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. The resistance of the variable resistor is adjusted until the galvanometer’s arm is at a full-scale deflection position. The circuit is to be used to find the resistance of a resistor that has an unknown resistance. The resistor with the unknown resistance must be connected to the circuit. In which of the following ways should the resistor be connected? (A) In parallel with the galvanometer, (B) in parallel with the fixed resistor, (C) in parallel with the variable resistor, (D) in parallel with the direct current source, (E) in series with all the other components.
In the circuit shown, we have a galvanometer, a device for measuring current; a fixed resistor; a variable resistor; and a direct current source. This source provides voltage to the circuit, and we want to use this circuit overall as an ohmmeter, a device for measuring resistance. The idea then is that we could have a resistor of unknown resistance whose resistance we can measure using this circuit. What we want to figure out there is how, among these answer options, we should connect this resistor to the circuit.
Before we think about that though, let’s consider just how this circuit could work as an ohmmeter in the first place. Recall that our galvanometer is a device for measuring current. If we were to look at the galvanometer this way, it could be a scale that looks like this. The device is capable of measuring currents from zero all the way up to some maximum current we’ve called 𝐼 sub 𝑔. The current reading is indicated by the position of this pink needle, also called an arm. The amount of current measured by the galvanometer, and therefore the amount of current in this circuit, is related to the total resistance, we’ll call it 𝑅 in the circuit, by an equation known as Ohm’s law. This law says that the potential difference across the circuit equals the current in the circuit multiplied by the circuit’s total resistance.
Note that it’s not just the fixed resistor and the variable resistor that have resistance in the circuit. The galvanometer also has some nonnegligibly small resistance. Our cell on the other hand we’re treating as an idealized cell that has zero internal resistance. That leaves us then with three components in this circuit that do have some meaningful amount of electrical resistance. When we combine these resistances, we get the overall resistance in the circuit. And as Ohm’s law shows us, the product of that resistance with the current in the circuit 𝐼 is equal to the voltage in the circuit 𝑉. For this particular circuit, that potential difference 𝑉 is fixed. That means it’s a constant and that whatever 𝐼 and 𝑅 are at any given moment, their product must equal that constant 𝑉.
Notice what this means about how the current 𝐼 and the resistance 𝑅 will relate to one another. If, for example, the overall resistance on our circuit increases, that must mean that the overall current in our circuit will decrease because 𝐼 times 𝑅 is a constant value. In the same way, if the resistance decreases, then that must mean that 𝐼 will increase. Because we have a variable resistor in our circuit, either one of those changes to the overall resistance, whether increasing or decreasing, can occur.
We’re told in our problem statement that the value of the variable resistor is set so that the arm of the galvanometer moves to what’s called its full-scale deflection position. This means that the galvanometer at this moment is measuring the maximum current it can measure given its scale. When the overall resistance 𝑅 in the circuit is tuned so that the galvanometer arm appears in this position, then our overall circuit is set up to function as an ohmmeter.
Let’s now look again at our answer options and notice that we can basically divide them into two types of answers. Options (A) through (D) suggest that we connect our resistor with unknown resistance in parallel with some component in the circuit. On the other hand, option (E) describes putting this resistor in series with the components. We can look at these types of answers in turn, and let’s start by considering answer option (D). This option proposes that we put a resistor of unknown resistance in parallel with a direct current source. Recall that our goal is to be able to use the reading on the galvanometer to measure the resistance of our resistor. This will require that some amount of current exists in the resistor passing through it as part of a circuit.
But then, remember that we’ve said that this cell is an idealized cell that we can treat it as having zero resistance. We know that when electrical charge passes through a circuit and comes to parallel branches of that circuit, it tends toward the branch that has the lower resistance. When we connect our resistor of unknown resistance in parallel with this zero-resistance cell, that tells us that when charge reaches this branch point in the circuit, it will always tend to pass through the cell and none of it will tend to pass through the resistor.
In this way, there will be effectively zero charge moving through this resistor. This means that by connecting it in parallel with our direct current source, we really haven’t changed the circuit at all. Electric charge still flows the exact same way it flowed before we connected the resistor. Since no charge flows through the resistor, we can’t measure its resistance. This tells us that answer option (D) is not a way that we can connect our resistor to the circuit to measure its resistance.
Knowing this, what if instead we connect our resistor in parallel with some other component in the circuit, say, in parallel with the galvanometer? Because the galvanometer does have some amount of resistance, unlike our idealized cell, in this setup, some charge in the circuit will indeed pass through the resistor.
The question still is though, can we use this approach to measure the resistor’s resistance? If we call the resistance of our galvanometer 𝑅 sub 𝑔 and the resistance of our unknown resistor 𝑅 sub 𝑢, in order to work out the resistance of our unknown resistor, we’ll need to be able to compare the overall or equivalent resistance of this parallel branch of resistors with the original resistor of the galvanometer. The difference between these two resistances will be reflected in the overall resistance 𝑅 of our circuit, and that will then impact the current measured by the galvanometer.
We don’t know the actual value of the galvanometer’s resistance or the resistance of our unknown resistor. But because these two resistors are connected in parallel, we can recall that in general any two resistors, we’ll call them 𝑅 one and 𝑅 two, that are connected this way have an effective or an equivalent resistance of 𝑅 one times 𝑅 two divided by 𝑅 one plus 𝑅 two. In other words, if we were to replace these two resistors with one equivalent resistor, that equivalent resistor would have this resistance. The interesting thing about this resistance is that it’s actually less than either 𝑅 one or 𝑅 two individually. That is, we could write that 𝑅 one is greater than this fraction, and we can also write that 𝑅 two is greater than the fraction.
To help us see that this really is true, that the equivalent resistance of two resistors in parallel is less than either one of the resistors by themselves, let’s say we give specific values to 𝑅 one and 𝑅 two. 𝑅 one we’ll let equal two ohms, and 𝑅 two we’ll let equal 10 to the six, or one million, ohms. We have then two resistors of very different values in parallel. And if we substitute these values into our expression, in our numerator, we have two ohms times 10 to the six ohms. That equals two times 10 to the six, or two million, ohms squared, while in our denominator, we have two ohms plus one million ohms. That’s one million and two ohms. And notice that in this fraction, one factor of ohms cancels from numerator and denominator.
What we have then is two million divided by one million and two ohms. Notice that this fraction is ever so slightly less than two ohms. In other words, the equivalent resistance of these two resistors in parallel is less than the value of either one. What we’ve seen then is that the resistance of our galvanometer by itself is actually greater than the resistance of our galvanometer in parallel with our unknown resistor. This means then that by adding our unknown resistor in parallel with a galvanometer, we have decreased the overall resistance in our circuit. That may seem strange, but we’ve seen that it’s true.
Since we’ve decreased the overall resistance of the circuit but the voltage in the circuit is a constant, that must mean that the current in the circuit would tend to increase when we put our unknown resistor in parallel with a galvanometer. And here is where we run into a problem. Recall that the arm on our galvanometer is already at its full deflection. The galvanometer can accurately measure current of a greater magnitude than this.
When we decrease our overall circuit resistance, thereby increasing the current in the circuit, that means our galvanometer is no longer able to measure this current accurately. The arm will either swing past 𝐼 sub 𝑔 off the galvanometer scale or even if it’s blocked from doing this by some physical barrier within the galvanometer, the reading on the display will no longer be accurate. That means that we can’t use this reading to measure the resistance of our unknown resistor.
What we’re seeing then is that putting our unknown resistor in parallel with a galvanometer is not a viable way to measure that unknown resistor’s resistance. And actually, the same problem would occur if we put our unknown resistor in parallel with a fixed resistor or with a variable resistor. In all three of these scenarios, we will be decreasing the overall resistance in the circuit, and this will cause an increase in the current 𝐼. This increase though cannot be measured by our galvanometer, which is already reading its maximum current before this change in resistance. What we’ve seen then is that none of our answer options, which suggest putting the unknown resistor in parallel with some component in our circuit, will work.
Let’s see what happens when we try out option (E), connecting our unknown resistor in series with the other components. The first thing we can note is that by doing this, we have increased the overall resistance in our circuit. This is because resistors in series with one another, as all our resistors now are, add together so that each additional resistor adds resistance to the overall circuit. Since our overall resistance increases, that means our overall current will decrease. This means that the measurement arm on our galvanometer will deflect back toward zero. This deflection represents some change in current that we can call Δ𝐼.
By Ohm’s law, that change in current corresponds to a change in resistance; we’ll call this Δ𝑅. That change in resistance in our circuit is equal to the resistance of the unknown resistor 𝑅 sub 𝑢. This then is how this circuit, consisting of a device for measuring current, can actually work as an ohmmeter to measure resistance. For this to happen though, the unknown resistor must be connected in series with the other components. This answer corresponds to answer option (E).
