# Question Video: Determining the Ratio of the Current in the Galvanometer to That in the Multiplier Resistor Physics

The circuit diagram represents a galvanometer connected with a multiplier resistor. The multiplier resistor has a resistance fifty times that of the galvanometer. What is the ratio of the current in the galvanometer, 𝐼_G, to the current in the multiplier resistor, 𝐼_M?

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### Video Transcript

The circuit diagram represents a galvanometer connected with a multiplier resistor. The multiplier resistor has a resistance 50 times that of the galvanometer. What is the ratio of the current in the galvanometer, 𝐼 G, to the current in the multiplier resistor, 𝐼 M?

So in this question, we’ve been given a circuit diagram that shows a galvanometer and a resistor called a multiplier resistor connected in series with a cell. Let’s start by recalling that the term multiplier resistor describes a resistor used in the construction of a voltmeter. Specifically, it’s the name given to a resistor, which is connected in series with a galvanometer, as is the case in this circuit. This combination of a multiplayer resistor and a galvanometer creates a voltmeter. So effectively, this circuit diagram shows a voltmeter being used to measure the voltage of a cell. Now, it is possible to use a galvanometer on its own to measure a voltage. However, galvanometers are so sensitive that they’re only able to measure voltages within a very small range.

The function of the multiplier resistor in a voltmeter is that it greatly increases or multiplies the maximum voltage that the galvanometer can measure. In a voltmeter, we typically find that the resistance of the multiplayer resistor, which we can call 𝑅 M, is much greater than the resistance of the galvanometer, which we can call 𝑅 G. As we can see, the same is true in this question. We’re told that the multiplier resistor has a resistance 50 times that of the galvanometer.

We’re then asked to calculate the ratio of the current in the galvanometer 𝐼 G to the current in a multiplier resistant 𝐼 M. So let’s start by writing down an expression for each of these currents in terms of their resistances, which we’ve been given some information about. We can do this using Ohm’s law, which tells us that the current in a conductor is equal to the voltage across that conductor divided by the resistance of that conductor. So we can say that the current in the galvanometer 𝐼 G is equal to the voltage across the galvanometer, which we could call 𝑉 G, divided by the resistance of the galvanometer, which is 𝑅 G.

Similarly, we can say that the current in the multiplier resistor 𝐼 M is equal to the voltage across the multiplier resistor, which we’ll call 𝑉 M, divided by the resistance of the multiplier resistor 𝑅 M. Now it’s really important to note that 𝑉 G and 𝑉 M are not necessarily the same. It’s tempting to assume that each of these voltages is simply the same as the voltage supplied by the cell, which we could call 𝑉 C. However, this isn’t the case. When we have resistors in series connected to a cell, as we do in this question, then the voltage drops across each component will add up to the total voltage supplied by the cell.

Now, the question is asking us to find the ratio of the current in the galvanometer 𝐼 G to the current in the multiplier resistor 𝐼 M. And one way of expressing the ratio of 𝐼 G to 𝐼 M is to calculate 𝐼 G over 𝐼 M, which must be equal to 𝑉 G over 𝑅 G divided by 𝑉 M over 𝑅 M. Dividing this fraction by this fraction is the same as multiplying this fraction by the reciprocal of this fraction, which gives us 𝑉 G over 𝑅 G times 𝑅 M over 𝑉 M, which is equivalent to 𝑉 G 𝑅 M over 𝑉 M 𝑅 G.

Now, the question tells us that the multiplier resistor has a resistance 50 times that of the galvanometer. In other words, 𝑅 M equals 50 𝑅 G. This means that we can substitute 50 𝑅 G in place of 𝑅 M in this expression, which then enables us to cancel the common factor of 𝑅 G in the numerator and the denominator, leaving us with 50 𝑉 G over 𝑉 M. Okay, so this simplifies our expression. However, we still don’t have a numerical value for this ratio. And we’re not able to calculate the actual values of 𝑉 G and 𝑉 M without first knowing the voltage supplied by the cell.

However, to help us, we can recall that when we have resistors connected in series with a cell, the size of the voltage drop over each resistor is proportional to its resistance. In other words, a bigger resistor will use a bigger proportion of the total available voltage supplied by the cell. Now, because we’re told that the resistance of the multiplier resistor is 50 times that of the galvanometer, that means that the voltage drop over the multiplier resistor, 𝑉 M, is 50 times the voltage drop over the galvanometer, 𝑉 G. Substituting 50 𝑉 G in place of 𝑉 M in our expression tells us that the ratio of 𝐼 G to 𝐼 M is 50 𝑉 G over 50 𝑉 G, which is equal to one. And this is the answer to our question.

However, there is a simpler way of answering this question that doesn’t require us to use any algebra. In fact, we don’t even need to know how a voltmeter is designed. Nor do we need to know anything about the resistances of the galvanometer and the resistor. In fact, it’s enough just to see that these two components are connected together in a single series circuit. In a series circuit, the rate of flow of charge, in other words, the current, is the same at every point, which means the current in the galvanometer 𝐼 G must be the same as the current in the multiplier resistor 𝐼 M. And if 𝐼 G equals 𝐼 M, then 𝐼 G over 𝐼 M is one. If we have a multiplier resistor connected in series with a galvanometer, then the ratio of the current in the galvanometer to the current in the multiplier resistor is one.