Video Transcript
A circuit that can be used as an
ohmmeter is shown. The circuit uses a galvanometer
with a resistance of 50 ohms that has a full-scale deflection current of 0.5
milliamperes. The circuit also includes a direct
current source with a voltage of 3.5 volts, a fixed resistor with a resistance of
2.5 kiloohms, and a variable resistor. The resistance of the variable
resistor is adjusted until the galvanometer arm is at a full-scale deflection
position. What resistance is the variable
resistor set to? Answer to the nearest ohm.
We’re told that this circuit here,
consisting of a galvanometer, a fixed resistor, a variable resistor, and a cell, can
function as an ohmmeter, a device for measuring resistance. This can happen by looking at the
reading of the galvanometer, a device that measures current, and working backward
from that reading using Ohm’s law to solve for the resistance of an unknown resistor
connected to the circuit.
In order to set up this circuit so
that it can function as an ohmmeter, we adjust the resistance of our variable
resistor, as our problem statement tells us, until the galvanometer is reading the
maximum current that its scale is calibrated to. When this happens, the galvanometer
arm, the needle used to indicate the galvanometer’s readings, is at what is called
its full-scale deflection position. Visually, if this was the face of
our galvanometer at a zero reading, then the full-scale deflection position of the
galvanometer arm would look this way.
Let’s recall now that Ohm’s law
tells us that the voltage in a circuit 𝑉 is equal to the current in that circuit 𝐼
multiplied by the circuit resistance 𝑅. For our circuit, we can write that
𝑉, the potential difference supplied by the cell, is equal to 𝐼 sub g, the maximum
current readable by the galvanometer, multiplied by the total circuit resistance
capital 𝑅. In our question, we’re asked to
indicate the resistance that the variable resistor is set to. The resistance of the variable
resistor along with that of the galvanometer and that of the fixed resistor make up
the total resistance in this series circuit 𝑅. Here, we’re treating our cell as an
ideal voltage supply that has no resistance.
Clearing some space at the bottom
of our screen, we can write that the total resistance 𝑅 of our series circuit is
equal to the resistance of the variable resistor plus the resistance of the
galvanometer plus the resistance of the fixed resistor. The reason we can solve for the
overall resistance of the circuit by simply adding these individual resistances
together is because all of these resistors are connected to one another in
series. This then is the equation for the
overall resistance 𝑅 in our circuit. And as we’ve seen, it’s the
resistance of the variable resistor 𝑅 sub v that we want to solve for.
If we take this equation and we
subtract 𝑅 sub g and 𝑅 sub f, the resistances of the galvanometer and fixed
resistor, respectively, from both sides of the equation, then both of those
resistances cancel out entirely on the right side. We find then that the resistance of
our variable resistor equals the overall circuit resistance minus the resistance of
the galvanometer minus the resistance of the fixed resistor.
Note that in our problem statement,
we’re told the resistance of the galvanometer and we’re also told the resistance of
the fixed resistor. However, we don’t yet know the
overall circuit resistance 𝑅. We can solve for it though using
our application of Ohm’s law. If we divide both sides of this
equation by the maximum current measurable by the galvanometer 𝐼 sub g, that factor
cancels on the right. And we find that the resistance 𝑅
equals 𝑉 divided by 𝐼 sub g. 𝑉, we can note, is the voltage
supplied by our cell. That’s given in our problem
statement as 3.5 volts. And 𝐼 sub g, the maximum
measurable current by the galvanometer, is a current of 0.5 milliamperes.
If we substitute 𝑉 divided by 𝐼
sub g in for capital 𝑅 in our equation for 𝑅 sub v, we arrive at this expression
for the resistance of the variable resistor 𝑅 sub v. Note that the values of all the
variables on the right side of this equation are known. 𝑉 equals 3.5 volts; 𝐼 sub g
equals 0.5 milliamperes; 𝑅 sub g, the resistance of the galvanometer, is 50 ohms;
and the resistance of the fixed resistor, 𝑅 sub f, is 2.5 kiloohms.
Substituting all these values into
our expression, we’re not quite yet ready to calculate 𝑅 sub v, because before we
can do that we’ll want to change our current from units of milliamperes to amperes
and the resistance of our fixed resistor from kiloohms to ohms. Note that this conversion will not
involve changing the actual physical value of these quantities. We only want to express them in a
different set of units. One milliampere is 10 to the
negative three or one one thousandth of an ampere, which means that 0.5 milliamperes
equals 0.5 times 10 to the negative three amperes. Likewise, one kiloohm equals 10 to
the three or 1,000 ohms. This means that 2.5 kiloohms equals
2.5 times 10 to the three ohms.
At this point, the units in all
three of our terms for 𝑅 sub v either are or simplify to ohms. We’re free then to calculate 𝑅 sub
v. And when we do, we get a result of
4,450 ohms. This is the resistance of the
variable resistor that we’ve used to calibrate the circuit.
