# Question Video: Solving For the Resistance of a Multiplier Resistor Physics

Every division of the scale of a voltmeter that has a resistance of 2,000 Ω denotes 1 V. What is the required resistance of the multiplier resistor that should be connected in series to the voltmeter to make every division of its scale denote 10 V? [A] 200 Ω [B] 1,800 Ω [C] 18,000 Ω [D] 20,000 Ω

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### Video Transcript

Every division of the scale of a voltmeter that has a resistance of 2,000 ohms denotes one volt. What is the required resistance of the multiplier resistor that should be connected in series to the voltmeter to make every division of its scale denote 10 volts? (A) 200 ohms, (B) 1,800 ohms, (C) 18,000 ohms, (D) 20,000 ohms.

In this example, we’re talking about a voltmeter, which we know to actually be based on a galvanometer. A galvanometer directly measures current using a scale that looks something like this. If the view we see now represents a zero reading on the galvanometer, then when the current in our circuit increases enough so that the galvanometer arm moves one division on the scale, that corresponds, we’re told, to a potential difference of one volt across our circuit. We can use Ohm’s law, that the potential difference across a circuit equals the current in that circuit multiplied by the circuit’s resistance, to relate the quantities we’ve been given.

Let’s say that the current required to move the galvanometer arm one division is 𝐼 sub one. This galvanometer we know makes up a voltmeter with a resistance of 2,000 ohms. If we multiply 𝐼 one and 2,000 ohms as per Ohm’s law, then we get a voltage of one volt. This then is the voltage that corresponds to the current value shown on the scale. We can solve for 𝐼 one by dividing both sides of this equation by 2,000 ohms. That causes that resistance to cancel out on the right. And we see that 𝐼 one equals one volt divided by 2,000 ohms, or 0.0005 amperes. This is equivalent by the way to 0.5 milliamperes and helps underscore the fact that in general the scale of a galvanometer extends only to very small current values.

But anyway, we imagine at this point adding a resistor in with our voltmeter, which again is a galvanometer. This is called a multiplier resistor. So we’ll call its resistance 𝑅 sub m. Considering our circuit now that it has these two elements in it, we’d like to maintain the current level in the circuit. That is, we want to keep supplying the galvanometer with a current that deflects its measurement arm by one division. Because we’ve added this multiplier resistor though, this single division on the voltmeter scale no longer will represent one volt like it did before. That’s because the resistance in our circuit is now 2,000 ohms plus the resistance of the multiplier resistor.

What we want to require is that as we add this multiplier resistor to the circuit, now the movement of our galvanometer arm across one division on this scale corresponds to a potential difference of 10 volts. What we’re seeing is that the same change in current as before, from zero up to a current of 𝐼 one, now indicates a greater potential difference than before because of the increased resistance in our circuit due to the multiplier resistor. Our goal is to solve for the value of that resistor. We can start to do this by dividing both sides of our equation by 𝐼 one, canceling out 𝐼 one on the right.

And then if we subtract 2,000 ohms from both sides so that 2,000 ohms on the right minus 2,000 ohms equals zero and then finally swap the two sides of this equation, we get this expression for the resistance of the multiplier resistor. Knowing that the current 𝐼 one equals 0.0005 amperes, we substitute this value in. And when we divide 10 volts by this quantity, that fraction equals 20,000 ohms. This means that 𝑅 sub m is 20,000 ohms minus 2,000 ohms, or 18,000 ohms, which agrees with answer choice (C). In order to make every division of the scale of the voltmeter denote 10 volts, the resistance of the multiplier resistor must be 18,000 ohms.