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.
