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Question Video: Multiple Resistor Size in a Voltmeter Physics

A voltmeter is used to measure the voltage of a direct-current source that is estimated to have a voltage of several volts. The galvanometer in the voltmeter has a resistance of a few milliohms. Which of the following correctly explains why the multiplier resistor in such a voltmeter must have a resistance much greater than the resistance of the galvanometer that the multiplier is connected in series with? [A] If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the voltage of the source will be significantly increased. [B] If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the direction of the deflection of the galvanometer arm will reverse and no reading will be displayed on the voltmeter. [C] If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the resistor will generate a magnetic field that significantly changes the deflection of the galvanometer arm. [D] If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the current through the galvanometer will be greater than the current that would produce a full-scale deflection of the galvanometer arm.

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Video Transcript

A voltmeter is used to measure the voltage of a direct-current source that is estimated to have a voltage of several volts. The galvanometer in the voltmeter has a resistance of a few milliohms. Which of the following correctly explains why the multiplier resistor in such a voltmeter must have a resistance much greater than the resistance of the galvanometer that the multiplier is connected in series with? (A) If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the voltage of the source will be significantly increased. (B) If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the direction of the deflection of the galvanometer arm will reverse and no reading will be displayed on the voltmeter. (C) If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the resistor will generate a magnetic field that significantly changes the deflection of the galvanometer arm. And lastly, (D) if the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the current through the galvanometer will be greater than the current that would produce a full-scale deflection of the galvanometer arm.

In this situation, we’re thinking about a voltmeter, which is made up of a galvanometer connected in series with a resistor. This voltmeter we’re told is meant to measure the voltage of a direct-current source, where that voltage is on the order of a few volts. Along with this, we’re told that the galvanometer in our voltmeter has a resistance of a few milliohms. Based on this setup, our question wants us to identify the best reason why the resistance of this resistor here, called a multiplier resistor, must be significantly greater than the resistance of the galvanometer in order for this voltmeter to successfully measure voltages on the order of several volts.

As we consider our answer choices, in order to shorten them down a bit so we can fit them all on screen at the same time, let’s note that all four of the possible answers begin with this phrase: “If the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer.” We can effectively condense then answer options (A), (B), and (C) by assuming that they begin with this phrase and then finish up in the following ways. Answer option (A) says that if this condition is met, the source voltage will significantly increase. (B) says that if the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, then the direction of the galvanometer arm deflection will reverse and no voltmeter reading will result.

And then lastly option (C) says that if this condition is met, the resistor will generate a magnetic field significantly changing the galvanometer arm deflection. Knowing all this, let’s now consider the display of our galvanometer, which we know is part of our voltmeter. This display may look like this. We recall that a galvanometer measures current. And the scale of the galvanometer is fairly small, meaning that the maximum current it can accurately read may be best expressed in milliamperes.

Another trait of the galvanometer is that it can differentiate current direction. What we see here is the reading that would result from zero current existing in the circuit, whereas if current were traveling in one direction, this would deflect the needle, say to the right, while current in the other direction would deflect the needle in the opposite way. A galvanometer then is indeed sensitive to the direction of current in the circuit.

We can understand how a galvanometer indirectly can serve as a voltmeter by recalling Ohm’s law. This law says that if we know the current 𝐼 in a circuit as well as the total circuit resistance 𝑅, then the product of these values is equal to the voltage across the circuit 𝑉. Our particular circuit we could say has a design specification. We want to be able to use this voltmeter to measure voltages of several volts. This means that if 𝐼 is the current in the circuit, and we know that current is the same everywhere because this is a series circuit, and if 𝑅 is the total circuit resistance, then 𝐼 times 𝑅 based on Ohm’s law must be approximately equal to a few volts. That’s the requirement if we want to accurately measure this source voltage.

Considering the current and resistance in the circuit, we can say that based on the scale on our galvanometer, even the greatest current magnitude that is readable on this device will be expressed in milliamperes, perhaps say tens of milliamperes. In any case, we expect it to be much less than one ampere. So that’s the maximum current we can accurately measure. And then if we think about the overall circuit resistance, that resistance is equal to the sum of the resistance of the galvanometer and the resistance of our multiplier resistor that we’ll call 𝑅 sub m. We’re told in our problem statement that the resistance of the galvanometer is several milliohms.

Now, for a moment let’s imagine that there is no multiplier resistor in the circuit. In that case, by Ohm’s law, we’re multiplying a current in milliamperes by a resistance in milliohms. And we can see that in doing that, the product will never be as large as a few volts. Rather, it might be expressed better in millivolts or even microvolts. To meet our design goal then of measuring a few volts of potential difference, we indeed do need the resistance from the multiplier resistor, and in fact this resistance has to be quite large. Its resistance has to overcome we could say the smallness of the resistance of the galvanometer combined with the relative smallness of the current that the galvanometer accurately can measure.

With this in mind, let’s now consider our answer options, starting with choice (D). This choice says that if the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, in other words, on the order of milliohms or even smaller, then the current through the galvanometer will be greater than the current that would produce a full-scale deflection of the galvanometer arm. Let’s note that with our circuit set up this way, whether we can measure it accurately or not, there is a potential difference of several volts across the circuit. And saying that the current in our circuit is on the order of milliamperes, we’ve already maxed out, so to speak, the measurement range of our galvanometer.

If we then added a multiplier resistor with a resistance also, say, on the order of milliohms, then that would fix the overall resistance of our circuit at a relatively low value, which would mean that the only way for the current 𝐼 multiplied by the resistance 𝑅 to equal the voltage 𝑉 would be for the current 𝐼 in the circuit to significantly increase. The problem with this is we’ve already assumed a current which fully deflects the measurement arm of our galvanometer. So any increase in current will push the arm past that point or otherwise lead to an inaccurate current measurement.

On that basis, answer option (D) looks correct. If the multiplier resistor did have a resistance of a magnitude comparable with or less than that of the galvanometer, the current through the galvanometer would indeed be greater than the maximum current it can accurately measure.

Let’s now consider our remaining answer options, starting with option (A). This response claims that if the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the source voltage, that is, the voltage here supplied by our direct-current source, will significantly increase. We know though that physically that can’t happen. Nothing about the rest of the circuit will impact the voltage supplied by our direct-current source. Answer option (A) then won’t be the one that we choose.

Next, answer option (B) says that if our multiplier resistor meets the condition named, that is, its resistance is comparable to or less than that of the galvanometer, then the direction of the galvanometer arm deflection will reverse. That would mean, for example, that rather than deflecting to the right as we have that arm currently, the arm would deflect to the left. As we’ve seen, that indicates current pointing in the opposite direction. However, the direction of current in the circuit depends on the orientation of our direct-current source. It doesn’t depend on the relative resistances of our multiplier resistor and our galvanometer. Therefore, the multiplier resistor having a resistance comparable to or less than that of the galvanometer will not have the effect described in answer choice (B).

Lastly, in answer choice (C), we read that if the multiplier resistor has a resistance of a magnitude comparable with or less than that of the galvanometer, the resistor, the multiplier resistor, will generate a magnetic field and that this will significantly change the galvanometer arm deflection. If our multiplier resistor were not a resistor but in fact were an inductor, we could expect that largely increasing the current through it would indeed generate a large magnetic field. But because we’re working with a resistor rather than an inductor, we have no such expectation of a magnetic field being formed by this component. For that reason then, we also won’t choose answer option (C).

For our final answer then, we will choose option (D). The best explanation for why the resistance of our multiplier resistor must be much greater than that of our galvanometer is that if the multiplier resistor had a resistance of a magnitude comparable with or less than that of the galvanometer, the current through the galvanometer will be greater than the current that would produce a full-scale deflection of the galvanometer arm.

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