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 for the ohmmeter 𝜙 equals 60 degrees. The angle of deflection of the ohmmeter arm 𝜃 equals 45 degrees. What is the unknown resistance? Answer to the nearest kilohm.
Our diagram shows us a representation of the measurement scale of our ohmmeter. For a full-scale reading, the ohmmeter arm will be deflected all the way through this angle 𝜙 of 60 degrees. Even though this reading can be used to determine the resistance of some component, it’s actually a direct measurement of the current in the circuit. The current is measured by a device called a galvanometer. When the galvanometer is put in series with a fixed resistor and a variable resistor, then this combination of components forms an ohmmeter. This can happen by using Ohm’s law.
This law tells us that the potential difference across the circuit equals the current in the circuit multiplied by the circuit’s resistance. So in a given circuit, if we know the voltage 𝑉 supplied to the circuit and if we know the galvanometer reading, which we recall is a reading of current, then we can use information about 𝑉 and 𝐼 to allow us to solve for the resistance 𝑅.
If we call the total resistance of our ohmmeter 𝑅 sub Ω, we can solve for this value by noting that it’s a particular tuned resistance which corresponds to full-scale deflection of the ohmmeter arm. This indicates that the galvanometer is reading the maximum current it is scaled to read. Let’s call that current 𝐼 sub G. Ohm’s law tells us that if we multiply 𝐼 sub G and 𝑅 sub Ω, then we get the potential difference 𝑉 across our circuit. If we were to then divide both sides of this equation by 𝐼 sub G so that that factor cancels on the right, we would see that we now have an expression for 𝑅 sub Ω, the resistance of our ohmmeter.
In our problem statement, we were told that our ohmmeter has a resistance of 25 kilohms. An ohmmeter, of course, isn’t designed to measure its own resistance, but rather the resistance of some other component in the circuit. Here, we’ve put an unknown resistor into our circuit and we’ve called it 𝑅 sub u. It’s the resistance of this resistor that we want to solve for.
Returning for a moment to our measurement arm, we know that without the unknown resistor in the circuit, the ohmmeter’s arm was fully deflected to this position here. However, when we do add the unknown resistor to the circuit, the arm moves back so that now it’s only deflected an angle called 𝜃 from the zero point. That angle 𝜃, we’re told, is 45 degrees, while the angle 𝜙, the full-scale deflection angle, is 60 degrees. And we can see that the difference between these two angles is 60 minus 45 degrees or 15 degrees.
An important thing to note about this measurement scale is that the scale is linear. This means, for example, that doubling the deflection of the needle indicates a doubling of current. We can say then that the current indicated by the measurement arm being located here, we’ll call that current 𝐼 sub u to help us remember that it applies when the resistor 𝑅 sub u is in the circuit, is equal to three-quarters of the current when the measurement arm is fully deflected. This is because 45 degrees is three-quarters of 60 degrees. By adding our unknown resistor to the circuit then, we’ve decreased our current by one-quarter. That information can help us solve for 𝑅 sub u.
Clearing some space at the top of our screen, let’s write out an Ohm’s law equation for when our unknown resistor 𝑅 sub u is a part of our circuit. Just like before, the potential difference across the circuit is still 𝑉. The current in the circuit is now 𝐼 sub u. And the resistance in the circuit overall is the resistance of the ohmmeter plus the resistance of the unknown resistor. If we multiply this overall resistance by 𝐼 sub u, Ohm’s law tells us it is equal to 𝑉. Let’s now recall that 𝐼 sub u we found to be equal to three-quarters 𝐼 sub G. If we make that substitution into this equation, then we can continue solving for 𝑅 sub u by rearranging this equation so that 𝑅 sub u is the subject.
First, let’s multiply both sides of the equation by four divided by three times 𝐼 sub G. That way, on the right-hand side, the factors of three cancel out as do the factors of 𝐼 sub G and the factors of four. Then, with the resulting equation, we can subtract 𝑅 sub Ω from both sides so that on the right 𝑅 sub Ω minus 𝑅 sub Ω adds up to zero. If we then reverse the right and left sides of the remaining equation, we get this expression where indeed 𝑅 sub u is the subject.
At this point, let’s note that 𝑅 sub Ω, the resistance of the ohmmeter, is known to be 𝑉 divided by 𝐼 sub G. When we make that substitution, we can note that 𝑉 divided by 𝐼 sub G appears in both of these terms and so can be factored out. When we do this, we have 𝑉 divided by 𝐼 sub G multiplied by the quantity four-thirds minus one. We can replace the one with three divided by three, which equals one, which helps us see that four-thirds minus three-thirds is one-third. So then, the unknown resistor is equal to 𝑉 divided by three times 𝐼 sub G. And as we’ve seen, 𝑅 sub Ω is equal to 𝑉 divided by 𝐼 sub G. And so 𝑅 sub u is equal to 𝑅 sub Ω divided by three.
In our problem statement, we’re told that 𝑅 sub Ω is 25 kilohms, and 25 kilohms divided by three is equal to 8.3 repeating kilohms. Our question though asks us to solve for 𝑅 sub u to the nearest kilohm. So rounding to that level of precision, we find that the resistance of the unknown resistor in our circuit is eight kilohms.
