across the loop must be equal to the emf of the source, which is 3.0 V.

Again assuming that the wires have no resistance, this means that all of the resistance along the orange loop is in the galvanometer.

Therefore, all of the 3.0 V potential difference that is across the orange loop is across the galvanometer. So, we know that the potential difference across the galvanometer is equal to 3.0 V.

As illustrated in this example, when we connect a shunt resistor in parallel with the galvanometer, the potential difference across the shunt resistor will be equal to the potential difference across the galvanometer.

We also know that the current splits into the two parallel branches. To find out how much current is in each branch, we can apply Ohm’s law to each branch separately.

We will label the resistance of the galvanometer, the current through the galvanometer, and the potential difference across it with a subscript 𝐺, and we will label these same quantities for the shunt resistor with a subscript 𝑆.

Then, Ohm’s law for each branch gives us the following two equations: 𝐼=𝑉𝑅,𝐼=𝑉𝑅.

We know that both branches have the same potential difference. That is, in these equations, we know that 𝑉=𝑉.

Let’s consider what happens we choose a shunt resistor with a resistance much smaller than that of the galvanometer, that is, when we pick a shunt resistor such that 𝑅𝑅.

For two resistors connected in parallel, the overall resistance is lower than the lowest of the two individual resistances. The total resistance of the galvanometer and shunt resistor in parallel is therefore smaller than 𝑅 and, hence, much smaller than 𝑅.

This means that the overall effect of the combination of galvanometer and shunt resistor on the current in the circuit is very small. In other words, by adding a shunt resistor in parallel with the galvanometer, we have overcome the problem that the resistance of the galvanometer would affect the current in the circuit that it was trying to measure. Any effect on the current in the circuit will now be much smaller.

We can also notice from the two Ohm’s law equations that the expressions for 𝐼 and 𝐼 have the same values in the numerator, but the denominator in the 𝐼 expression is much larger than in the 𝐼 expression. What this means is that if 𝑅𝑅, then 𝐼𝐼.

In other words, most of the current goes through the path containing the shunt resistor. Meanwhile, there is a small, constant proportion of the current through the galvanometer. This means that the deflection of the galvanometer’s needle will be proportional to the current in the circuit. Hence, the combination of the galvanometer and shunt resistor may be used to measure the current in the circuit.

So, everything within the orange box in the diagram below together functions as an ammeter.

When building an ammeter in this way, it is important to carefully choose the resistance 𝑅 of the shunt resistor to give the best results. Remember that the resistance 𝑅 of the galvanometer has a fixed value. Changing the value of 𝑅 changes the fraction of the current that passes through the galvanometer. We want a value of 𝑅 such that the current through the galvanometer is high enough that the needle displays a clear reading, but also low enough that the needle does not quite hit full-scale deflection.

To work out the best value for 𝑅, we can once again make use of Ohm’s law, which states that, for a potential difference 𝑉, resistance 𝑅, and current 𝐼, 𝑉=𝐼𝑅.

Since we are trying to find a resistance, we want to make 𝑅 the subject. Dividing both sides by 𝐼 gives us 𝑅=𝑉𝐼.

We are trying to work out what value we should use for the shunt resistance 𝑅. So, let’s replace 𝑅 in Ohm’s law with this shunt resistance 𝑅. We should also use the potential difference across the shunt resistor, 𝑉, in place of 𝑉 and the current through the shunt resistor, 𝐼, in place of 𝐼.

Making these substitutions gives us 𝑅=𝑉𝐼.

We can make some substitutions in this equation in order to make it a bit more useful. First off, we have already said that the potential difference across the shunt resistor, 𝑉, is equal to the potential difference across the galvanometer, 𝑉. So, we can replace 𝑉 by 𝑉 in our equation: 𝑅=𝑉𝐼.

We also know that the total current in the circuit, 𝐼, splits into the two currents 𝐼 and 𝐼 such that 𝐼=𝐼+𝐼. Alternatively, subtracting 𝐼 from both sides of this, we have that 𝐼=𝐼𝐼.

Substituting this expression for 𝐼 into our Ohm’s law equation, we get 𝑅=𝑉𝐼𝐼.

Now, we can use Ohm’s law again in order to replace 𝑉, the potential difference across the galvanometer. We know that the current through the galvanometer is 𝐼 and the resistance of the galvanometer is 𝑅. Ohm’s law tells us that 𝑉=𝐼𝑅.

Substituting this in for 𝑉 in our expression for 𝑅 gives us 𝑅=𝐼𝑅𝐼𝐼.

One way to understand what this equation means is to consider a specific value of current in the galvanometer. In particular, there will be a value of this current 𝐼 that gives a full deflection of the galvanometer’s needle. This value is the full-scale deflection current.

We also know that 𝐼 is a small but constant proportion of the total current 𝐼. So, this maximum value of 𝐼 that can be recorded corresponds to a maximum value of the current 𝐼 that can be measured by the ammeter.

With this in mind, we arrive at the following interpretation of our equation for 𝑅.

Equation: The Resistance of a Shunt Resistor in an Ammeter

Suppose we have an ammeter consisting of a galvanometer with a full-scale deflection current 𝐼 and a resistance 𝑅 connected in parallel with a shunt resistor.

In order to be able to measure a maximum current of 𝐼 using this ammeter, we must use a shunt resistor with a resistance 𝑅 given by 𝑅=𝐼𝑅𝐼𝐼.

Now let’s have a look at an example problem.

Example 2: Finding the Required Resistance of a Shunt Resistor in an Ammeter

A galvanometer has a resistance of 15 mΩ. A current of 125 mA produces a full-scale deflection of the galvanometer. Find the resistance of a shunt that when connected in parallel with the galvanometer, allows it to be used as an ammeter that can measure a maximum current of 12 A. Answer to the nearest microhm.

Answer

Let’s begin by drawing a circuit diagram and labeling the values that we have been given.

We have explicitly drawn in a resistor representing the resistance of the galvanometer. This resistance is 𝑅=15mΩ. We know that the full-scale deflection current of the galvanometer is 𝐼=125mA, and we want to use this setup as an ammeter to measure a maximum current of 12 A.

We are asked to work out what value of the shunt resistance, which we have labeled 𝑅, will allow us to measure this maximum current.

We can recall that we know an equation for the required shunt resistance 𝑅 in order to be able to measure a maximum current 𝐼, given an ammeter using a galvanometer with full-scale deflection current 𝐼 and resistance 𝑅: 𝑅=𝐼𝑅𝐼𝐼.

Before substituting in our values for the quantities on the right-hand side, we need to convert them so that they all have consistent units. If we measure the currents in amperes and the galvanometer’s resistance in ohms, then we will get a shunt resistance with units of ohms.

The value of 𝐼=12A is already given in amperes, so we just need to convert 𝐼 and 𝑅.

We have that 𝐼=125=0.125mAA and 𝑅=15=1.5×10mΩΩ.

Substituting in our values for 𝐼, 𝑅, and 𝐼, we get that 𝑅=(0.125)×1.5×10120.125.AΩAA

Evaluating the right-hand side of this expression, we find that 𝑅=1.5789×10.Ω

Here, an ellipsis is used to indicate that there are further decimal places.

Finally, we note that the question asks us for our answer in units of microhms, to the nearest microhm. Recalling that 1=10µΩΩ, we can give our answer for the required shunt resistance as 𝑅=158.µΩ

We can also take our equation for the shunt resistance 𝑅 and look at it in another way.

If we rearrange the equation to make the current 𝐼 the subject, then we get an equation that tells us the maximum current we can measure with an ammeter, given the properties of its components.

Let’s see how we can make 𝐼 the subject. Recall that we start with the following equation: 𝑅=𝐼𝑅𝐼𝐼.

We want to get the current 𝐼 out of the denominator of the fraction, so we will begin by multiplying both sides of the equation by 𝐼𝐼: 𝑅(𝐼𝐼)=𝐼𝑅(𝐼𝐼)𝐼𝐼𝑅(𝐼𝐼)=𝐼𝑅.

In the second line, we have canceled the 𝐼𝐼 term that appears in the numerator and denominator on the right-hand side.

Next, we divide both sides of the equation by 𝑅: 𝐼𝐼=𝐼𝑅𝑅.

Finally, we add 𝐼 to both sides: 𝐼=𝐼𝑅𝑅+𝐼.

Equation: The Measurement Range of an Ammeter

Suppose we have an ammeter consisting of a galvanometer with a full-scale deflection current 𝐼 and a resistance 𝑅 connected in parallel with a shunt resistor of resistance 𝑅.

The maximum current that can be measured using this ammeter, also known as the measurement range of the ammeter, is then given by 𝐼=𝐼𝑅𝑅+𝐼.

Let’s finish up by taking a look at a couple more example questions.

Example 3: Calculating the Measurement Range of an Ammeter

A galvanometer has a resistance of 12 mΩ. A current of 150 mA produces full-scale deflection of the galvanometer. A shunt is connected in parallel with the galvanometer to convert it into an ammeter. The resistance of the shunt is 70 µΩ. What is the greatest current that the ammeter can measure? Answer to one decimal place.

Answer

Let’s begin by drawing a circuit diagram and labeling the values that we have been given

We have explicitly drawn in a resistor representing the resistance of the galvanometer. This resistance is 𝑅=12mΩ. We know that the full-scale deflection current of the galvanometer is 𝐼=150mA and that the resistance of the shunt resistor is 𝑅=70µΩ.

We are asked to work out what the greatest current is that this ammeter can measure.

We can recall that we have an equation for the maximum current that an ammeter can measure given the properties of its components: 𝐼=𝐼𝑅𝑅+𝐼.

We know the values of all of the quantities on the right-hand side of this equation.

However, before substituting in our values, we need to make the units compatible. If we use units of amperes for 𝐼 and units of ohms for both resistances 𝑅 and 𝑅, then we will get a current 𝐼 in units of amperes.

Converting our values into these units, we have that 𝐼=150=0.15mAA, 𝑅=12=1.2×10mΩΩ, and 𝑅=70=7×10µΩΩ.

Substituting in these values into our expression for 𝐼, we get that 𝐼=(0.15)×1.2×107×10+0.15.AΩΩA

Evaluating the right-hand side gives us that 𝐼=25.864,A where the ellipsis indicates that the value has further decimal places.

Finally, we note that we are asked to give our answer to one decimal place. Rounding to one decimal place, we have that the greatest current that the ammeter can measure is given by 𝐼=25.9.A

Example 4: Calculating the Currents through the Galvanometer and Shunt Resistor in an Ammeter

The current 𝐼 in the circuit shown is 2.5 mA, which is the greatest current that can be measured using the ammeter connected to the circuit. The resistance of the galvanometer is ten times the resistance of the shunt.

  1. Find 𝐼, the current through the galvanometer. Answer to the nearest microampere.
  2. Find 𝐼, the current through the shunt. Answer to two decimal places.

Answer

Part 1

We are given a circuit that acts as an ammeter. We are asked to work out the current 𝐼 through the galvanometer, given that the maximum current that this ammeter can measure is 𝐼=2.5mA.

We are not told the actual values of the resistances of the galvanometer and shunt resistor. However, we are told that the galvanometer’s resistance, which we will call 𝑅, is ten times the shunt resistance, which we will call 𝑅. In other words, we have that 𝑅=10𝑅.

We can recall that we know an equation linking the current 𝐼 in a circuit, such as the one in this question, and the current 𝐼 through the galvanometer: 𝐼=𝐼𝑅𝑅+𝐼.

We also know that, for this ammeter, 𝑅=10𝑅. So, we can replace 𝑅 by 10𝑅 in our expression for 𝐼 to give us that 𝐼=𝐼×(10𝑅)𝑅+𝐼.

We can now cancel 𝑅 in the numerator of the fraction on the right-hand side with 𝑅 in the denominator of this fraction: 𝐼=10𝐼+𝐼=11𝐼.

Since in this case we know the current 𝐼 and we want to find the value of 𝐼, we can make 𝐼 the subject by dividing both sides of this equation by 11: 𝐼=𝐼11.

Finally, substituting in 𝐼=2.5mA gives us our result for 𝐼, the current through the galvanometer: 𝐼=2.511=0.227𝐼=227.mAmAμA

Here, we have given our answer in microamperes to the nearest microampere, as requested by the question.

It is worth noticing that, since we were told that 𝐼 is the maximum current that can be measured by this ammeter, then we know that 𝐼 is the full-scale deflection current of the galvanometer.

Part 2

We were told in the question that the current in the circuit is 𝐼=2.5mA. We have also now worked out that the current in the galvanometer is 𝐼=227=0.227μAmA.

This second part of the question is asking us to find the current 𝐼 through the shunt resistor.

We can recall that the total current into a junction must be equal to the current out of it.

Looking back at the diagram in the question, we see that we have a current 𝐼 into the junction, as well as currents 𝐼 and 𝐼 each coming out of the junction along different branches. So, the current 𝐼 is getting split at the junction into the currents 𝐼 and 𝐼.

In other words, we know that we must have 𝐼=𝐼+𝐼.

We are trying to find the value of 𝐼, so we should rearrange this equation by subtracting 𝐼 from both sides to make 𝐼 the subject: 𝐼=𝐼𝐼.

When substituting in values for 𝐼 and 𝐼, we need to make sure we use the same units for both. Using units of milliamperes, we have that 𝐼=2.5mA and 𝐼=0.227mA. Substituting these values in gives us our result for the current 𝐼 through the shunt resistor: 𝐼=2.50.227𝐼=2.27.mAmAmA

We have given our answer to two decimal places, as requested by the question.

Finally, let’s summarize what we have learned in this explainer.

Key Points

  • An ammeter can be made by connecting a galvanometer and a resistor, known as a shunt resistor, in parallel.
  • To make an ammeter in this way, the resistance 𝑅 of the shunt resistor must be much less than the resistance 𝑅 of the galvanometer: 𝑅𝑅.
  • If we have an ammeter consisting of a galvanometer with a full-scale deflection current of 𝐼 and a resistance of 𝑅 connected in parallel with a shunt resistor, then in order to be able to measure a current 𝐼, the required value of the shunt resistance 𝑅 is given by 𝑅=𝐼𝑅𝐼𝐼.
  • We can rearrange this equation to give us an expression for the maximum current 𝐼 that we can measure with a particular ammeter, given the properties of its components: 𝐼=𝐼𝑅𝑅+𝐼.

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