A galvanometer has a resistance of
175 milliohms. A current of 20 milliamperes
produces a full-scale deflection of the galvanometer. Find the resistance of a multiplier
resistor that when connected in series with the galvanometer allows it to be used as
a voltmeter that can measure a maximum voltage of 15 volts. Answer to the nearest ohm.
We can start out here by picturing
a galvanometer, a device for measuring current. And specifically, we’ll think of
the scale used by the galvanometer to display the measured current. We’re told that when a current of
20 milliamperes, we’ll call this current 𝐼 sub G, exists in the galvanometer, it
causes a full-scale deflection of the galvanometer measurement arm. In other words, based on this
scale, 20 milliamperes is the maximum measurable current magnitude. Any current of greater magnitude
would not be able to be measured accurately by this galvanometer.
Maintaining the same current in the
circuit, we then imagine adding a resistor, called a multiplier resistor, in series
with the galvanometer. Our problem statement tells us that
together these two components can be used as a voltmeter. This is possible because of Ohm’s
law. This law says that the potential
difference across a circuit equals the current in that circuit multiplied by the
circuit’s total resistance. In our circuit, we would say that
the maximum measurable voltage, we’ll call it 𝑉 sub m, is equal to the maximum
current the galvanometer can measure times the total resistance of our circuit. And this total resistance, we see,
has two parts. There’s the resistance of the
galvanometer, we’ve called this 𝑅 sub G, and the resistance of the multiplier
resistor, 𝑅 sub m.
Our question asks us to solve for
the resistance of the multiplier resistor. We can do this by dividing both
sides of this equation by 𝐼 sub G, canceling out that factor on the right, and then
in the equation that remains, subtracting the resistance of the galvanometer 𝑅 sub
G from both sides. On the right, 𝑅 sub G minus 𝑅 sub
G is equal to zero. From here, if we then switch the
sides of our equation, we find that the resistance of the multiplier resistor equals
the maximum voltage measurable by our voltmeter divided by the maximum current
measurable by our galvanometer all minus the galvanometer’s resistance.
Our problem statement tells us that
𝑉 sub m, the maximum measurable voltage by our voltmeter, is 15 volts. 𝐼 sub G, the current that produces
a maximum deflection in the galvanometer, is 20 milliamperes. And finally, 𝑅 sub G, the
resistance of the galvanometer, is 175 milliohms.
Before we use this expression to
calculate 𝑅 sub m, we’ll want to convert milliamperes into units of amperes and
milliohms into units of ohms. We can do this by noting that the
prefix milli- means that we’re taking 10 to the negative three or one one thousandth
of some quantity. So for example, 20 milliamperes is
the same as 20 times 10 to the negative three amperes. And likewise, 175 milliohms equals
175 times 10 to the negative three ohms.
We’re now ready to calculate 𝑅 sub
m. When we do, rounding our answer to
the nearest ohm, we find a result of 750 ohms. This is the resistance of the