### 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.