The diagram shows the scale of an ohmmeter that is being used to measure an unknown resistance. The resistance of the ohmmeter is 25 kilohms. The angle of full-scale deflection of the ohmmeter 𝜙 equals 60 degrees. The angle of deflection of the ohmmeter arm 𝜃 equals six degrees. What is the unknown resistance? Answer to the nearest kilohm.
Our diagram shows us a representation of the ohmmeter measurement scale. That scale we see goes from zero all the way up to this full-scale measurement, where the arm of the ohmmeter is deflected by an angle 𝜙. Even though an ohmmeter is a device for measuring resistance, what’s actually being directly measured here is current.
Electric current is being measured using a device called a galvanometer. When a galvanometer is connected in series with a fixed resistor as well as a variable resistor, then these components altogether serve as an ohmmeter. This happens, we can say, indirectly, in a way that makes use of Ohm’s law. This law tells us that the potential difference across a circuit equals the current in the circuit multiplied by the circuit’s resistance. If we call the potential difference supplied by our cell 𝑉 and the resistance of the ohmmeter 𝑅 sub Ω, then these two quantities are related to one another by Ohm’s law.
Whenever we have an ohmmeter designed like this, the variable resistor is tuned to a particular value so that when the circuit consists just of the ohmmeter and the voltage supply, like it does here, the current read out by the galvanometer is the maximum current that scale allows. We’ll call current of this magnitude 𝐼 sub G. 𝐼 sub G multiplied by 𝑅 sub Ω is equal to the potential difference across the circuit 𝑉. If we divide both sides of this equation by 𝐼 sub G so that that current cancels out on the right, we get an expression for the resistance of our circuit. At the moment, that’s just the resistance of the ohmmeter. But we can use a similar approach to solve for the resistance of an unknown resistor 𝑅 sub u in the circuit.
The idea is that without this unknown resistor in the circuit, our measurement arm was fully deflected. With this unknown resistor added in though, the deflection of this arm is now through an angle of 𝜃, where 𝜃 we’re told is equal to six degrees. That angle is to be compared to the full-scale deflection angle of 60 degrees. Six degrees divided by 60 degrees equals one-tenth. And this tells us that when our unknown resistor is present in the circuit, the current in the circuit is one-tenth what it was before that resistor was added. We’ve called the full-scale deflection current in the circuit 𝐼 sub G.
Let’s now say that the current when the unknown resistor is present in the circuit is 𝐼 sub u. The ratio of the angle 𝜃 to the angle 𝜙 tells us that 𝐼 sub u is equal to one-tenth 𝐼 sub G. This fact, along with our knowledge of 𝑅 sub Ω, which our problem statement tells us is 25 kilohms, will help us to solve for 𝑅 sub u.
To do that, let’s clear some space at the top of our screen and then apply Ohm’s law for the case where our circuit looks like this, with 𝑅 sub u included. Just like before, the potential difference across the circuit is 𝑉. Now though, the current in the circuit is not 𝐼 sub G, but it’s 𝐼 sub u. And the total circuit resistance is 𝑅 sub Ω, the resistance of the ohmmeter, plus the resistance of our unknown resistor. Let’s recall that it’s 𝑅 sub u that we want to solve for. To help us do that, let’s divide both sides of this equation by 𝐼 sub u, canceling that current out on the right. And then, with our remaining equation, we can subtract 𝑅 sub Ω from both sides so that on the right-hand side, 𝑅 sub Ω minus 𝑅 sub Ω adds up to zero.
With those terms canceling one another out, our right-hand side simplifies to 𝑅 sub u. And if we then swap the sides of our equation, we get an expression for the resistance of our unknown resistor in terms of the potential difference 𝑉, the current 𝐼 sub u, and the resistance of our ohmmeter 𝑅 sub Ω. At this point, let’s note first of all that 𝐼 sub u is equal to one-tenth 𝐼 sub G. Therefore, we can make that substitution in for 𝐼 sub u here. Having done that, we can multiply this fraction by 10 divided by 10. Since 10 over 10 is one, we’re not changing the value of this fraction. But notice that in the denominator, 10 and one-tenth cancel one another out. The fraction overall can be written as 10 times 𝑉 divided by 𝐼 sub G.
Now, let’s recall that 𝑅 sub Ω, the resistance of the ohmmeter, is equal to 𝑉 divided by 𝐼 sub G. Making that substitution, now on the right-hand side of our equation, we have two terms, both of which have 𝑉 divided by 𝐼 sub G in them. This means we can factor out that fraction. We get 𝑉 divided by 𝐼 sub G times the quantity 10 minus one, and this is equal to nine times 𝑉 divided by 𝐼 sub G. We can now use the fact that this fraction 𝑉 divided by 𝐼 sub G equals 𝑅 sub Ω, and 𝑅 sub Ω we know to be equal to 25 kilohms. What we found then is that the unknown resistor 𝑅 sub u has a value equal to nine times 25 kilohms, and that works out to 225 kilohms. This is the resistance of the resistor added to our circuit.