A sensitive galvanometer is connected to a shunt 𝑋, of resistance 0.2 ohms, which is then replaced by another shunt 𝑌, of resistance 0.02 ohms. In which case is the ammeter able to measure a higher range of current intensity? Why?
If we were to sketch out a galvanometer in a circuit, we could represent it this way by a resistance. And we call it 𝑅 sub g, the resistance of the galvanometer. If we call the current running through this circuit 𝐼, then right now all of the current 𝐼 goes through our galvanometer. In a way, this is good because our galvanometer is made to measure current. But this arrangement also means that whatever the maximum value of current the galvanometer can accurately measure, that’s the most current that we can send through our circuit. If we use current higher than that, the galvanometer would simply read off its maximum current value but this would be inaccurate. If we want to extend that measurement range though, we can make a modification to the circuit.
We can install a parallel branch in the circuit called a shunt, which has its own resistance we’ve called 𝑅 sub s. The presence of this parallel branch means that the current 𝐼 as it approaches this junction then splits into two parts. One part travels to the upper branch through the galvanometer and the other part splits off and goes through the shunt. Typically, the shunt resistance 𝑅 sub s is less than the galvanometer resistance 𝑅 sub g. This means that relatively much more current travel through the shunt than travels through the galvanometer. In this example, we consider two different shunt resistances: one called shunt 𝑋 resistance 0.2 ohms and the other shunt 𝑌 with one 10th that resistance. We’re told that these two shunts in succession are connected into our circuit as the shunt resistor. The question is which of these two shunts in the circuit allows for a greater maximum measurable current.
To figure this out, let’s add a few labels to our diagram. Let’s call the current that splits off and goes through the galvanometer 𝐼 sub g. And we’ll call the current that goes through the shunt resistance 𝐼 sub s. As we mentioned earlier, the current 𝐼 splits itself between these two parallel branches. So we can now write an equation that summarizes that. We can write that 𝐼 is equal to 𝐼 sub g plus 𝐼 sub s. We can write that the total current is equal to the current through the galvanometer plus the current through the shunt. In addition to this, there is a second way to mathematically relate the parameters of these two parallel branches.
Because each of these branches has a resistance to it, we know there will be some potential difference across them. And since the branches are in parallel, that tells us the potential difference across each one is the same. Moreover, we can find that potential difference by recalling Ohm’s law. This law tells us that potential difference across a circuit is equal to the current in the circuit multiplied by circuit resistance. This law also holds true for a component in a circuit.
Let’s consider the start here and the end here of this parallel section of our circuit. Let’s call the potential difference across this part of our circuit capital 𝑉. According to Ohm’s law, we can express this potential 𝑉 in terms of the currents and resistances we’ve written in this section of our circuit. In particular, 𝑉 is equal to the current through the galvanometer times the resistance of the galvanometer. And since the potential differences across the two parallel branches are the same, this is also equal to the current through the shunt times the resistance of the shunt.
Now let’s recall that in all this, our goal is to figure out for which value of our shunt resistance — either 0.2 ohms or 0.02 ohms — is our current 𝐼 going to be the maximum value. That is which shunt 𝑋 or 𝑌 will enable the maximum measurable current through this circuit. As we look at the different terms in these two equations, let’s consider which are fixed and which are variable. We know that the resistance of our galvanometer is a fixed value. That simply is whatever that value is. And additionally, we mention that our galvanometer has a maximum possible current that it can measure according to its scale. That’s 𝐼 sub g. So this means 𝐼 sub g is also a fixed quantity. It’s the maximum current that the galvanometer can measure. That leaves the current through the shunt and the resistance of the shunt.
We know the shunt resistance is variable. That’s what we’re switching out when we go from shunt 𝑋 to shunt 𝑌 and the shunt current will respond to that resistance. Using these two independent equations, we can substitute in for 𝐼 sub s replacing it with the term in terms of 𝐼 and 𝐼 sub g. The value of that is that we’ll then have an expression in terms of fixed quantities, our shunt resistance and the maximum measurable current. Considering our first equation, if we subtract the current running through the galvanometer from both sides, that term cancels from the right side of our expression. And we find that 𝐼 sub s the current through the shunt is equal to 𝐼 minus 𝐼 sub g. We’ll then substitute in 𝐼 minus 𝐼 sub g for 𝐼 sub s in our second equation.
And then having done that, we can rearrange this expression to find an equation with the current 𝐼 isolated on one side. If we divide both sides by 𝑅 sub s, that term cancels on the right. And then if we add 𝐼 sub g to both sides, that term cancels on the right. We’re left with an expression for the maximum measurable current in the circuit according to the galvanometer resistance as well as the shunt resistance. Using this expression, we can answer our question of which shunt 𝑋 or 𝑌 enables the greater measurable current.
Since the only appearance of 𝑅 sub s, the shunt resistance in the left-hand side of this expression, is in the denominator of this one term, that means the larger 𝑅 sub s is, the larger the shunt resistance, the lower the overall maximum measurable current. And vice versa, the smaller 𝑅 sub s is, the greater the current intensity is possible to measure. So which of our two shunts 𝑋 or 𝑌 allows the measurement of a higher range of current intensity? That’s shunt 𝑌 because it has a smaller resistance, 0.02 ohms compared to shunt 𝑋’s 0.2 ohms.
Along with that question, we’re also asked to explain why this is. We saw from our formula for the maximum measurable current 𝐼 why this is the case. It’s because as the shunt resistance decreases as it gets lower, the measuring range increases. That’s why the smaller resistance of shunt 𝑌 allows for the measurement of a higher range of current intensity.