### Video Transcript

A voltmeter is used to measure the
voltage of a direct-current source that is estimated to have a voltage of several
volts. The galvanometer in the voltmeter
has a resistance of a few milliohms. Which of the following correctly
explains why the multiplier resistor in such a voltmeter must have a resistance much
greater than the resistance of the galvanometer that the multiplier is connected in
series with? (A) If the multiplier resistor has
a resistance of a magnitude comparable with or less than that of the galvanometer,
the voltage of the source will be significantly increased. (B) If the multiplier resistor has
a resistance of a magnitude comparable with or less than that of the galvanometer,
the direction of the deflection of the galvanometer arm will reverse and no reading
will be displayed on the voltmeter. (C) If the multiplier resistor has
a resistance of a magnitude comparable with or less than that of the galvanometer,
the resistor will generate a magnetic field that significantly changes the
deflection of the galvanometer arm. And lastly, (D) if the multiplier
resistor has a resistance of a magnitude comparable with or less than that of the
galvanometer, the current through the galvanometer will be greater than the current
that would produce a full-scale deflection of the galvanometer arm.

In this situation, we’re thinking
about a voltmeter, which is made up of a galvanometer connected in series with a
resistor. This voltmeter we’re told is meant
to measure the voltage of a direct-current source, where that voltage is on the
order of a few volts. Along with this, we’re told that
the galvanometer in our voltmeter has a resistance of a few milliohms. Based on this setup, our question
wants us to identify the best reason why the resistance of this resistor here,
called a multiplier resistor, must be significantly greater than the resistance of
the galvanometer in order for this voltmeter to successfully measure voltages on the
order of several volts.

As we consider our answer choices,
in order to shorten them down a bit so we can fit them all on screen at the same
time, let’s note that all four of the possible answers begin with this phrase: “If
the multiplier resistor has a resistance of a magnitude comparable with or less than
that of the galvanometer.” We can effectively condense then
answer options (A), (B), and (C) by assuming that they begin with this phrase and
then finish up in the following ways. Answer option (A) says that if this
condition is met, the source voltage will significantly increase. (B) says that if the multiplier
resistor has a resistance of a magnitude comparable with or less than that of the
galvanometer, then the direction of the galvanometer arm deflection will reverse and
no voltmeter reading will result.

And then lastly option (C) says
that if this condition is met, the resistor will generate a magnetic field
significantly changing the galvanometer arm deflection. Knowing all this, let’s now
consider the display of our galvanometer, which we know is part of our
voltmeter. This display may look like
this. We recall that a galvanometer
measures current. And the scale of the galvanometer
is fairly small, meaning that the maximum current it can accurately read may be best
expressed in milliamperes.

Another trait of the galvanometer
is that it can differentiate current direction. What we see here is the reading
that would result from zero current existing in the circuit, whereas if current were
traveling in one direction, this would deflect the needle, say to the right, while
current in the other direction would deflect the needle in the opposite way. A galvanometer then is indeed
sensitive to the direction of current in the circuit.

We can understand how a
galvanometer indirectly can serve as a voltmeter by recalling Ohm’s law. This law says that if we know the
current 𝐼 in a circuit as well as the total circuit resistance 𝑅, then the product
of these values is equal to the voltage across the circuit 𝑉. Our particular circuit we could say
has a design specification. We want to be able to use this
voltmeter to measure voltages of several volts. This means that if 𝐼 is the
current in the circuit, and we know that current is the same everywhere because this
is a series circuit, and if 𝑅 is the total circuit resistance, then 𝐼 times 𝑅
based on Ohm’s law must be approximately equal to a few volts. That’s the requirement if we want
to accurately measure this source voltage.

Considering the current and
resistance in the circuit, we can say that based on the scale on our galvanometer,
even the greatest current magnitude that is readable on this device will be
expressed in milliamperes, perhaps say tens of milliamperes. In any case, we expect it to be
much less than one ampere. So that’s the maximum current we
can accurately measure. And then if we think about the
overall circuit resistance, that resistance is equal to the sum of the resistance of
the galvanometer and the resistance of our multiplier resistor that we’ll call 𝑅
sub m. We’re told in our problem statement
that the resistance of the galvanometer is several milliohms.

Now, for a moment let’s imagine
that there is no multiplier resistor in the circuit. In that case, by Ohm’s law, we’re
multiplying a current in milliamperes by a resistance in milliohms. And we can see that in doing that,
the product will never be as large as a few volts. Rather, it might be expressed
better in millivolts or even microvolts. To meet our design goal then of
measuring a few volts of potential difference, we indeed do need the resistance from
the multiplier resistor, and in fact this resistance has to be quite large. Its resistance has to overcome we
could say the smallness of the resistance of the galvanometer combined with the
relative smallness of the current that the galvanometer accurately can measure.

With this in mind, let’s now
consider our answer options, starting with choice (D). This choice says that if the
multiplier resistor has a resistance of a magnitude comparable with or less than
that of the galvanometer, in other words, on the order of milliohms or even smaller,
then the current through the galvanometer will be greater than the current that
would produce a full-scale deflection of the galvanometer arm. Let’s note that with our circuit
set up this way, whether we can measure it accurately or not, there is a potential
difference of several volts across the circuit. And saying that the current in our
circuit is on the order of milliamperes, we’ve already maxed out, so to speak, the
measurement range of our galvanometer.

If we then added a multiplier
resistor with a resistance also, say, on the order of milliohms, then that would fix
the overall resistance of our circuit at a relatively low value, which would mean
that the only way for the current 𝐼 multiplied by the resistance 𝑅 to equal the
voltage 𝑉 would be for the current 𝐼 in the circuit to significantly increase. The problem with this is we’ve
already assumed a current which fully deflects the measurement arm of our
galvanometer. So any increase in current will
push the arm past that point or otherwise lead to an inaccurate current
measurement.

On that basis, answer option (D)
looks correct. If the multiplier resistor did have
a resistance of a magnitude comparable with or less than that of the galvanometer,
the current through the galvanometer would indeed be greater than the maximum
current it can accurately measure.

Let’s now consider our remaining
answer options, starting with option (A). This response claims that if the
multiplier resistor has a resistance of a magnitude comparable with or less than
that of the galvanometer, the source voltage, that is, the voltage here supplied by
our direct-current source, will significantly increase. We know though that physically that
can’t happen. Nothing about the rest of the
circuit will impact the voltage supplied by our direct-current source. Answer option (A) then won’t be the
one that we choose.

Next, answer option (B) says that
if our multiplier resistor meets the condition named, that is, its resistance is
comparable to or less than that of the galvanometer, then the direction of the
galvanometer arm deflection will reverse. That would mean, for example, that
rather than deflecting to the right as we have that arm currently, the arm would
deflect to the left. As we’ve seen, that indicates
current pointing in the opposite direction. However, the direction of current
in the circuit depends on the orientation of our direct-current source. It doesn’t depend on the relative
resistances of our multiplier resistor and our galvanometer. Therefore, the multiplier resistor
having a resistance comparable to or less than that of the galvanometer will not
have the effect described in answer choice (B).

Lastly, in answer choice (C), we
read that if the multiplier resistor has a resistance of a magnitude comparable with
or less than that of the galvanometer, the resistor, the multiplier resistor, will
generate a magnetic field and that this will significantly change the galvanometer
arm deflection. If our multiplier resistor were not
a resistor but in fact were an inductor, we could expect that largely increasing the
current through it would indeed generate a large magnetic field. But because we’re working with a
resistor rather than an inductor, we have no such expectation of a magnetic field
being formed by this component. For that reason then, we also won’t
choose answer option (C).

For our final answer then, we will
choose option (D). The best explanation for why the
resistance of our multiplier resistor must be much greater than that of our
galvanometer is that if the multiplier resistor had a resistance of a magnitude
comparable with or less than that of the galvanometer, the current through the
galvanometer will be greater than the current that would produce a full-scale
deflection of the galvanometer arm.