Question Video: Calculating the Measurement Range of an Ammeter | Nagwa Question Video: Calculating the Measurement Range of an Ammeter | Nagwa

# Question Video: Calculating the Measurement Range of an Ammeter Physics • Third Year of Secondary School

A galvanometer has a resistance of 12 mΩ. A current of 150 mA produces full-scale deflection of the galvanometer. A shunt is connected in parallel with the galvanometer to convert it into an ammeter. The resistance of the shunt is 70 𝜇Ω. What is the greatest current that the ammeter can measure? Answer to one decimal place.

07:59

### Video Transcript

A galvanometer has a resistance of 12 milliohms. A current of 150 milliamperes produces full-scale deflection of the galvanometer. A shunt is connected in parallel with the galvanometer to convert it into an ammeter. The resistance of the shunt is 70 microhms. What is the greatest current that the ammeter can measure? Answer to one decimal place.

In this situation, we can imagine starting out with a galvanometer by itself. This is a device that measures current. The measured current is displayed on a scale that looks like this. We’re told that when a current of 150 milliamperes exists in the circuit, the measurement arm on the galvanometer scale is fully deflected. That is, it indicates the maximum current that this scale allows. If we want to convert this galvanometer into an ammeter, which is also a device for measuring current, we’d like to extend the ammeter’s range beyond that of the galvanometer by itself. That way, we can measure currents that are larger than 150 milliamperes.

The way we do this is by connecting what’s called a shunt in parallel with the galvanometer. The effect of this parallel branch is that now some of the total circuit current does indeed pass through the galvanometer, but some is split off and passes through the shunt. This means that the total circuit current, we’ll call that 𝐼 sub t, and the current that passes through the galvanometer, we’ll call this 𝐼 sub G, are no longer the same. Rather, 𝐼 sub G is less than the total circuit current because some of that total current passes through the shunt and doesn’t pass through the galvanometer.

Interestingly, it’s possible to predict how the current will divide over these two parallel branches. That division depends on the relative resistances of these branches. Let’s imagine a general situation where we have two resistors, we’ll call them 𝑅 one and 𝑅 two, arranged in parallel. If the total current passing through this circuit overall is 𝐼 sub t, then the current passing through the branch with resistance 𝑅 one, we’ll call this current 𝐼 one, is given by this equation. It’s equal to the total circuit current 𝐼 sub t multiplied by this ratio, the resistance of the other branch of the parallel circuit 𝑅 two divided by the sum of the resistances 𝑅 one and 𝑅 two.

Notice that according to this equation, the higher the resistance 𝑅 two, the greater the current that will exist in the other parallel branch, 𝐼 one. This is because charge tends to flow towards the branch of a parallel circuit that has the least resistance. If the resistor 𝑅 two has a very large resistance, then charge in this circuit will tend towards the other branch with resistance 𝑅 one according to this ratio. As a side note, if we were to think instead about the current in the other parallel branch, let’s call it 𝐼 two, the equation telling us that current is very similar to the equation for 𝐼 one. All that’s changed is that now in the numerator of our fraction we’re considering the resistance of the opposite branch, in this case 𝑅 one.

If we return now to our ammeter made out of a galvanometer and a shunt, we can see that we have a similar situation. The galvanometer, after all, does have a resistance. We’re told it’s 12 milliohms. The shunt has a resistance as well of 70 microhms. Let’s call these resistances 𝑅 sub G and 𝑅 sub S, respectively. According to our equation here, it’s a ratio involving these resistances that can tell us just how much current exists in either one of our branches, for example, the current in the branch with the galvanometer. In this question, we want to solve for the greatest current that our ammeter can measure. In other words, we want to solve for the largest possible value of 𝐼 sub t that will not overrun the scale of our galvanometer.

Our problem statement tells us that a current of 150 milliamperes fully deflects the measurement arm of the galvanometer. That full deflection corresponds to the maximum current the galvanometer can accurately measure. Since we want to solve here for the maximum allowable value of 𝐼 sub 𝑡, we’ll assume that 𝐼 sub G, the current in the galvanometer, is 150 milliamperes. If we apply this relationship to our scenario, the equation reads 𝐼 sub G, the current in the galvanometer branch is equal to 𝐼 sub t, the total current in our circuit, multiplied by the shunt resistance, 𝑅 sub S, divided by the resistance of the galvanometer plus the resistance of the shunt.

As we’ve seen, it’s not 𝐼 sub G we want to solve for — in fact, we know that value — but rather it’s 𝐼 sub t, the total measurable current in the circuit. To make 𝐼 sub t the subject of this equation, we multiply both sides by this ratio, the sum of the resistances 𝑅 sub G and 𝑅 sub S divided by the shunt resistance 𝑅 sub S. This means that over on the right-hand side, 𝑅 sub S cancels out in numerator and denominator, as do these two resistances added together.

If we then take the remaining equation and switch the sides, we find that 𝐼 sub t is given by this expression. Notice that it’s equal to 𝐼 sub G times what we could call this multiplier. We want this multiplier to be greater than one because that will mean the maximum measurable current in our circuit 𝐼 sub t exceeds the maximum measurable current through the galvanometer itself.

As a side note, if we think about this scenario from the perspective of designing a circuit, we can see that depending on our choice of the resistance of the shunt 𝑅 sub S, we can change the maximum measurable current in our circuit 𝐼 sub t without changing the maximum current measurable by the galvanometer 𝐼 sub G. In this case, since we know values for 𝐼 sub G, 𝑅 sub G, and 𝑅 sub S, we can start to substitute those in. 𝐼 sub G, the maximum current in the galvanometer branch is 150 milliamperes. 𝑅 sub G, the resistance of the galvanometer, is given as 12 milliohms. And 𝑅 sub S, the shunt resistor’s resistance, is equal to 70 microhms.

Before we calculate 𝐼 sub t, we’ll want to convert our current from units of milliamperes to units of amperes, and we’ll want to convert the units of all our resistors to those of ohms. Let’s recall that the prefix milli- before a unit indicates 10 to the negative three or one one thousandth of some value. This means for example that 150 milliamperes, if we divide this quantity by 1000, shifting the decimal point three spots to the left, will give us an equivalent quantity in units of amps. 150 milliamps is equal to 0.150 amperes. Something similar happens with this value of milliohms. If we divide 12 by 1000, we get 0.012. So that’s the resistance of our galvanometer in units of ohms.

When it comes to the resistance of our shunt resistor, we know that the prefix micro- indicates 10 to the negative six or one one millionth of some quantity. To express these resistances in ohms then, we’re going to divide each one by one milliohm. Rather than write out lots of decimal places, it’s easiest to express this as 70 times 10 to the negative sixth ohms. This quantity is equal to 70 microhms.

We’re now ready to calculate the total measurable current in our circuit 𝐼 sub t. As one last step before we do though, let’s calculate this fraction right here. This is what we’re going to use to multiply 𝐼 sub G to solve for 𝐼 sub t. So the value of this fraction will tell us by what we’re able to multiply the maximum measurable current through our galvanometer. Entering this fraction on our calculator, we get a result of about 172. And note that the units of ohms have canceled out completely from this result. This is telling us that 𝐼 sub t is about 172 times greater than 𝐼 sub G. That’s how much we’ve increased the measurement range of our ammeter beyond that of our galvanometer by adding this shunt into the circuit.

So then, just what is 𝐼 sub t, that total measurable circuit current, to one decimal place? It’s equal to 25.9 amperes. This is a much better maximum measurable current than our galvanometer’s maximum current of 0.150 amperes. And in fact if we wanted to make 𝐼 sub t even larger than it is, then in designing our circuit we could’ve decreased the value of 𝑅 sub S even more. That would’ve drawn relatively more charge towards this branch in the circuit. And so long as we know 𝑅 sub S and 𝑅 sub G, we can figure out by what factor we’re multiplying the maximum measurable current through the galvanometer to solve for the maximum measurable circuit current 𝐼 sub t. What we found is that the greatest current that this ammeter can measure is 25.9 amperes.

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