### Video Transcript

The diagram shows the scale of an ohmmeter that is being used to measure an unknown resistance. The resistance of the ohmmeter is 25 kiloohms. The angle of full-scale deflection for the ohmmeter Φ equals 60 degrees. The angle of deflection of the ohmmeter arm 𝜃 equals 15 degrees. What is the unknown resistance? Answer to the nearest kiloohm.

In our diagram, we see a representation of the ohmmeter display. Though an ohmmeter does indeed measure resistance, what’s directly being measured here is actually current. Current is measured by a component known as a galvanometer. If the measurement arm on our galvanometer, which in this case is being used indirectly as an ohmmeter, were fully deflected, it would appear as we see it here. This deflection is through an angle called Φ of 60 degrees. In the instance we’re considering though, the measurement arm is only deflected through an angle 𝜃 of 15 degrees. We can use the ratio of 𝜃 to Φ to solve for the unknown resistance being measured by our ohmmeter.

To start off, let’s fill out the circuit that our galvanometer is a part of. If we put a fixed as well as a variable resistor in series with a galvanometer, then, combined together, this is the ohmmeter that we’re using to measure resistance. We do it by using Ohm’s law, which tells us that the potential difference across a circuit equals the current in that circuit multiplied by the circuit’s resistance. We can see that all by itself, even without another resistor being placed in the circuit, the ohmmeter has some resistance. We can label that resistance 𝑅 sub Ω. This resistance is typically tuned so that the galvanometer in the ohmmeter will experience a full-scale deflection of its measurement arm. We can refer to the current indicated by that full-scale deflection as 𝐼 sub 𝐺.

By Ohm’s law, if we multiply 𝐼 sub 𝐺 and 𝑅 sub Ω, then that product will equal the potential difference across the circuit, that is, the voltage supplied by this cell. If we divide both sides of this equation by 𝐼 sub 𝐺 so that that factor cancels on the left, we can see how, if we know the voltage across our circuit and we know the reading of the galvanometer, we can calculate the resistance of our ohmmeter. But of course an ohmmeter is designed to measure the resistance of some other component in the circuit. So let’s say that we indeed open up our circuit and add in a resistor of unknown resistance that we’ll call 𝑅 sub 𝑢. To solve for this resistance, we can use Ohm’s law once again. Note that this first application only applies when the only resistance in our circuit is due to the presence of the ohmmeter.

Now that we’ve added an unknown resistance, we’ll want to make a new application of this law. Because nothing about our cell has changed, there’s still a potential difference 𝑉 across this circuit. The resistance of the circuit though is now equal to the resistance of the ohmmeter plus the resistance of the unknown resistor. Along with this, we know that the current in this circuit is no longer 𝐼 sub 𝐺. Rather, the ohmmeter measurement arm is now deflected through a different angle. Let’s say that the current in this circuit when the unknown resistor is present is 𝐼 sub 𝑢. An important point about this measurement scale is that in this case it is linear. That means, for example, that if the angle through which our measurement arm deflects doubles, say, then that indicates a doubling of our current.

Now when we divide 60 degrees by 15 degrees, that’s equal to four. This means that the maximum current readable by the galvanometer, 𝐼 sub 𝐺, is equal to four times the current in the circuit when our unknown resistor is present. Dividing both sides of this equation by four so that that factor cancels out on the right, we find that 𝐼 sub 𝑢 is equal to 𝐼 sub 𝐺 divided by four. And we can make that substitution over in this relationship. 𝐼 sub 𝐺 over four multiplied by this overall resistance of 𝑅 sub Ω plus 𝑅 sub 𝑢 is equal to 𝑉. Notice that in both of these equations, there is the resistance of the ohmmeter 𝑅 sub Ω. Our problem statement tells us that this resistance is 25 kiloohms, or 25000 ohms.

Clearing some space to work at the top of our screen, let’s now work on rearranging this equation we’ve developed so we can solve for 𝑅 sub 𝑢, the unknown resistance. First, we’ll multiply both sides by four divided by 𝐼 sub 𝐺 so that on the right-hand side the factors of four and the factors of 𝐼 sub 𝐺 cancel. Then, as a next step, we’ll subtract the resistance of the ohmmeter, 𝑅 sub Ω, from both sides so that on the right 𝑅 sub Ω minus 𝑅 sub Ω is zero.

Now, let’s rewrite this equation so that 𝑅 sub 𝑢 is on the left. Let’s now remember that 𝑅 sub Ω, the resistance of the ohmmeter, was found earlier to be equal to 𝑉 divided by 𝐼 sub 𝐺. If we make this substitution, replacing 𝑉 over 𝐼 sub 𝐺 with 𝑅 sub Ω, we find that 𝑅 sub 𝑢 equals four times 𝑅 sub Ω minus 𝑅 sub Ω, or three times 𝑅 sub Ω. Since 𝑅 sub Ω is 25 kiloohms, 𝑅 sub 𝑢 equals three times 25 kiloohms, or 75 kiloohms. This is the resistance of the until now unknown circuit resistor.