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 kilohms. The angle of full-scale deflection of the ohmmeter 𝜙 equals 60 degrees. The angle of deflection of the ohmmeter arm 𝜃 equals 54 degrees. What is the unknown resistance? Answer to the nearest kilohm.
Our diagram here shows us the measurement scale of the ohmmeter. The scale goes from zero, where the measurement arm is in this position, all the way up to a full-scale deflection reading, where the arm has moved through an angle 𝜙. Even though we’re talking about an ohmmeter, what’s being measured directly here is actually current. This measurement is made by a device called a galvanometer. When a galvanometer is connected in series with a fixed resistor as well as a variable resistor, altogether these components function as an ohmmeter, a device for measuring resistance.
The ohmmeter itself has some resistance. We’ll call this resistance 𝑅 sub Ω. And this value is specifically set so that when current is read through the galvanometer in the circuit, the measurement arm is deflected to a full-scale scale reading. We’ll call the current represented by this full-scale deflection 𝐼 sub 𝐺. If we multiply this maximum current measurable in the circuit by the resistance of the ohmmeter, then Ohm’s law tells us that this product is equal to the potential difference across the circuit. We’ll call that potential difference 𝑉.
Working from this equation, if we divide both sides by 𝐼 sub 𝐺 so that 𝐼 sub 𝐺 cancels out on the right, we get an expression for the resistance of our ohmmeter. It’s equal to the known voltage across our circuit divided by the current measured in the circuit. This is how our galvanometer that directly measures current can be used to indirectly indicate resistance. Of course, an ohmmeter is not designed to measure its own resistance, but rather the resistance of some other resistor.
Our problem statement tells us that our ohmmeter is being used to measure a resistor of unknown resistance. Let’s say we represent that resistor this way as 𝑅 sub 𝑢 in our circuit. When this resistor is present, our galvanometer no longer reads a full-scale deflection, but instead the measurement arm is now deflected through an angle 𝜃, which is less than 𝜙.
Specifically, we’re told that 𝜃 is 54 degrees, while 𝜙 is 60 degrees. Let’s say we call the magnitude of the measured current here when our unknown resistor is part of our circuit 𝐼 sub 𝑢. The ratio of 𝐼 sub 𝑢 to 𝐼 sub 𝐺 equals the ratio of the angles 54 degrees to 60 degrees. This is because we’re assuming that our measurement scale is linear. That means, for example, that if we double the amount of current in the circuit, then the deflection of the measurement arm will double as well. So, 𝐼 sub 𝑢 to 𝐼 sub 𝐺 is equal to 54 degrees divided by 60 degrees or nine-tenths.
Clearing some space on screen to work, if we take this equation for our currents and we multiply both sides by 𝐼 sub 𝐺 causing that current to cancel out on the left, then we find an expression for the current in our circuit when the unknown resistor is present in terms of the circuit current when that resistor is not present. Earlier on, we referred to Ohm’s law, which tells us that the potential difference across a circuit equals the current in the circuit multiplied by the circuit’s resistance. Let’s now make an application of this law in the case where our circuit involves all these components.
We know first of all that the potential difference across the circuit is 𝑉. That doesn’t change whether or not our unknown resistor is present. The current in the circuit with this resistor present is called 𝐼 sub 𝑢. And the total circuit resistance is the resistance of the ohmmeter plus the resistance of the unknown resistor. Ohm’s law tells us that this total resistance multiplied by this total circuit current equals 𝑉. Since it’s 𝑅 sub 𝑢 that we want to solve for, let’s divide both sides by 𝐼 sub 𝑢, canceling that current out on the right. And then with the remaining equation, we’ll subtract 𝑅 sub Ω from both sides so that on the right 𝑅 sub Ω minus 𝑅 sub Ω is equal to zero.
Then, finally, if we switch the left- and right-hand sides of this expression, we arrive at an expression where 𝑅 sub 𝑢 is the subject. In this equation, we don’t know the value for 𝑉 and we also don’t know the value for 𝐼 sub 𝑢. We do, however, know that 𝐼 sub 𝑢 is equal to nine-tenths 𝐼 sub 𝐺. If we make that substitution, it may not seem to help us because after all, we also don’t know 𝐼 sub 𝐺. Let’s recall at this point though that 𝑅 sub Ω can be expressed in terms of 𝐼 sub 𝐺. It’s equal to 𝑉, the same potential difference we’re working with here, divided by 𝐼 sub 𝐺. And notice now that in both of the terms on the right-hand side, we have 𝑉 divided by 𝐼 sub 𝐺. This means we can factor out this fraction, and that gives us this expression.
And now considering this first term, let’s multiply both its numerator and its denominator by 10. Since 10 divided by 10 equals one, we won’t be changing the value of this fraction. But notice that in the denominator, 10 multiplied by one over 10 equals one so that the fraction simplifies to ten-ninths. We can note now that one is equal to nine divided by nine. So, we find that 𝑅 sub 𝑢 is equal to 𝑉 divided by 𝐼 sub 𝐺 times one-ninth. It seems like we’ve made progress. But have we really? Because we still don’t know 𝑉 or 𝐼 sub 𝐺.
We can now once again recall that 𝑉 divided by 𝐼 sub 𝐺 is equal to 𝑅 sub Ω. And we do know the value for 𝑅 sub Ω; it’s given to us. We can then replace 𝑉 over 𝐼 sub 𝐺 with 𝑅 sub Ω. And making one last substitution, we know that 𝑅 sub Ω is equal to 25 kilohms. If we calculate this fraction, we find a result of 2.7 repeating kilohms. Our question, though, asks us to give our result rounded to the nearest kilohm. That is equal to three kilohms. Based on how its addition affected the deflection of our ohmmeter arm then, the resistance of the resistor we added to the circuit is three kilohms.
