Question Video: Using Experimental Results to Find the Resistance of a Resistor Physics • 9th Grade

A student has a resistor of unknown resistance. She places the resistor in series with a source of variable potential difference. Using an ammeter, she measures the current through the resistor at different potential differences and plots her results on the graph as shown in the diagram. What is the resistance of the resistor?

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Video Transcript

A student has a resistor of unknown resistance. She places the resistor in series with a source of variable potential difference. Using an ammeter, she measures the current through the resistor at different potential differences and plots her results on the graph as shown in the diagram. What is the resistance of the resistor?

Looking at our graph, we see it’s a plot of the current, in amperes, running through this resistor plotted against the voltage, in volts, running across it. And based on the description in the problem statement, we can make a little sketch of the circuit that generated the data plotted here.

Let’s say that this is our resistor of unknown value. We’re told that this resistor is connected to a variable potential difference supply and that also in this circuit is an ammeter for measuring current. The idea then is that we use this variable supply of potential difference to apply two, four, six, and eight volts across this resistor. And then using our ammeter, we read out corresponding current values of 0.4, 0.8, 1.2, and 1.6 amperes.

With these values plotted on the graph, we see they’ve been fit with a line of best fit that runs directly through all four points and also passes through the origin. Now this line is indeed a line that has a constant slope. And it’s that slope that will help us answer this question of what is the resistance of our unknown resistor.

To see how, let’s recall Ohm’s law. This law tells us that, for a resistor of constant value, that resistance multiplied by the current running through the resistor is equal to the voltage across it. In our case, we want to rearrange this equation to solve for 𝑅. And we see that that’s equal to the potential difference divided by the current. We aren’t given explicit values for the potential difference or the current. But we can get those from the data plotted in our graph.

Recall that those data points are the basis for the line of best fit that passes through all of them. This means that, in order to supply the voltage and current we need to solve for the resistance, 𝑅, we can choose from among any of our four data points plotted in this graph. In fact, we could choose from any point along this line of best fit line because it so happens to pass perfectly through all these data points. But just to make things easier, we may as well constrain our choice to these four. It doesn’t matter which of the four we choose. Any of them will give the same ratio and therefore the same overall result for the resistance of the resistor.

And just to pick one of the points then, let’s choose the one at four volts. That voltage corresponds to a current running through the resistor of 0.8 amps. So then to solve for the resistance of the resistor, we’ll divide four volts by 0.8 amps. When we do this, we find a result of five ohms, where ohm is the unit of resistance. Based on our graph and Ohm’s law, we find the resistance of the resistor to be five ohms.

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