Video: Identifying Equivalent Circuit Diagrams

The diagram shows four circuits with components connected in parallel. Which two circuits are equivalent?


Video Transcript

The diagram shows four circuits with components connected in parallel. Which two circuits are equivalent?

So here, we’ve got four circuit diagrams: a, b, c, and d. And we need to find out which two represent the same thing so one way to do this is to start with circuit a and then compare it to b, c, and d. And if none of those work, then we can go on to circuit b, compare them to c and d, so on and so forth. This way we can try out every single possibility if necessary and then we can find the two that are equivalent.

Now let’s first describe circuit a in words so that we know what we’re looking for in the other circuits. So circuit a consists of a cell here in parallel with a 10-ohm resistor. And that is in parallel with a system that consists of a lamp and a 20-ohm resistor.

The lamp and 20-ohm resistor are in series. And that whole lamp slash 20- ohm resistor combo is in parallel with the other two branches. Also the cell is lined up such that the positive terminal meets the lamp first in the branch which contains the lamp and the resistor.

In other words, conventional current, which is positive charge, flows this way and meets the lamp first. So that’s basically what we’re looking for in all of the other circuits, assuming of course that a is one of the two equivalent circuits. It may not be, but for now we’re checking if it is. So let’s compare it to circuit b.

We’ve also got a cell, a 20-ohm resistor, a 10-ohm resistor and a lamp. In fact, that’s true for all four circuits. So we can’t eliminate any of them based on just the components that they have. But we need to show some understanding of how parallel circuits work here.

So in circuit b, we’ve got a cell and it’s in parallel with a 20-ohm resistor that’s by itself. Now this automatically means that t’s not identical to circuit a because in circuit a, the cell was parallel to a 10-ohm resistor that was by itself. So automatically we know that a and b is not the solution.

Let’s now move on to circuit c. Well circuit c immediately stands out because it looks different to the others, because it’s got a bit that sticks out. But does that mean that it’s not identical to any of the others? Let’s check.

Let’s look at the stickie outie bit. We’ve got this 10-ohm resistor that’s in parallel with only the lamp. In other words, it’s not in parallel with both the 20-ohm resistor and the lamp. That’s not what it is. And in none of the other circuits is anything parallel to only the lamp.

Therefore, we can say that a and c are not equivalent circuits. In fact, we can confidently say that none of the other circuits are equivalent to c. So even though we were comparing circuit a to all the other circuits, we’ve also just managed to completely eliminate circuit c from the picture.

So let’s move on to circuit d and compare that to circuit a. So in circuit d, we’ve got a cell that’s in parallel with a system that consists of a lamp and a 20-ohm resistor in series. And that whole thing is in parallel with the 10-ohm resistor.

That sounds familiar, doesn’t it? That sounds a little bit like circuit a, especially if we were to rotate the diagram of circuit d 90 degrees anticlockwise just so it looks similar to circuit a. And this is what it looks like rotated 90 degrees anticlockwise.

So it kind of looks like circuit a, but it looks like these two branches here have been swapped. In circuit a, we’ve got the 10-ohm resistor in the middle and the lamp and the 20 ohm-resistor at the bottom.

However, because these circuits are connected in parallel, it doesn’t matter. This is because any parts of a circuit connected in parallel have the same voltage across it. So in other words, the voltage across this bit is the same as the voltage across this bit.

Also in a parallel circuit, we know that the current splits. And in this case, it’s going to split in a certain ratio. That ratio depends on the total resistances of the branches along which it’s going. So it depends on the resistance, once again, of this branch and of this branch.

However, these two branches are exactly the same as they are in circuit a. They’re just wind up the wrong way round. So the current is still gonna split in the same ratio because the ratio of the resistance of this branch in total to the resistance of this branch in total is exactly the same as in circuit a.

That makes sense. We’ve got a 10-ohm resistor in parallel with a system that consists of a lamp and a 20-ohm resistor. And that’s true for circuit d as well. So although we’re not going to do any calculations, let’s make some assumptions for argument’s sake.

Let’s say that the current in the circuit is I don’t know 100 amps. Now let’s say that the current splits along this branch so that it goes as 20 amps along the 10-ohm resistor and the rest of the 80 amps go along this bit.

And the same thing happens in circuit d. Let’s say we’ve got a 100-amp current again. Well because we’ve got the branch with the lamp and the 20-ohm resistor, 80 of those ohms are going to go along that branch and 20 along the branch with the 10- ohm resistor.

Now obviously these numbers mean nothing, but it only goes to illustrate that the splitting of the current along these branches is going to be exactly the same. So as far as a circuit is concerned, it doesn’t really matter which way round we connect the two branches.

Because they’re in parallel, it doesn’t make a difference. And therefore, circuits a and d do appear to be equivalent. Now, here we can see we’ve compared circuit a to d, a to b, a to c. We’ve also compared b to c and d to c.

So the only combination that’s left is circuits b to d. Now if we think a and d are identical and a and b are different, then by that logic b and d must be different also. But we can- we can show that.

So in circuit d, we’ve got a system where the conventional current, the positive current, flows to two branches. And one of these branches has a lamp and a 20-ohm resistor. The conventional current first meets the lamp. However, in circuit b, the conventional current flows this way around and first meets the resistor.

Also there’s a small matter that the resistors in this case are swapped, exactly the same problem as we had with circuit a. We’ve got the lamp here in circuit b with a 10-ohm resistor rather than the 20-ohm resistor, which is what we see in circuit a and d. So overall, the two equivalent circuits are circuit a and circuit d.

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