Video: Eg17S1-Physics-Q36



Video Transcript

An ohmmeter has a circuit with resistance 3750 ohms, where the maximum current that can pass through it is 400 microamps. Calculate the value of the external resistance that makes its pointer deflect halfway through the scale.

Here, we can say, is a sketch of this ohmmeter which is a device used to measure electrical resistance. The ohmmeter has an analogue scale that we used to read out the measure resistance. And we see there are markings on the scale for the minimum resistance, zero, the maximum, and then values in between that. Now in our problem statement, it says that the ohmmeter itself has a circuit with a certain resistance of 3750 ohms. At first, that may be confusing. Doesn’t the ohmmeter measure resistance rather than have a resistance?

It’s true that the ohmmeter measures resistance. But the way it does it is by creating its own little closed circuit. When we connect the two electrical leads that typically come from an ohmmeter, we create a closed circuit for current to flow through. This circuit has everything. It has current, it has resistance, and it has potential difference. That potential is supplied by the battery that typically powers the ohmmeter. We can call the potential then 𝑉 sub 𝐵.

In the problem statement, we’re not told what that potential is. But we are told the ohmmeter’s resistance, 3750 ohms, as well as the maximum current it can handle, 400 microamps. Knowing this let’s recall Ohm’s law which tells us that the potential difference across a circuit is equal to the current in the circuit multiplied by its resistance. This means that the battery potential 𝑉 sub B is equal to the maximum current through our ohmmeter, we will call 𝐼 sub m, multiplied by the resistance of the ohmmeter, we’ll call it 𝑅 sub ohm. Remember, this resistance is an internal resistance provided by the ohmmeter itself. It’s not an externally measured resistance.

That means that whatever resistance we measure using the ohmmeter, this value 𝑅 sub ohm will be in addition to that resistance. This is just the value that comes built-in with our measurement device, Now, here’s an important point about how an ohmmeter works. This potential difference 𝑉 sub B is a constant. And this internal resistance 𝑅 sub ohm is a constant too. So when we add an external resistor to our ohmmeter to be measured, what we actually end up measuring is not the resistance directly but rather the current in the circuit.

This is important because we wanna calculate the value of an external resistor that makes the pointer on our ohmmeter which we know indicates resistance deflect halfway through its scale. But here’s the thing, if that pointer deflects halfway through its scale, if it comes to a position marked out here, then what that really means is that the current running through the circuit that includes the external resistor is half the maximum value that the ohmmeter can handle. And that half maximum value corresponds to deflection of the pointer halfway through its scale.

Let’s say then that we take an external resistor, we’ll call it 𝑅 sub 𝑒, and we hook it up to our ohmmeter circuit. When we do that, when we connect up this external resistance to the ohmmeter circuit, we’re told that the pointer deflects to half its maximum value. As we said, this means that a current of half our maximum value is being measured by this ohmmeter. Now that we’ve added a resistor to our ohmmeter circuit, we know that that doesn’t change the fact that Ohm’s law still applies to it. And in fact, the potential of our circuit is the same as it was before 𝑉 sub 𝐵. This means then that we can write a second expression for Ohm’s law for ohmmeter circuit, now that we have an external resistor added in.

Because our pointer deflected halfway through the scale, that tells us that the current in our circuit is half its maximum value. That is, 𝐼 sub 𝑚 divided by two. And now as far as the resistance of the circuit, it’s the sum of the ohmmeter’s internal resistance, 𝑅 sub ohm, plus our external resistor 𝑅 sub 𝑒. That is, these two resistances are in series with one another. Now that we have these two, in some sense, separate equations for Ohm’s law, we can ignore the fact that we don’t know 𝑉 sub 𝐵, the potential difference supplied by the battery, and realise that we now have an equation between the before and after of inserting our external resistor. And recall that it’s that value we want to solve for.

Here is how we’ll do that. Let’s start out algebraically by subtracting 𝐼 sub 𝑚 divided by two multiplied by 𝑅 sub ohm from both sides of this equation. Now, if we do that, notice what happens? We have the factor 𝐼 sub 𝑚 divided by two on both sides of our equation. So algebraically, it cancels out.

That means, perhaps surprisingly, that in order for this condition to be true, in order for our pointer to deflect halfway through the scale when we connect up our external resistor, it must be the case that the internal resistance of our ohmmeter is equal to that external resistor. And what is our internal resistance? We’re given that as 3750 ohms which means then that must be the value of 𝑅 sub 𝑒, our external resistor. This is the external resistance value which meets the condition of deflecting the pointer just as described.

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