A galvanometer has a resistance of 12 milliohms. A multiplier resistor is connected in series with the galvanometer to convert it into a voltmeter. The resistance of the multiplier is 1.1 kiloohms. What percent of the greatest voltage that the voltmeter can measure is the voltage across the galvanometer? Answer to four decimal places.
In this scenario, we have a galvanometer connected in series with a resistor. The galvanometer has its own resistance; we’ll call this 𝑅 sub G. And the resistor we’ve put in series with the galvanometer is called a multiplier resistor. So we’ll refer to it as 𝑅 sub m. By combining the multiplier resistor with a galvanometer like this, we’ve created a voltmeter. We do this by taking the current measured by the galvanometer, let’s call that 𝐼 sub G, and multiplying that current by the total circuit resistance, which is equal to 𝑅 sub G plus 𝑅 sub m.
Ohm’s law tells us that if we take the current in a circuit 𝐼 and multiply it by the circuit resistance 𝑅, that product is equal to the voltage in the circuit 𝑉. Indirectly then, by measuring the current in our circuit and knowing the overall circuit resistance, we measure the voltage in our circuit. Our question says, what percent of the greatest voltage that the voltmeter can measure is the voltage across the galvanometer? Now, this measured voltage here is the total voltage in our circuit. We want to compare this to the voltage or the potential difference across just the galvanometer in the circuit. Here, we can once again apply Ohm’s law. If we take the current in the galvanometer and multiply this by the resistance just of the galvanometer itself, then by Ohm’s law, that product will equal the voltage across the galvanometer. We’ll call it 𝑉 sub G.
So, when our question says, “What percent of the greatest voltage that the voltmeter can measure is the voltage across the galvanometer?,” it’s asking us to solve for 𝑉 sub G divided by 𝑉 multiplied by 100 percent. So our answer is a percent.
Let’s do a quick sanity check that this equation makes sense. Let’s say that 𝑉 is 100 volts and 𝑉 sub G is one volt. In that case, this fraction would give us one volt divided by 100 volts, or one one hundredth of a volt. And then multiplying this by 100 percent would give us one percent. And indeed one volt is one percent of 100 volts. To help us on our way to a solution, let’s note that the total voltage in our circuit is equal to the sum of the voltage across the galvanometer and the voltage across the multiplier resistor. We already have an expression for the voltage across the galvanometer. We can create a similar one for the voltage across the multiplier resistor. We’ll call it 𝑉 sub m. It’s equal to the current in the circuit, which is 𝐼 sub G, the current measured by the galvanometer, multiplied simply by the resistance of the multiplier resistor.
Considering this pair of equations, we see that 𝐼 sub G appears in both of them. This means that we could, for example, divide both sides of the second equation by the resistance of the multiplier resistor 𝑅 sub m, which cancels that factor on the left, which gives us an expression for 𝐼 sub G that we could then substitute in for 𝐼 sub G in the other equation so that now we have an expression for 𝑉 sub G, the potential difference across the galvanometer, in terms of the potential difference across the multiplier resistor, as well as the resistances of our galvanometer and the multiplier resistor.
At this point, let’s recall that this is the expression that we want to solve for and that 𝑉, the potential difference across our entire circuit, is equal to the potential difference across our galvanometer plus the potential difference across the multiplier resistor. Seeing this, we now recognize that we have an expression in terms of 𝑉 sub m for the voltage across the galvanometer 𝑉 sub G. We can take the entire left-hand side of this equation then and substitute it in for 𝑉 sub G. That gives us this big result.
Notice though that the voltage across the multiplier resistor 𝑉 sub m can be factored out of the denominator, which means that since we’re both multiplying and dividing by 𝑉 sub m, we can effectively cancel it out. That leaves us with an expression entirely in terms of the resistances of our two circuit components. If we multiply both the numerator and denominator of this fraction by 𝑅 sub m, the resistance of our multiplier resistor, then 𝑅 sub m cancels out from our numerator. And we’re left with this much simplified expression. Mathematically, this is equal to the expression we had originally. We’ve just now rearranged it so that it’s written in terms of values we know. According to our problem statement, the galvanometer has a resistance of 12 milliohms, and that’s 𝑅 sub G. And the resistance of the multiplier is given as 1.1 kiloohms. That is 𝑅 sub m.
Regarding these prefixes milli- and kilo-, we know that milli- indicates 10 to the negative three or one one thousandth of a unit, while kilo- indicates 10 to the positive three or 1000 of them. 12 milliohms then, if we divide 12 by 1000, gives us 0.012 ohms. And 1.1 kiloohms, if we multiply this by 1000, gives us 1100 ohms. When we enter this expression on our calculator and round the answer to four decimal places, we get a result of 0.0011 percent. This is the voltage across the galvanometer compared as a percent to the greatest voltage measurable by the voltmeter. We see then that adding this multiplier resistor to our circuit has greatly increased the maximum measurable voltage using our voltmeter. And indeed, the larger the resistance of the multiplier resistor, the more this range is extended.