### Video Transcript

A circuit that can be used as an
ohmmeter is shown. The circuit uses a galvanometer, a
direct-current source with a known voltage, a fixed resistor, and a variable
resistor. Which of the following most
correctly states how to calibrate the circuit to directly measure the circuit’s
total resistance? (A) Adjust the resistance of the
variable resistor until it is equal to the mean of the resistance of the fixed
resistor and the resistance of the galvanometer. (B) Adjust the resistance of the
variable resistor until it is equal to the sum of the resistance of the fixed
resistor and the resistance of the galvanometer. (C) Adjust the resistance of the
variable resistor until it is equal to the difference between the resistance of the
fixed resistor and the resistance of the galvanometer. (D) Adjust the resistance of the
variable resistor until the galvanometer’s arm is at a full-scale deflection
position. (E) Adjust the resistance of the
variable resistor until the galvanometer’s arm is at the zero-deflection
position.

In this example, our circuit
consisting of a galvanometer, a device for measuring current, a fixed resistor here,
a variable resistor here, and a constant voltage supply here is intended to work as
an ohmmeter, a device for measuring resistance.

The idea then is if we had a
resistor of unknown resistance, we can connect it to this circuit and then solve for
its resistance. For that to work though, as our
problem statement tells us, this circuit needs to first be calibrated. This is possible because the
circuit involves a variable resistor. This is a resistor whose resistance
can be changed. And we see that all of our answer
options, including option (A) that we can’t see on the screen right now, depend on
this step of adjusting the resistance of this variable resistor.

So the question is, how do we
adjust this resistor’s resistance so that the circuit does indeed function as an
ohmmeter? The answer has to do with our
galvanometer, the device in this circuit for measuring current. Typically, a galvanometer reads out
the measured current on a scale, like this. The scale has a minimum value of
current, zero, and some maximum value that we’ve called 𝐼 sub 𝑔. The galvanometer is capable of
accurately measuring current anywhere between and including these values.

As we’ve seen though, we don’t want
to use our circuit to measure current, but rather resistance. We can still do this using the
current setup by making use of Ohm’s law. This law tells us that the
potential difference across a circuit is equal to the current in that circuit
multiplied by the total circuit resistance.

In our circuit, we have a fixed
potential difference supply due to this cell. Given that the voltage 𝑉 is fixed,
we can think of the current 𝐼 and the total resistance 𝑅 in this circuit as
balancing one another out. That is, if due to a change in the
resistance of the variable resistor, the overall resistance of our circuit was
increased, then Ohm’s law would dictate that the current in the circuit must
decrease so that 𝐼 times 𝑅 is still equal to this fixed voltage 𝑉. The same thing holds true if the
total resistance decreases; then the current 𝐼 would increase in response.

Now, if we look back at our answer
options, several of them, including options (B) and (C), describe adjusting the
resistance of the variable resistor based on a relationship between the resistance
of the fixed resistor and that of the galvanometer. This shows us that in our circuit,
it’s not just the fixed and the variable resistors that have some resistance. The galvanometer does as well,
while for the cell, we’re treating it as an ideal voltage supply with effectively no
resistance.

Speaking though of the resistances
of the fixed resistor and the galvanometer, option (C) says that the resistance of
the variable resistor should be adjusted until it is equal to the difference between
these two resistances. Option (B) says that the variable
resistor should be adjusted until its resistance is equal to the sum of these two
resistances. And if we were to import a
shortened version of answer option (A), this option says that we should adjust the
resistance of the variable resistor until it is equal to the mean or the average of
the resistances of a fixed resistor and the galvanometer. However, the value to which we
should adjust the resistance of our variable resistor actually has nothing to do
with the resistance of our fixed resistor and rather depends only on the current
that is measured by our galvanometer.

Our circuit, recall, is intended to
measure the resistance of an unknown resistor. The only way you can do this is if
there’s some kind of measurement scale we can use that will indicate this
resistance. Only answer choices (D) and (E)
describe adjusting the resistance of the variable resistor with reference to the
galvanometer’s scale. These then are our only viable
options for developing a measurement device for resistance. We can cross out then answer
choices (A), (B), and (C) and then clear them from the screen.

We’ve seen that both answer choices
(D) and (E) involve adjusting the resistance of the variable resistor. Let’s look now at what makes these
choices different. Answer choice (E) involves
adjusting the resistance of the variable resistor until the galvanometer’s arm is at
the zero-deflection position. That’s the position that we see
here in our sketch. An important thing to realize about
a zero-deflection position is that that means that effectively zero current exists
in the circuit. But for zero current to exist in
the circuit, that would mean that 𝑅 would need to be effectively infinitely
big. In other words, we would have to
adjust the resistance of our variable resistor to a very, very high value. This may not be practical.

But actually there’s another reason
why we wouldn’t want our galvanometer’s arm to be at a zero-deflection position when
we use it for measuring resistance. Let’s say that we somehow had
adjusted the resistance of our variable resistor so that the arm on our galvanometer
indeed indicated zero current. If we were to take our unknown
resistor whose resistance we want to measure and added this resistor to the circuit,
then that addition, which would cause the overall resistance of the circuit
theoretically to increase, would have no measurable impact on the current measured
by our galvanometer. That current was already zero, so
it couldn’t possibly go any lower.

This would prevent us from working
back from the galvanometer’s reading to solve for the resistance of our unknown
resistor. By calibrating our circuit
according to answer option (E) then, we couldn’t use it as a device for measuring
resistance, an ohmmeter.

But let’s say that instead we
adjusted the resistance of the variable resistor in our circuit so that the arm on
our galvanometer was moved to its full-scale deflection position. Note that we would do this without
the resistor whose resistance we want to measure being inserted into the
circuit. So now, prior to serving as an
ohmmeter, we’ve tuned the overall resistance of our circuit 𝑅 so that the current
in the circuit 𝐼 is equal to the maximum measurable current according to our
galvanometer. Set up this way, if we now inserted
our unknown resistor into the circuit, that would cause the overall circuit
resistance to increase since we’ve added our resistor in series, which would result
in some measurable decrease in the overall current in the circuit, where that
decrease is measured by our galvanometer.

If the change in measured current
in the circuit due to the addition of our unknown resistor is Δ𝐼, then we could use
that measured change along with the voltage in our circuit to calculate a change in
resistance in the circuit Δ𝑅. It’s this change in resistance Δ𝑅
that is equal to the resistance of our unknown resistor. This is how our circuit can
function as an ohmmeter. And so for our answer, we choose
option (D).

The question may come up
though. When we tune the resistance of our
variable resistor, why do we do it so that the arm on the galvanometer is at a
full-scale deflection position? Why don’t we do it, for example, so
that the galvanometer’s arm is at a half-scale deflection position or some other
value? The reason we set up the resistance
of the variable resistor so that the galvanometer’s arm is at the full-scale
deflection position is that this allows us a maximum measurement range for the
resistance of an unknown resistor inserted into the circuit. If instead our galvanometer arm
were calibrated so that it’s at half the full-scale position, then we would only
have one-half of the galvanometer’s scale to work with as we measured unknown
resistances.

Better than this is having the full
scale of the galvanometer to work with. And this is why, as answer option
(D) says, we adjust the resistance of the variable resistor until the galvanometer’s
arm is at a full-scale deflection position.