### Video Transcript

Analog meters use a galvanometer,
which essentially consists of a coil of wire with a small resistance and a pointer
with a scale attached. When there is current in the coil,
the pointer turns. The amount the pointer turns is
proportional to the amount of current in the coil. Galvanometers can be used to make a
voltmeter if a resistor is placed in series with the galvanometer. Consider a galvanometer that has a
resistance of 25.00 ohms and gives a full scale reading for a 50.00 microamp
current. The galvanometer is used to make a
voltmeter that has a full scale reading of 10.00 volts, as shown. Recall that a voltmeter is
connected in parallel with a component of interest. So the meter must have a high
resistance or it will change the magnitude of current in the component. What is the potential drop across
the series resistor in the meter? What is the resistance of the
parallel resistor?

In part one, we want to solve for
the potential drop across the series resistor. Weβll call this π sub π
. And in part two, we want to solve
for the resistance value of the resistor in parallel. Weβll call this value π
sub
π. To start on our solution, we can
recall that the current in this circuit, which we can call πΌ, is given as 50.00
microamps. And that the resistance of the
galvanometer, which we can call π
sub π, is given as 25.00 ohms.

To solve first for our voltage drop
over our resistor π
sub π, we can see from our diagram that a potential difference
of 10.00 volts is applied across the circuit. We know that some of that voltage
is lost over π
sub π and some over π
sub π. When we recall Ohmβs law that the
voltage dropped over a particular component is equal to the current through that
component multiplied by its resistance, we see that π sub π
is equal to Ξπ, the
total potential difference applied across the circuit, minus the product of the
current in the circuit times the resistance of the meter π
sub π. When we plug in for these three
values, making sure to use units of amperes for our current πΌ, and enter these
values on our calculator; we find that π sub π
, to four significant figures, is
9.999 volts. So π
sub π, the resistor setup in
series, uses up nearly all of this circuitβs potential difference.

Now we move on to solving for the
value of that resistor over which π sub π
volts drops. Again, working off of Ohmβs law, we
can write that Ξπ is equal to the current in the circuit, πΌ, plus the sum of the
two resistors, as shown in the diagram. When we rearrange this equation to
solve for π
sub π, we find itβs equal to Ξπ, the potential difference across the
whole circuit, divided by πΌ minus π
sub π. Weβre now ready to plug in for
these three values and solve for π
sub π.

When we do, again converting the
units of our current to amps, and enter these values on our calculator, we find that
π
sub π, the resistor placed in parallel with our meter, to four significant
figures, has a resistance of 200.0 kiloohms. Thatβs the resistance of the
resistor we add to make a voltmeter.