### Video Transcript

In this video, weโre 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 ๐ inside a circle. And in this circuit, the voltmeter
is being used to measure the voltage drop across this resistor. In this video, weโll see how we can
construct a voltmeter by using a galvanometer and a resistor connected in
series. Weโll 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, letโs
consider a cell. And letโs say that this cell has a
certain voltage ๐ 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. Letโs 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 ๐ผ. This causes the needle on the
galvanometer to deflect. And as long as the current isnโt
too big, the deflection will be proportional to the current.

Now, Ohmโs 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 ๐
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 Ohmโs 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 ๐ผ G, then it means it will reach maximum deflection for a
voltage equal to ๐ผ G times ๐
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 galvanometerโs needle from reaching maximum deflection.

Fortunately, thereโs 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
๐
M. The reason itโs 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 Ohmโs law to our voltmeter as a whole. Ohmโs 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, itโs 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, ๐
V equals ๐
M
plus ๐
G. So overall we can write ๐ equals ๐ผ G times ๐
M plus ๐
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 ๐
๐ the subject. To do this, we start by multiplying
out the brackets on the right-hand side of the expression to give us ๐ equals ๐ผ G
๐
M plus ๐ผ G ๐
G. We can then subtract ๐ผ G times ๐
G from both sides of the
equation and then finally dividing both sides of the equation by ๐ผ G.

Finally, weโll just swap around the
left and right sides of this expression to give us ๐
M equals ๐ over ๐ผ G minus ๐
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 ๐, using a galvanometer with a
resistance ๐
G and a full deflection current ๐ผ 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 weโre 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 weโre 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, weโve shown in this
equation that if our galvanometer has a maximum deflection current ๐ผ G, then the
voltmeter that we build using this galvanometer will have a maximum deflection
voltage ๐. We can use this expression to
calculate the value of ๐ that we would write in place of ๐ผ G on the galvanometerโs
dial.

For example, if weโre using a
galvanometer that have a resistance ๐
G of 100 ohms and a full deflection current
๐ผ G of 100 microamps and weโre using a multiplier resistor with a resistance equal
to five kiloohms, then the range of our voltmeter ๐ would be given by 100 times 10
to the negative six amps, thatโs ๐ผ G, multiplied by 5000 ohms, thatโs ๐
M, plus
100 ohms. Thatโs ๐
G, which works out at
0.51 volts, which we could then write at the maximum deflection position on the
voltmeterโs dial. So once weโve chosen the value of
our multiplier resistor, connected it in series with the galvanometer, and
calibrated the scale, our voltmeterโs ready to be used.

Of course, measuring the voltage of
a cell isnโt 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 letโs say that these resistors
have resistances of ๐
one and ๐
two, respectively. Now, in this circuit, the cell
provides a voltage, which weโll call ๐, and this creates a current, which weโll
call ๐ผ. Now, if we were analyzing a circuit
that looked like this, it might be useful to measure the voltage thatโs dropped
across each of these resistors.

And to measure the voltage drop
across ๐
one, for example, we would attach a voltmeter in parallel with ๐
one. And of course, we now know that a
voltmeter is essentially comprised of a multiplier resistor with a resistance ๐
M
and a galvanometer with a resistance ๐
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.

Letโs say that the current flowing
through the voltmeter is called ๐ผ V and the current going through resistor ๐
one
is called ๐ผ 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 ๐
one. And once again, Ohmโs 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 weโre trying
to measure, which obviously isnโt 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 ๐ผ V is very small, this means that the current through the resistor, ๐ผ
R, is approximately equal to the current in the rest of the circuit, ๐ผ. The upshot of this is that
connecting a voltmeter in parallel with ๐
one makes only a negligible change to the
current in ๐
one, therefore making a negligible change to the voltage across ๐
one. So now that weโve seen how a
voltmeter is designed and used, letโs 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, ๐ผ G, to the current in the multiplier resistor, ๐ผ M?

So in this question, weโve been
given a circuit diagram that shows a galvanometer and a resistor called a multiplier
resistor connected in series with a cell. Letโs start by recalling that the
term multiplier resistor describes a resistor used in the construction of a
voltmeter. Specifically, itโs 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 theyโre 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 ๐
M, is much
greater than the resistance of the galvanometer, which we can call ๐
G. As we can
see, the same is true in this question. Weโre told that the multiplier
resistor has a resistance 50 times that of the galvanometer.

Weโre then asked to calculate the
ratio of the current in the galvanometer ๐ผ G to the current in a multiplier
resistant ๐ผ M. So letโs start by writing down an expression for each of these
currents in terms of their resistances, which weโve been given some information
about. We can do this using Ohmโs 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 ๐ผ G is equal to the voltage across the galvanometer, which we
could call ๐ G, divided by the resistance of the galvanometer, which is ๐
G.

Similarly, we can say that the
current in the multiplier resistor ๐ผ M is equal to the voltage across the
multiplier resistor, which weโll call ๐ M, divided by the resistance of the
multiplier resistor ๐
M. Now itโs really important to note that ๐ G and ๐ M are
not necessarily the same. Itโs tempting to assume that each
of these voltages is simply the same as the voltage supplied by the cell, which we
could call ๐ C. However, this isnโt 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 ๐ผ G to the current in the
multiplier resistor ๐ผ M. And one way of expressing the ratio of ๐ผ G to ๐ผ M is to
calculate ๐ผ G over ๐ผ M, which must be equal to ๐ G over ๐
G divided by ๐ M over
๐
M. Dividing this fraction by this fraction is the same as multiplying this
fraction by the reciprocal of this fraction, which gives us ๐ G over ๐
G times ๐
M over ๐ M, which is equivalent to ๐ G ๐
M over ๐ M ๐
G.

Now, the question tells us that the
multiplier resistor has a resistance 50 times that of the galvanometer. In other words, ๐
M equals 50 ๐
G. This means that we can substitute 50 ๐
G in place of ๐
M in this expression,
which then enables us to cancel the common factor of ๐
G in the numerator and the
denominator, leaving us with 50 ๐ G over ๐ M. Okay, so this simplifies our
expression. However, we still donโt have a
numerical value for this ratio. And weโre not able to calculate the
actual values of ๐ G and ๐ 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 weโre 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, ๐ M, is 50 times the
voltage drop over the galvanometer, ๐ G. Substituting 50 ๐ G in place of ๐ M in
our expression tells us that the ratio of ๐ผ G to ๐ผ M is 50 ๐ G over 50 ๐ 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 doesnโt require us to use any algebra. In fact, we donโt 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, itโs 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 ๐ผ G must be the same as the current in the
multiplier resistor ๐ผ M. And if ๐ผ G equals ๐ผ M, then ๐ผ G over ๐ผ 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.

Letโs finish by recapping the key
points that weโve 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 thatโs being measured. Weโve also seen that to build a
voltmeter with a voltage range ๐, using a galvanometer with a resistance ๐
G and a
full deflection current ๐ผ G, the required resistance ๐
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.