# Video: Determining the Circuit Diagram That Represesnts the Ammeter

Which of the following circuit diagrams most correctly represents a galvanometer combined with a shunt resistor being used as an ammeter to measure the current through a circuit that has a direct-current source? [A] Circuit A [B] Circuit B [C] Circuit C

06:34

### Video Transcript

Which of the following circuit diagrams most correctly represents a galvanometer combined with a shunt resistor being used as an ammeter to measure the current through a circuit that has a direct current source?

So here we can see we’ve been given three different circuit diagrams to choose between. And each one contains a galvanometer which is represented by a G inside a circle. Each of the circuits also contains a cell, which is a type of direct current source. And each of the circuits also contains either one or two resistors connected in various ways. The question asks us to identify which one of these circuits shows a galvanometer combined with a shunt resistor being used as an ammeter. And an ammeter is, of course, a device which measures current.

The term shunt resistor refers to a resistor with a special function inside an ammeter. But it’s important to note that a shunt resistor is really just a normal resistor. So let’s start by recalling that a galvanometer is a device which indicates the magnitude and direction of a current within it. Galvanometers generally have a dial with a zero in the middle. And a current passing through the galvanometer will cause the needle to deflect. Now up to some maximum current, the deflection of the galvanometer’s needle is proportional to the current passing through it. So reducing the size of the current will reduce the deflection of the needle. And reversing the direction of the current will cause the needle to deflect in the other direction.

Since the galvanometer responds to current in a predictable way, it seems reasonable to suggest that we could just use a galvanometer as an ammeter. However, there are two main problems with this. The first problem is that galvanometers are very sensitive, which means that the maximum current that they can indicate in either direction tends to be in the microamp or the milliamp region. The second problem with trying to use a galvanometer as an ammeter is that galvanometers have their own internal resistance, which is why in circuit diagrams, we often see galvanometers represented, not just by a G in a circle, but by an additional resistor as well with a resistance 𝑅 G.

The presence of this additional resistance means that when we connect a galvanometer to a circuit, it can drastically affect the resistance of the circuit as a whole, which means it changes the current that we’re trying to measure. We can see this effect in action if we consider circuit diagram (A). Now, in this circuit diagram, we simply have a cell, a galvanometer, and a resistor connected in series.

Now, since any circuit that we’re trying to measure using a galvanometer must have initially had some resistance of its own before we added the galvanometer, we can assume that the circuit must have looked like this before the galvanometer was introduced. So there’s a cell applying some voltage, which we’ll call 𝑉, to a resistor, which we’ll say has resistance 𝑅 one. And we know that this must produce a current 𝐼 in accordance with Ohm’s law, which says that the current through a resistor 𝐼 is equal to the voltage across that resistor 𝑉 divided by the resistance of that resistor 𝑅.

But if we want to measure the size of this current, then wiring in a galvanometer like this is actually a bad way of doing it. This is because of the fact that a galvanometer has its own resistance 𝑅 G. When we connect to resistance in series, the total resistance of those resistors is equal to the sum of that individual resistances. This means that before we connected the galvanometer to our circuit, the total resistance of the circuit was just 𝑅 one. But after we’ve connected the galvanometer, the total resistance is 𝑅 one plus 𝑅 G. This means that the current changes as well according to Ohm’s law. So just connecting a galvanometer in series with the other components will change the current that we’re trying to measure. So we know that (A) is not the correct answer.

Now, we can actually prevent the problem of the galvanometer’s resistance contributing to the overall resistance of the circuit by introducing another resistor known as a shunt resistor. Connecting this resistor in parallel with the galvanometer means that the incoming current is split between the two parallel branches. We could say that some of the current is shunted around the galvanometer, which is why this resistor is known as the shunt resistor. And we can say that it has a resistance 𝑅 S. Now, the relative magnitudes of the current through the galvanometer, which we can call 𝐼 G, and the current in the shunt resistor, which we can call 𝐼 S, depends on the relative resistances of the galvanometer and the shunt resistor.

The majority of the current will go down the path of least resistance. So if the shunt resistor had a larger resistance compared to the galvanometer, then we’d find most of the current goes through the galvanometer, which wouldn’t really change how the circuit behaved. However, if we make sure that the resistance of the shunt resistor is much smaller than the resistance of the galvanometer, then we find that the majority of the current passes through the shunt resistor. And only a small amount of current passes through the galvanometer.

This is ideal for two reasons. Firstly, because the majority of the current goes through the shunt resistor, but the shunt resistor only has a very small resistance, this means that the current 𝐼 is barely affected. And secondly, because a small proportion of the current now passes through the galvanometer, we can use the galvanometer to indicate the overall current 𝐼 in the circuit. And by carefully choosing the resistance of the shunt resistor, we can make sure that the current in the galvanometer never exceeds the maximum deflection current, which actually solves our original problem of the galvanometer being too sensitive for high currents.

Indeed, connecting a shunt resistor in parallel with a galvanometer like this is how ammeters are made. So we could consider everything within this pink box to be equivalent to an ammeter. Now, if we look at the two remaining answer options (B) and (C), we can see that both of these circuit diagrams effectively contain an ammeter, that is, as long as the resistances of these resistors are significantly lower than the resistances of the galvanometers. However, if we consider option (B), we can see that if everything in the pink box is an ammeter, then this is just an ammeter connected to a cell as there are no other resistors in the circuit diagram.

Now, having a cell in a circuit with no resistance isn’t a very realistic situation, since all circuits have some resistance. So we have to assume that the original circuit looked like this. And in an attempt to measure the current in this resistor, a galvanometer was wired in, in parallel like this. However, connecting a galvanometer in this way would split the current into the two parallel branches, once again changing the current that we were trying to measure in the first place. In fact, the only sensible answer to this question is shown in circuit diagram (C), where an assembly consisting of a galvanometer and a resistor connected in parallel appears to have been connected to an original circuit consisting of a cell and a resistor connected in series.

Attaching a galvanometer and a shunt resistor to the circuit like this means that the current in the original circuit is basically unchanged. But a very small proportion of this current passes through the galvanometer which causes a deflection of the needle which we can use to infer the current 𝐼. So the correct answer to this question is option (C).