### Video Transcript

In this video, we’re going to be
talking about the design of the ammeter, which is a device we use to measure current
in a circuit. We’ll see how we can make an
ammeter using a galvanometer and a resistor known as a shunt resistor. And we’ll see how to calculate the
required resistance of the shunt resistor based on the properties of the
galvanometer and the size of the current that we want to measure. Since ammeter design is based
around a galvanometer, let’s start by reminding ourselves how a galvanometer
behaves.

A galvanometer is an
electromechanical device which responds to the direction and magnitude of
current. If we apply a potential difference
to a galvanometer, then the resulting current causes the needle on the galvanometer
to deflect towards one end of the scale. If we then reverse the polarity of
this potential difference, that’s reversing the direction of the current, then the
needle on the galvanometer deflects toward the other end of the scale. At the moment, we can see that the
needle is pointed fully toward one end of the scale. We can say that it has maximum
deflection.

But if we reduce the size of the
potential difference, that’s reducing the size of the current, then we see that the
needle backs away from the end of the scale. We actually find that for small
currents, the deflection of the galvanometer needle is proportional to the magnitude
of the current. This means that a galvanometer can
effectively measure the size of a current as long as that current is small
enough. Typically, we find that a
galvanometer will reach full-scale deflection for a current in the microamp or
milliamp region.

So if this galvanometer reached
full deflection for a current of magnitude 100 microamps in either direction, then
it would be able to accurately measure the size and direction of the current as long
as it was smaller than 100 microamps. So it looks like for small
currents, we could just use a galvanometer as an ammeter. And if we could find a way to
extend the range of the galvanometer, then we could use it for larger currents as
well. However, if we tried to just use a
galvanometer as an ammeter, we run into a problem due to the fact that a
galvanometer has its own internal resistance.

To show why we can’t just use a
galvanometer to act as an ammeter, let’s consider this simple circuit containing
just a cell and a resistor. Here, the cell provides some
voltage 𝑉 across our resistor, which we can say has resistance 𝑅. And this creates a current which
we’ll call 𝐼. And Ohm’s law tells us that the
size of our current is equal to the size of our voltage 𝑉 divided by the size of
our resistance 𝑅. So now, let’s see what happens if
we introduce a galvanometer into a circuit and try to use that to measure the
current 𝐼. And to make things easy for
ourselves, we’ll say that 𝑉 and 𝑅 are such that 𝐼 is less than the full-scale
deflection current of the galvanometer.

Okay, so here’s our galvanometer
connected in series with the other components. Now, unfortunately, we run into a
problem here because the galvanometer has its own resistance. This means that as soon as we
connect it to the circuit, it changes the overall resistance of the circuit. And so it changes the current that
we were trying to measure in the first place. So this obviously isn’t very
useful. It’s almost like having a ruler
that changes the length of an object when we try to measure it. We can make this explicit in our
circuit diagram by drawing in an additional resistor with resistance 𝑅 G, that is,
the resistance of our galvanometer. And this makes it clear that
effectively there are now two resistors in our circuit.

At this point, it’s useful to
recall that when we have resistors connected in series, the total resistance, which
we could call 𝑅 T, is equal to the sum of the individual resistances involved. For clarity, let’s say that this
original resistor has a resistance 𝑅 one. We can then write that the total
resistance of the circuit 𝑅 T is given by 𝑅 one plus 𝑅 G. And we can see how this
affects the current in the circuit by applying Ohm’s law to the circuit as a
whole. Now, the potential difference
applied to the circuit hasn’t changed; that’s still 𝑉. But the resistance is now equal to
𝑅 one plus 𝑅 G. And since 𝑅 one plus 𝑅 G is bigger than just 𝑅 one, this means
that we’re dividing 𝑉 by a bigger number, which means that the current 𝐼 is
smaller.

Fortunately, there is a way to deal
with the problem of the galvanometer changing the current in a circuit. The way that we do this is to add a
second resistor in parallel with the galvanometer. When we add a parallel branch to a
circuit in this way, we find that the current is split between the two branches. Some of the current will still pass
through the galvanometer, but the rest of it will go around the galvanometer through
this new resistor. By giving charge an alternative
route to follow around the circuit, we actually reduce the resistance of the circuit
as a whole. So adding this resistor decreases
the overall resistance of everything within this yellow box.

Because the effect of this new
resistor is that some of the current is now shunted around the galvanometer, we call
it the shunt resistor. And we can say it has a resistance
of 𝑅 S. If we choose a shunt resistance so that its resistance is much less than
that of the galvanometer, that is, we choose 𝑅 S so that it’s much less than 𝑅 G,
then we find that the majority of the current goes through the shunt resistor
because it has less resistance and only a small amount of current goes through the
galvanometer. Because most of the charge now
flows along a path with a very low resistance, this means that everything in the
yellow box has an overall negligible effect on the circuit’s resistance, which means
that the current is barely affected.

In addition, because a small but
fixed proportion of the current goes through the galvanometer, that means the
galvanometer needle is proportional to the current in the circuit. So by carefully choosing the
resistance of this shunt resistor, we control how the current is split between the
two parallel branches. This means we can make sure that
the current in the galvanometer is high enough that the needle displays a clear
reading, but low enough that the needle doesn’t deflect fully. And this is the basic design
principle of an ammeter. When building an ammeter in this
way, it’s important that the resistance of the shunt resistor is carefully chosen to
get the best results. And to work out the best value for
this resistance, we can use Ohm’s law.

So because we’re trying to find a
resistance, let’s rearrange Ohm’s law to make 𝑅 the subject. This gives us 𝑅 equals 𝑉 over
𝐼. And because we want to find the
resistance of the shunt resistor 𝑅 S, that means that 𝑉 in our equation should be
the voltage that’s applied across the shunt resistor — we can call this 𝑉 S — and
𝐼 in our equation will be the current in the shunt resistor; we can call this 𝐼 S.
Let’s label 𝐼 S in our circuit diagram and we can also label 𝐼 G, the current in
the galvanometer. Now, we can make some substitutions
into this expression to make it a bit more useful.

Firstly, we can recognize that the
voltage across the shunt resistor, that’s 𝑉 S, is the same as the voltage applied
across the galvanometer. We know this because the two
parallel branches are connected to the rest of the circuit at the same points. So there’s a fixed potential
difference between these two points. So in our expression, we can
replace 𝑉 S with 𝑉 G, the voltage across the galvanometer. Secondly, because the incoming
current 𝐼 is split into two parts, 𝐼 G and 𝐼 S, we can say that 𝐼 equals 𝐼 G
plus 𝐼 S. Subtracting 𝐼 G from both sides of this expression gives us 𝐼 minus 𝐼
G equals 𝐼 S. And we can substitute this into our expression to give us 𝑅 S equals
𝑉 G over 𝐼 minus 𝐼 G.

Our final step in this derivation
is to make a substitution for 𝑉 G, the voltage across the galvanometer. Rearranging Ohm’s law to make 𝑉
the subject gives us 𝑉 equals 𝐼 𝑅, which means that 𝑉 G, the voltage across the
galvanometer, is equal to 𝐼 G, the current in the galvanometer, multiplied by 𝑅 G,
the resistance of the galvanometer. Substituting this in place of 𝑉 G
in our expression, we’re left with 𝑅 S equals 𝐼 G 𝑅 G divided by 𝐼 minus 𝐼 G,
where 𝑅 S is the resistance of the shunt resistor, 𝐼 G is the current in the
galvanometer, 𝑅 G is the resistance of the galvanometer, and 𝐼 is the current in
the circuit.

Now, in practice, ammeters have a
maximum value of 𝐼 that they can measure. And for this value of 𝐼, we expect
the needle on the ammeter to be fully deflected. In other words, for our ammeter’s
maximum rated current of 𝐼, we want the current in the galvanometer 𝐼 G to be the
full deflection current of the galvanometer. Considering this specific case
gives us a different way of interpreting the variables in this equation. We can think of 𝐼 as being the
maximum current which our ammeter can measure, in other words, the range of our
ammeter. And we can think of 𝐼 G as the
full deflection current of our galvanometer, in other words, the range of our
galvanometer.

So the equation now tells us the
shunt resistance that we need to use in order to be able to measure a current 𝐼,
given a galvanometer with a maximum deflection current of 𝐼 G and the resistance of
𝑅 G. We can also rearrange this equation to give us a formula for the maximum range
of an ammeter based on the properties of its components. To do this, we just need to make 𝐼
the subject. So we can start by multiplying both
sides of the equation by the denominator on the right-hand side, then divide both
sides by 𝑅 S, and, finally, add 𝐼 G to both sides to give us 𝐼 equals 𝐼 G times
𝑅 G over 𝑅 S plus 𝐼 G.

This formula tells us the range of
our ammeter 𝐼 in terms of the range of our galvanometer 𝐼 G, the resistance of the
galvanometer 𝑅 G, and the resistance of the shunt resistor 𝑅 S. Now, there’s one
more important thing to note when we build an ammeter in this way. The circuit we’re shown within the
yellow box is a direct current or DC ammeter, meaning it measures the magnitude of a
current in one direction. However, most galvanometers are
made to indicate the magnitude of a current flowing in either direction. That means that the zero tends to
be in the middle and the needle will deflect either to the left or the right,
depending on the direction of current.

Because here we’re building a DC
ammeter, we’re only interested in current going in one direction. That means we effectively only want
to use half of the scale. We can then calculate the maximum
deflection current, which goes here on the scale, using the formula at the top left
of the screen. This results in an ammeter which
accurately measures current in one direction. Okay, now that we’ve looked at the
principles behind ammeter design, let’s have a go at answering a question.

Which of the following circuit
diagrams most correctly represents a galvanometer combined with a shunt resistor
being used as an ammeter to measure the current through a circuit that has a direct
current source?

So here we can see we’ve been given
three different circuit diagrams to choose between. And each one contains a
galvanometer which is represented by a G inside a circle. Each of the circuits also contains
a cell, which is a type of direct current source. And each of the circuits also
contains either one or two resistors connected in various ways. The question asks us to identify
which one of these circuits shows a galvanometer combined with a shunt resistor
being used as an ammeter. And an ammeter is, of course, a
device which measures current.

The term shunt resistor refers to a
resistor with a special function inside an ammeter. But it’s important to note that a
shunt resistor is really just a normal resistor. So let’s start by recalling that a
galvanometer is a device which indicates the magnitude and direction of a current
within it. Galvanometers generally have a dial
with a zero in the middle. And a current passing through the
galvanometer will cause the needle to deflect. Now up to some maximum current, the
deflection of the galvanometer’s needle is proportional to the current passing
through it. So reducing the size of the current
will reduce the deflection of the needle. And reversing the direction of the
current will cause the needle to deflect in the other direction.

Since the galvanometer responds to
current in a predictable way, it seems reasonable to suggest that we could just use
a galvanometer as an ammeter. However, there are two main
problems with this. The first problem is that
galvanometers are very sensitive, which means that the maximum current that they can
indicate in either direction tends to be in the microamp or the milliamp region. The second problem with trying to
use a galvanometer as an ammeter is that galvanometers have their own internal
resistance, which is why in circuit diagrams, we often see galvanometers
represented, not just by a G in a circle, but by an additional resistor as well with
a resistance 𝑅 G.

The presence of this additional
resistance means that when we connect a galvanometer to a circuit, it can
drastically affect the resistance of the circuit as a whole, which means it changes
the current that we’re trying to measure. We can see this effect in action if
we consider circuit diagram (A). Now, in this circuit diagram, we
simply have a cell, a galvanometer, and a resistor connected in series.

Now, since any circuit that we’re
trying to measure using a galvanometer must have initially had some resistance of
its own before we added the galvanometer, we can assume that the circuit must have
looked like this before the galvanometer was introduced. So there’s a cell applying some
voltage, which we’ll call 𝑉, to a resistor, which we’ll say has resistance 𝑅
one. And we know that this must produce
a current 𝐼 in accordance with Ohm’s law, which says that the current through a
resistor 𝐼 is equal to the voltage across that resistor 𝑉 divided by the
resistance of that resistor 𝑅.

But if we want to measure the size
of this current, then wiring in a galvanometer like this is actually a bad way of
doing it. This is because of the fact that a
galvanometer has its own resistance 𝑅 G. When we connect to resistance in series,
the total resistance of those resistors is equal to the sum of that individual
resistances. This means that before we connected
the galvanometer to our circuit, the total resistance of the circuit was just 𝑅
one. But after we’ve connected the
galvanometer, the total resistance is 𝑅 one plus 𝑅 G. This means that the current
changes as well according to Ohm’s law. So just connecting a galvanometer
in series with the other components will change the current that we’re trying to
measure. So we know that (A) is not the
correct answer.

Now, we can actually prevent the
problem of the galvanometer’s resistance contributing to the overall resistance of
the circuit by introducing another resistor known as a shunt resistor. Connecting this resistor in
parallel with the galvanometer means that the incoming current is split between the
two parallel branches. We could say that some of the
current is shunted around the galvanometer, which is why this resistor is known as
the shunt resistor. And we can say that it has a
resistance 𝑅 S. Now, the relative magnitudes of the current through the
galvanometer, which we can call 𝐼 G, and the current in the shunt resistor, which
we can call 𝐼 S, depends on the relative resistances of the galvanometer and the
shunt resistor.

The majority of the current will go
down the path of least resistance. So if the shunt resistor had a
larger resistance compared to the galvanometer, then we’d find most of the current
goes through the galvanometer, which wouldn’t really change how the circuit
behaved. However, if we make sure that the
resistance of the shunt resistor is much smaller than the resistance of the
galvanometer, then we find that the majority of the current passes through the shunt
resistor. And only a small amount of current
passes through the galvanometer.

This is ideal for two reasons. Firstly, because the majority of
the current goes through the shunt resistor, but the shunt resistor only has a very
small resistance, this means that the current 𝐼 is barely affected. And secondly, because a small
proportion of the current now passes through the galvanometer, we can use the
galvanometer to indicate the overall current 𝐼 in the circuit. And by carefully choosing the
resistance of the shunt resistor, we can make sure that the current in the
galvanometer never exceeds the maximum deflection current, which actually solves our
original problem of the galvanometer being too sensitive for high currents.

Indeed, connecting a shunt resistor
in parallel with a galvanometer like this is how ammeters are made. So we could consider everything
within this pink box to be equivalent to an ammeter. Now, if we look at the two
remaining answer options (B) and (C), we can see that both of these circuit diagrams
effectively contain an ammeter, that is, as long as the resistances of these
resistors are significantly lower than the resistances of the galvanometers. However, if we consider option (B),
we can see that if everything in the pink box is an ammeter, then this is just an
ammeter connected to a cell as there are no other resistors in the circuit
diagram.

Now, having a cell in a circuit
with no resistance isn’t a very realistic situation, since all circuits have some
resistance. So we have to assume that the
original circuit looked like this. And in an attempt to measure the
current in this resistor, a galvanometer was wired in, in parallel like this. However, connecting a galvanometer
in this way would split the current into the two parallel branches, once again
changing the current that we were trying to measure in the first place. In fact, the only sensible answer
to this question is shown in circuit diagram (C), where an assembly consisting of a
galvanometer and a resistor connected in parallel appears to have been connected to
an original circuit consisting of a cell and a resistor connected in series.

Attaching a galvanometer and a
shunt resistor to the circuit like this means that the current in the original
circuit is basically unchanged. But a very small proportion of this
current passes through the galvanometer which causes a deflection of the needle
which we can use to infer the current 𝐼. So the correct answer to this
question is option (C).

Now, let’s recap some of the key
points we’ve learned in this lesson. Firstly, we’ve seen that an ammeter
can be made by connecting a galvanometer and a resistor, known as a shunt resistor,
in parallel to each other. The resistance of the shunt
resistor 𝑅 S must be much smaller than the resistance of the galvanometer 𝑅 G as
represented by this inequality. Finally, we’ve seen that if we have
a galvanometer with a maximum measurable current 𝐼 G and a resistance 𝑅 G, then to
construct an ammeter with a range 𝐼, we’d need to use a shunt resistor with a
resistance 𝑅 S given by this expression. We can also rearrange this
expression like this to calculate the range of a given ammeter. This is a summary of the design of
the ammeter.