# Video: Understanding Types of Electric Current

The graph shows the potential difference against time for three current sources. Which two current sources have a potential difference with a magnitude that is always nonnegative? Which current source has a potential difference that is constant over time? What is the peak voltage of source (c)? What is the peak voltage of source (b)? Which of the current sources shown in the graph is most likely to be produced by a dynamo containing a commutator?

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### Video Transcript

The graph shows the potential difference against time for three current sources.

Okay, before we get to our questions, let’s take a quick look at this graph. We see it shows us potential difference versus time for these three current sources. One is marked out a in green. The other is b in blue and c in red. Knowing that, on to our first question.

Which two current sources have a potential difference with a magnitude that is always nonnegative?

Nonnegative means that a value can be zero or it can be positive. But it just can’t be less than zero. Looking at our three current sources plotted out, we see that source c, the one in red, definitely falls below the horizontal axis. It takes on negative potential difference values. On the other hand, sources a, which is always positive, and b, which is either positive or zero, never do attain negative values. So that’s our answer to this question. It’s sources a and b which are always nonnegative.

The next question says which current source has a potential difference that is constant over time?

Looking back at our graph, we see source a, the green line, always has the same value. It looks to be 140 volts. This is the only plot of the three that doesn’t change with time. So a is our answer. The potential difference of this source stays constant over the timescale shown. Okay, on to a few more questions about this graph.

The next thing we want to know is what is the peak voltage of source c?

From our graph, we can tell that source c is the red curve. And the peak voltage is the maximum voltage that curve attains as it oscillates up and down. To find that value, we’ll mark out the highest part of the curve for this source. And then we’ll trace a horizontal line over to the vertical axis.

We see this line crosses that axis between 150 and 100 volts. In particular, we can divide that span up into five evenly marked-out parts. Each part then represents a difference of 10 volts. Our horizontal line crosses the axis two of those parts above 100 volts. Since each one of the two parts represents a difference of 10 volts, that means our line crosses the axis at 120 volts. We’ll write that down as our answer, the peak voltage of source c.

Next, we ask the same question but this time about source b. To find this out, we’ll start by locating the highest point, the peak, on the blue curve. That’s for source b. We see that this source has a maximum value in line with the maximum value for source c. In other words, if we drew a horizontal line over from this maximum value, it would be on the same vertical level as the maximum for source c. This tells us that if we follow this line all the way to the vertical axis, we’ll find the same result as we did earlier. Source b has the same peak or maximum voltage as source c, 120 volts. Then, finally, we move on to our last question.

Which of the current sources shown in the graph is most likely to be produced by a dynamo containing a commutator?

The important thing to know here is what a commutator does. A commutator looks a bit like this, like a metal ring that’s been split into two halves. When a split ring like this is integrated into a generator that works by electromagnetic induction, the effect of this is to make all of the current generated by the generator to flow in the same direction. In other words, the potential difference always has the same sign. And, by convention, that’s a positive sign.

A commutator, in other words, would take a sinusoidally varying potential difference. And it would take all of the negative values and invert them about the horizontal axis. That is, if we had a graph that started out looking like this, then the effect of putting a commutator into the circuit would be to invert the negative parts of this curve. This dashed line we’ve just left for comparison. The real part of the curve, once the commutator is involved, is all what’s above the horizontal axis.

We can now turn back to our graph and see if any of the three current sources generate a plot that looks like what we’ve seen here. And, indeed, we see that current source b, the one in blue, does look like this. This plot looks like, originally, it might’ve had half of these data points falling below the horizontal axis. But, then, a commutator was inserted into the system, making the generator a dynamo. And all of these negative values were inverted or rectified above that horizontal axis.

All this to say, current source b is most likely to have been produced by a dynamo containing a commutator. That’s because this voltage-versus-time curve looks very much like a sinusoidal curve that has been rectified or all brought above the horizontal axis by a commutator.