### Video Transcript

A galvanometer and a resistor are connected in parallel to form an ammeter. How does adding the resistor make the resistance of the ammeter compare to the resistance of the galvanometer? (A) The ammeter has greater resistance than the galvanometer. (B) The ammeter has less resistance than the galvanometer. (C) The ammeter and galvanometer have the same resistance.

Our problem statement tells us how to make an ammeter, a device for measuring current. We do it by connecting a galvanometer, another current measurement device, and a resistor in parallel. The circuit symbol for a galvanometer looks like this. Connecting this component in parallel with a resistor, we get what functions overall as an ammeter. The question is, which has greater resistance, the ammeter or the galvanometer all by itself?

Itβs important to note that even though thereβs no resistor circuit symbol thatβs part of the galvanometer symbol, the galvanometer nonetheless has some resistance. We can represent that resistance using π
sub g. Then typically, the resistor thatβs in parallel with a galvanometer is called a shunt resistor. This is so because itβs used to shunt current away from the branch with a galvanometer in it. We can say then that the resistance of this parallel branch of our circuit is π
sub s.

What we want to do then is to compare the overall resistance of π
sub g and π
sub s in parallel with the resistance of π
sub g by itself. Even though weβre talking here about galvanometers and ammeters, this specific question weβre asked really has a general application to it. We could say that that question is, if we have two resistors in parallel, weβll call them π
one and π
two, how does the overall resistance or equivalent resistance of these two resistors arranged this way compare to either one of the resistor values individually? In other words, if we had a single resistor, weβll call it π
sub e for the effective resistance, that was equal to the combined resistances of π
one and π
two in parallel, then how would π
sub e compare with π
one and π
two?

We can begin answering this question by noting that, in general, when two resistors are arranged in parallel, here weβre using the names π
one and π
two for those resistors, then the equivalent resistance of those resistors is equal to their product, π
one times π
two, divided by their sum, π
one plus π
two. Thereβs something very interesting about this result. It turns out that this fraction π
one times π
two divided by π
one plus π
two is always less than either π
one or π
two individually. Just by way of example, letβs say that π
one has a value of three ohms. And letβs say that π
two has a resistance of 10 to the seven, or 10 million, ohms. If we substitute these values for π
one and π
two into our expression, we get three ohms times 10 million ohms divided by three ohms plus 10 million ohms.

Here, weβve just picked these two resistor values so that theyβre very different from one another. Other than that, thereβs no particular significance to three ohms and 10 million ohms. Anyway, three ohms times 10 million ohms gives us 30 million ohms squared, while three ohms plus 10 million ohms in the denominator gives us 10 million and three ohms. Note that a factor of ohms cancels from numerator and denominator. And so, the overall or effective resistance of these two resistors in parallel is 30 million divided by 10 million and three ohms. If we calculate out this fraction, weβll find that it is ever so slightly less than three ohms. The point is our equivalent resistance is less than the resistance of either of the resistors individually. This is just one example of that, but itβs true generally.

Returning then to our resistors with resistances π
one and π
two, we can write that the equivalent resistance of these two resistors in parallel is less than π
one and itβs less than π
two. If we take this result back to our application of the resistance of the ammeter compared to the resistance of the galvanometer, we see we can think of π
sub g as either one of our resistors π
one or π
two. Our result tells us that that resistance is greater than the effective resistance of π
sub g arranged in parallel with π
sub s.

Our conclusion then is that the resistance of the galvanometer by itself is greater than the resistance of the galvanometer in parallel with the resistor. Or put another way, the ammeter has less resistance than the galvanometer. We choose answer option (B).