# Video: Finding Pairs of Resistances that can be used in a Potential Divider

Isabella has a light bulb that requires a 12 V potential difference across it in order to work. She only has an 18 V battery to power it. Which of the following pairs of resistors can be used to create a potential divider to power the light bulb? [A] 8 kΩ and 6 kΩ [B] 12 kΩ and 2 kΩ [C] 200 Ω and 500 Ω [D] 10 kΩ and 5 kΩ [E] 48 Ω and 36 Ω

10:22

### Video Transcript

Isabella has a light bulb that requires a 12-volt potential difference across it in order to work. She only has an 18-volt battery to power it. Which of the following pairs of resistors can be used to create a potential divider to power the light bulb? A) Eight kilo ohms and six kilo ohms, B) 12 kilo ohms and two kilo ohms, C) 200 ohms and 500 ohms, D) 10 kilo ohms and five kilo ohms, E) 48 ohms and 36 ohms.

Okay, so in this question, what we’ve been told is that Isabella has a light bulb requiring a 12-volt potential difference across it. But the problem that she has is that she’s only got an 18-volt battery. However, this is not really a problem because she can create a potential divider circuit that will allow her to get 12 volts of potential difference across her light bulb, even with an 18-volt battery. So let’s first recall what a potential divider circuit actually is.

Let’s first start with Isabella’s 18-volt battery. Now what we’ve drawn here are two cells placed end to end because remember, a battery is two or more cells placed end to end. But that’s not really relevant here. We can attach wires to both terminals of the battery. So here’s the wire coming out of the positive terminal and here’s the wire coming out of the negative terminal. Additionally, a potential divider circuit will consist of a pair of resistors. And these resistors will be connected in series with each other.

So here’s the first resistor. We will say it has a resistance 𝑅 one. And here is the second resister. We will say it has a resistance 𝑅 two. And then, all we need to do to finish off our potential divider circuit is to connect wires across the two terminals of one of these resistors. So let’s say we connect a wire of one terminal of resistor 𝑅 two and the other terminal of resistor 𝑅 two. What this allows us to do is to place a component here in this part of the circuit — let’s say Isabella’s light bulb. and the reason we do this is the following.

Let’s first start by recalling that resistor 𝑅 one and 𝑅 two are placed in series with each other. And this, therefore, means that the potential difference provided by our battery, which happens to be 18 volts will be shared between resistors 𝑅 one and 𝑅 two. In other words, some of those 18 volts will be dropped as potential difference across 𝑅 one and the rest of those volts will be dropped across resistor 𝑅 two. Then, we can also remember that components placed in parallel with each other will have the same potential difference across them. So whatever the potential difference across resistor 𝑅 two is, that is the same as the potential difference across our light bulb here because the resistor 𝑅 two and our light bulb here are placed in parallel with each other.

So this is how a potential divider circuit works. We start with a battery that generates a certain potential difference. And then, we share that potential difference over, say, two or more resistors. Finally, we connect a component in parallel with one of the resistors so that the potential difference across the component is the same as the potential difference across that resistor. Now, we’ve been told that the question that the light bulb only works if there are 12 volts of potential difference across it. This means that in our diagram we can label that there are 12 volts of potential difference across the light bulb.

And then because resistor 𝑅 two is in parallel with the light bulb, there must also be 12 volts of potential difference across it, which means that out of the 18 volts of potential difference generated by the battery, 12 of them are across resistor 𝑅 two and the rest must be across resistor 𝑅 one. In other words, the potential difference across resistor 𝑅 one is going to be 18 volts minus 12 volts, and that is equal to six volts, which we can label on our diagram.

So, to recap quickly, we’ve set up a circuit such that these 18 volts of potential difference generated by the battery are split across the resistors as six volt across one of them and 12 volts across the other. And we connect the lamp in parallel with the resistor that has 12 volts across it. So we need to find a combination of the values 𝑅 one and 𝑅 two the resistance is of the resistors that would result in six volts of potential difference across this resistor and 12 volts across this resistor. To do this, we’re first going to disconnect our light bulb from the circuit. This allows us to simplify things a little bit and only focus on the resistors 𝑅 one and 𝑅 two. And more importantly, it will allow us to simplify the current flow through the circuit, which we’ll need to consider in a moment.

But before we do, we’re going to recall something known as Ohm’s law. Ohm’s law tells us that the potential difference across a component in the circuit — let’s say this resistor here — is equal to the current through that component, 𝐼, multiplied by the resistance of that component, which in this particular cases 𝑅 one. And of course, we can apply Ohm’s law to the other resistor 𝑅 two, as well as to the entire circuit. And here’s where we find out why we momentarily disconnected the light bulb from our circuit. Because now, we can recall that in a series circuit, the current through all of the components is the same.

In other words, if we consider conventional current, which is the flow of positive charges moving from the positive terminal of the cell to the negative terminal, we see that the current flow through the six-volt resistor is the same as the current flow through the 12-volt resistor. And let’s say that the magnitude or size of this current is 𝐼, which means we can now apply Ohm’s law to resistor one. We can say that the potential difference across resistor one, which we’ll call 𝑉 one, is equal to the current through that resistor, which we know is 𝐼, multiplied by the resistance of that resistor, which is 𝑅 one.

But then, we already know that we want the value of 𝑉 one to be six volts. And so, we can say that six volts is equal to 𝐼 times 𝑅 one. Then, we can apply Ohm’s law to resistor 𝑅 two. We can say that the potential difference across 𝑅 two, which is 𝑉 two, is equal to the current through that resistor, which is once again 𝐼, multiplied by the resistance of that resistor, 𝑅 two. And once again, we do the same thing as before. We can see that we want 𝑉 two to be 12 volts. And hence, our equation becomes 12 volts is equal to 𝐼 times 𝑅 two.

Now what we have are two simultaneous equations. And we can notice that both of these have the quantity 𝐼, the current through the resistors, in them. And the value of 𝐼 is the same in both of these equations. So what we’re going to do now is to call the first equation equation number one and the second equation equation number two. Very creative names, I know! But let’s go with that for now. Then, what we’re going to do is to take equation one, which is that six volt is equal to 𝐼 times 𝑅 one, and then we’re going to rearrange it.

Specifically, we’re going to solve for 𝐼. And we do this by dividing both sides of the equation by the resistance 𝑅 one. Because this way, on the right-hand side, we have a fraction that is 𝑅 one divided by 𝑅 one. And that just cancels to give us one. And so, on the right-hand side, we’re left with 𝐼 times one or simply 𝐼. Whilst on the left-hand side, we’ve got six volts divided by 𝑅 one. Now, what we’re going to do is to take what we know to be the value of 𝐼, which is six volts divided by 𝑅 one, and substitute that into equation number two because we know that the current 𝐼 is equal to six volts divided by 𝑅 one. And so, we’ve now got an equation that tells us that 12 volts is equal to six volts divided by 𝑅 one multiplied by 𝑅 two.

And at this point, we’re almost there. We then rearrange this equation by dividing both sides by six volts, which allows us on the right-hand side to see that we’ve got a six volts in the numerator and a six volts in the denominator. We’re dividing six volts by itself, which ends up just leaving us with 𝑅 two in the numerator and 𝑅 one in the denominator. Cleaning things up a bit on the right-hand side, we see that we’re left with 𝑅 two divided by 𝑅 one. And on the left, we’re left with 12 volts divided by six volts.

Now, looking at the unit, we see that volts in the numerator will cancel with volts in the denominator. And then, we’re left with a fraction that says 12 divided by six, which is two, which therefore is telling us that the value of 𝑅 two divided by 𝑅 one is equal to two. Or rearranging the equation by multiplying both sides by 𝑅 one, we see that two lots of 𝑅 one is equal to 𝑅 two. And the reason that this is important is because we’ve now got a relationship between the resistances 𝑅 one and 𝑅 two. That’s these resistances here.

In other words, we need the resistance 𝑅 two to be twice as large as the resistance 𝑅 one in order for the potential difference across the resistor 𝑅 two to be 12 volts. And we need the potential difference across the resistor 𝑅 two to be 12 volts so that we can connect our lamp across it and the lamp will work. However, it was important that we disconnected our lamp whilst we were doing all these calculations just to make things a little bit simpler. Because if we hadn’t, if the lamp was still there on the circuit, then as the current flowed from the positive terminal to the first resistor and then through the first resistor, it would arrive at this junction here. And at junctions, current must split. So some would go this way and some would go this way. And so if we’d left the lamp in the circuit, we wouldn’t have been able to use the hand effect that the current through resistor one must be the same as the current through resistor two, which is what we called 𝐼 earlier.

But that doesn’t mean that everything changes when we do connect the light bulb to the circuit. Because even though the current will split at this junction here, that still doesn’t prevent there being 12 volts of potential difference across resistor 𝑅 two and so 12 lots of potential difference across the lamp. We don’t really care what the current is through the lamp. So why don’t we make our lives as easy as possible?!

But anyway, so to answer a question from the beginning, we’re looking for a pair of resistances, where one resistance is twice the value of the other. Looking at option A, the two possible values we’ve been given are eight kilo ohms and six kilo ohms. Well, in this case, the larger resistance of eight kilo ohms is not twice as large as six kilo ohms. So A is not the answer to our question. Similarly, for option B, we see that the larger resistance is 12 kilo ohms and the smaller one is two kilo ohms. This doesn’t have the relationship that we’re looking for. So we can move on from B.

And the same is true for C: 500 ohms is not twice as large as 200 ohms. It’s more than twice as large. And so, this is not the answer to our question. However, in option D, we’ve got 10-kilo ohms and five kilo ohms, which does show one resistance value to be twice as large as the other. And so, it looks like option D is the answer to our question. And we can confirm this by realizing that 48 ohms is not twice the largest 36 ohms. So E is not the answer either. And so, we found the answer to our question.

Isabella can make a potential divider that would power our lamp. But in order to do so, she must have a pair of resistors where the resistance value of one of the resistors is twice as large as the resistance value of the other. And she must connect the lamp in parallel with the larger resistor. And out of the options that we’ve been given, the only pair of resistors that will form the correct potential divider is if one of the resistors had a resistance of 10 kilo ohms and the other had a resistance of five kilo ohms.