# Video: Identifying How Current Changes Depending on the Number of Parallel Paths in a Circuit

A student sets up the circuit shown in the diagram. Initially, the switch is open. When the student closes the switch, will the current flowing through the circuit increase or decrease?

08:17

### Video Transcript

A student sets up the circuit shown in the diagram. Initially, the switch is open. When the student closes the switch, will the current flowing through the circuit increase or decrease?

Taking a look at this diagram, we see in it a simple circuit. There is a cell providing voltage. And there are two resistors arranged in parallel with one another. On one of the resistor branches, we see there is a switch which currently is open. We want to figure out when the student closes the switch, so that this bottom section of the circuit is now a complete loop and current can follow through it, will that overall current in the circuit increase or decrease? It’s an interesting question. And before we figure out what happens to the current when the switch is closed, let’s first map out where the current can move when the switch is open.

Looking at our cell, we see that conventional current will move in a counterclockwise direction. So, it will come around here, and then it will reach this junction point in the circuit. If the switch at the bottom part of the circuit were closed, that would mean that the current has the option of travelling through this lower branch. But of course the switch is open. And the current has no such option. That means 100 percent of the current will flow through the upper branch, through that resistor, and then back around to the negative terminal of the cell. That’s the current in the circuit with the switch open.

But then we’re told the student closes the switch. When that happens, now it’s possible when current reaches this part of the circuit to continue on downward. There is a closed path for it to follow. So it’s able to traverse the other resistor and then come up and join the other branch. And as we said, the question is, what happens to the current when the switch is closed? Does it go up overall or does it go down overall?

Let’s start answering this question by recalling a law of electrical circuits, Ohm’s law. This law says that for a resistor of constant resistance, if we multiply that resistance value by the current running through it, then that’s equal to the potential difference across the resistor. Looking at our circuit, we see that it has virtually no labels. But that doesn’t mean we can’t apply some ourselves. What if we go to our cell and say that this cell produces a potential difference of 𝑉? And not only that, but we can also give labels to our unnamed resistors. Let’s call the top resistor 𝑅 sub one. And we’ll call the bottom resistor 𝑅 sub two. Notice that we haven’t specified what these values are or even how they relate to one another. We’re just giving them names.

We’re interested overall in two states of this circuit. One we could call the before state. That’s when the switch is still open and current only flows through the one resistor. If we apply Ohm’s law in this scenario before the switch is closed, then we can say that 𝑉, the potential difference created by the cell, is equal to 𝐼 — we’ll call it 𝐼 sub 𝑏, the current in this circuit before the switch is closed — multiplied by the only resistor in the circuit, which is 𝑅 sub one. Recall that when the switch was open, the current had no choice but to travel through this upper branch in the circuit through the resistor 𝑅 sub one. It didn’t go through 𝑅 sub two at all.

So then, before the switch was closed, this is our Ohm’s law equation. And we can rearrange it to solve for that current, 𝐼 sub 𝑏. 𝐼 sub 𝑏, the current in the circuit before the switch was closed, is equal to the potential provided by the cell divided by 𝑅 sub one. Of course we don’t know what 𝑉 and 𝑅 sub one are, but we don’t need to. We only need to solve for 𝐼 sub 𝑏 in relation to the current after the switch is closed, what we’ll call 𝐼 sub 𝑐. So, let’s look at that now. Let’s look at the current in the circuit after the switch is closed.

Before we apply Ohm’s law to the state after the switch is closed, there is one important point to make. When the switch is closed, we noted that the current divides up across the two parallel branches of the circuit. In other words, if we picked a point on either one of these two parallel branches, we wouldn’t get the total circuit current. But since it is the total circuit current that we want to solve for, we’re gonna analyze this circuit at a point where the current is undivided, before it splits between the branches. Let’s analyze the circuit right at this point there.

Applying Ohm’s law to this scenario, we can say once again that the potential difference supplied by the cell is equal to the total current in the circuit at that point, what we’ll call 𝐼 sub 𝑐, multiplied by the total resistance, right now we’ll just call it 𝑅 sub 𝑡, of our circuit. Briefly comparing this equation with our previous equation that we used to solve for 𝐼 sub 𝑏, we can see that the crux of the matter is 𝑅 sub 𝑡, the total resistance of the circuit, compared to 𝑅 sub one. It’s the comparison between those two values which will determine whether 𝐼 sub 𝑐 is greater than or less than 𝐼 sub 𝑏. That is, whether the current has increased or decreased.

So then, what is 𝑅 sub 𝑡? What is the total equivalent resistance of our circuit when the switch is closed? Under this condition, we have two resistors arranged in parallel. We’ve called them 𝑅 sub one and 𝑅 sub two. When resistors are arranged this way and there is exactly two of them, there is a particular mathematical relationship for their overall or equivalent resistance. We can call that equivalent resistance 𝑅 sub two 𝑝, when two resistors are arranged in parallel. In that case, their overall resistance is equal to the product of their individual resistances divided by their sum.

Let’s now apply this relationship to our scenario with 𝑅 sub one and 𝑅 sub two. When we do this, we see that the total resistance of our parallel circuit, when the switch is closed, is equal to 𝑅 one times 𝑅 two divided by 𝑅 one plus 𝑅 two. At first glance, substituting that in for 𝑅 sub 𝑡 may not look like it helps us very much. After all, when we rearrange to solve for 𝐼 sub 𝑐, the total circuit current after the switch is closed, all we get on the right-hand side is this expression. And remember, we don’t know what 𝑉, 𝑅 sub one, or 𝑅 sub two are. But thankfully, we don’t have to. All we need to know is which is bigger, 𝐼 sub 𝑐 or 𝐼 sub 𝑏.

The way we can figure that out is by rewriting this denominator expression in our 𝐼 sub 𝑐 equation. Notice that it has 𝑅 sub one multiplied by 𝑅 sub two divided by the sum of 𝑅 sub one plus 𝑅 sub two. That means we could write this a different way. We could express that as 𝑅 sub one multiplied by the quantity, 𝑅 sub two divided by 𝑅 sub one plus 𝑅 sub two. The significance of doing this is that now we have 𝑉 divided by 𝑅 sub one, which is what we have over for 𝐼 sub 𝑏. That’s equal, simply, to 𝑉 divided by 𝑅 sub one.

So then, our whole question now hinges on this. Is this expression here, 𝑅 sub two divided by 𝑅 sub one plus 𝑅 sub two, is that greater than or less than one? If it’s greater than one, then that means when we multiply it by 𝑅 sub one, overall, compared to 𝐼 sub 𝑏, 𝐼 sub 𝑐 will be lower. On the other hand, if 𝑅 sub two divided by 𝑅 sub one plus 𝑅 sub two is less than one, then that means when we multiply it by 𝑅 sub one, our overall current, 𝐼 sub 𝑐, will go up compared to 𝐼 sub 𝑏. So, which is it? Is 𝑅 sub two divided by 𝑅 sub one plus 𝑅 sub two greater than one? Or is it less than one?

We can think about it like this. 𝑅 sub two appears in both the numerator and the denominator. If 𝑅 sub one was equal to zero, then this whole fraction would simply be one. It would be 𝑅 sub two divided by 𝑅 sub two. But we know that 𝑅 sub one is there in the denominator. And we can assume that 𝑅 sub one is greater than zero. In other words, that resistor really is there in the circuit. And so, when we add that nonzero number for 𝑅 sub one to the value for 𝑅 sub two and then divide that sum into 𝑅 sub two, what we’ll get is a result which is less than one. And that’s because the denominator, 𝑅 sub two plus 𝑅 sub one, is greater than the numerator, 𝑅 sub two, by itself. So, if this fraction here is less than one, and we just found that it is, then when we multiply that by 𝑅 sub one and then divide it into 𝑉, what we’re doing is dividing into one numerator by a relatively smaller denominator compared to the denominator for 𝐼 sub 𝑏.

A way of clarifying this mathematically would be to take 𝐼 sub 𝑏, which is equal to 𝑉 divided by 𝑅 sub one, and substitute that in for the expression in 𝐼 sub 𝑐. When we do that, as well as rearrange the fraction in the denominator of 𝐼 sub 𝑐 as expression, then we find that 𝐼 sub 𝑐 is equal to 𝐼 sub 𝑏 multiplied by this value, which is greater than one. This tells us that the current after the switch is closed goes up compared to where it was before the switch was closed. That means we can answer this question by saying that the current will increase when the switch is closed.