Lesson Video: Using Potential Dividers

Video Transcript

In this video, we will be looking at a type of electronic circuit known as a potential divider, as well as how this circuit is commonly used. So let’s first see what we mean when we say potential divider. The most basic construction of this circuit consists of a power source, such as, for example, a cell, and two resistors connected in series. We’ll say that these resistors have a resistance of 𝑅 one and 𝑅 two, respectively. Then we connect wires across one of these resistors. So let’s say we connect these wires across 𝑅 two in this case. And these wires can be connected to other circuit components to complete this circuit.

Now the reason that we do this is because our cell or other power source can provide a certain potential difference, which we’ll call 𝑉, across this circuit. And because the resistors 𝑅 one and 𝑅 two are in series, that potential difference gets shared between those two resistors. In other words then, we can say that there’s a potential difference, 𝑉 one, across resistor one and a potential difference, 𝑉 two, across resistor two, such that the total potential difference across this circuit, 𝑉, is equal to 𝑉 one plus 𝑉 two.

Therefore, placing wires across the resistor 𝑅 two allows us to create a circuit here such that the potential difference across it is 𝑉 two. In other words, despite only having a power source that can generate a potential difference of 𝑉, we can create a current loop where the potential difference across it is 𝑉 two. And so the potential difference provided by the power source, 𝑉, is divided into 𝑉 one and 𝑉 two. This is why this circuit is known as a potential divider.

And we can actually tune the value of 𝑉 two by changing the ratio of the resistances 𝑅 one and 𝑅 two. In other words, we can choose how much of the potential difference, 𝑉, gets dropped across the resistor 𝑅 one and how much gets dropped across 𝑅 two, thus allowing us to have a part of the circuit in this region here where the potential difference 𝑉 two can be anything between zero volts and 𝑉.

Before we consider that in more detail though, it’s worth learning about a different way in which potential divider circuits are labeled. Often, the cell which we showed earlier on our diagram is not included. Instead, these two wires are left open-ended with the assumption that there’s a power source between them. And the way that this is done is to say that this lower wire is at a potential or voltage of zero volts. And this upper wire is at the voltage provided by the voltage source. Let’s call this voltage 𝑉 subscript in.

Now remember, we said that this lower wire was at a potential of zero volts. Well, this wire is still part of the same wire. And so they must both be at the same potential because, between this part of the wire and this part of the wire, there’s no resistor to decrease the potential of the wire. Nor is there a voltage source to increase it. Hence, this part of the wire is also at zero volts.

Now it’s also worth noting that zero volts is simply measured relative to the voltage of the earth. In other words, this wire is at the same voltage as the earth. And that is conventionally said to be zero volts. And so often this wire is also said to be earthed or grounded. And we may even see a symbol like this to suggest that the wire is grounded. All that tells us is that the wire is at zero volts.

Now for this wire, on the other hand, well, this wire is said to be at the output voltage, which we’ll call 𝑉 subscript out. And so now we know that the lower wire is at zero volts. This wire up here is at 𝑉 subscript in. And this wire is at 𝑉 subscript out. So we know the voltage of each one of these wires. And what’s more is that the potential difference between this top wire and this bottom wire is 𝑉 in minus zero volts, which is simply 𝑉 in. And similarly, the potential difference in this region is 𝑉 out minus zero volts, which is simply 𝑉 out.

Additionally, we can also calculate the potential difference in this region, which happens to be 𝑉 in, which is the potential of this wire, minus 𝑉 out, which is the potential of this wire. And so that becomes 𝑉 in minus 𝑉 out, which is also the potential difference dropped across this resistor here. Using our notation from earlier, we said that the potential difference across that resistor was 𝑉 one. And so we can see that 𝑉 in minus 𝑉 out is equal to 𝑉 one. And similarly, we said 𝑉 two was the potential difference across the second resistor, which is equal to 𝑉 out. So this is just a different way of drawing or symbolizing potential dividers.

However, it’s important that we know how to read these types of circuit diagrams as well as being able to draw them. But anyway, so the important thing about a potential divider is that the potential difference across this part of the circuit is 𝑉 out. And if we say that the top resistor has a resistance of 𝑅 one and the bottom resistor has a resistance of 𝑅 two, then we can actually calculate this value of 𝑉 out in terms of 𝑅 one, 𝑅 two, and the input voltage 𝑉 in.

To do this, we need to realize that resistor 𝑅 one and resistor 𝑅 two are placed in series with each other. Therefore, whatever potential difference source is present in this region, it’s trying to push a conventional current this way around the circuit. And because at the moment we’ve got nothing connected here, therefore that part of the circuit acts as an open switch. So all of the current flowing through resistor 𝑅 one must be the same as the current flowing through resistor 𝑅 two. Let’s say that current is 𝐼.

And then we can recall Ohm’s law, which tells us that the potential difference across a component in a circuit is equal to the current through that component multiplied by the resistance of that component. And since we said earlier that the potential difference across the first resistor was 𝑉 one and the potential difference across the second resistor was equal to 𝑉 two. Which by the way is the same thing as 𝑉 out. Then we can use Ohm’s law on the second resistor say. We can say that 𝑉 two is equal to the current for the resistor, which is 𝐼, multiplied by the resistance of the resistor, which is 𝑅 two.

And then we can use Ohm’s law on the entire circuit. We can say that the total potential difference across this circuit, which we know is 𝑉 in, is equal to the current through the circuit, which since we’ve got a series circuit is 𝐼 once again, multiplied by the total resistance of the circuit. Now because the resistors are in series, we simply add their resistances to find the total resistance of the circuit. And so now what we’ve got is 𝑉 in is equal to 𝐼 multiplied by 𝑅 one plus 𝑅 two. And then we can substitute 𝑉 two for 𝑉 out because, as we said earlier, they’re the same thing. After which we can divide this equation by this equation.

So we’ve got the left-hand sides on this side. We’ve got 𝑉 out divided by 𝑉 in. And the right-hand sides are 𝐼𝑅 two divided by 𝐼𝑅 one plus 𝑅 two. So now that we’ve got this fraction, we can focus solely on it and see that the currents in the numerator and the denominator on the right-hand side cancel. And so what we’re left with is that the ratio 𝑉 out divided by 𝑉 in is equal to the ratio 𝑅 two — that’s the resistance of this resistor — divided by 𝑅 one plus 𝑅 two — that’s the total resistance of the circuit.

And so based on this, we can modify the values of either 𝑅 one or 𝑅 two or both 𝑅 one and 𝑅 two to give us any desired value of 𝑉 out between zero volts and 𝑉 in. So it’s important to know this equation but also know how to derive this equation. And in fact, tuning the value of 𝑉 out becomes really easy when we change the second resistor into a variable resistor, because this way we can change the value of 𝑅 two in a continuous fashion.

And so we can see that if we set the resistance of the variable resistor to be zero ohms, then the fraction on the right-hand side of this equation becomes zero divided by whatever 𝑅 one is plus zero. And the whole fraction on the right-hand side therefore becomes zero because we’ve got zero in the numerator. And so we find that 𝑉 out divided by 𝑉 in is equal to zero. In other words, whatever the value of 𝑉 in, 𝑉 out is still equal to zero. In other words then, if we set the resistance of the variable resistor to be zero ohms, then the potential difference across that resistor becomes zero volts. Conversely, if we set the resistance of the variable resistor to be very very large. Or another way to write this is that this resistance goes to infinity. Then, in the denominator of this fraction, 𝑅 one plus 𝑅 two, we see that 𝑅 two is so much larger than 𝑅 one that we can basically ignore 𝑅 one. And this becomes more true the larger the value of 𝑅 two becomes.

And hence, we can say that our fraction 𝑉 out divided by 𝑉 in goes to 𝑅 two divided by 𝑅 two, because we can ignore the plus 𝑅 one in the denominator. But then 𝑅 two divided by 𝑅 two is just one. And so we can see that the ratio of 𝑉 out to 𝑉 in goes to one. In other words, the larger this resistance 𝑅 two becomes, the more potential difference gets dropped across it until eventually 𝑉 out becomes the same as 𝑉 in. And none of that potential difference is dropped across the first resistor. And so having a variable resistor in the position of 𝑅 two allows us to easily tune the value of 𝑉 out. So that’s one potential use for a potential divider circuit.

There is another one that we can look at very quickly. We can take the resistance 𝑅 two and replace it with a light-dependent resistor or LDR. And we know that this circuit diagram is representing an LDR by these two little arrows that are meant to signify light falling on the resistor. Now with a light-dependent resistor, the amount of light falling on the resistor has an influence on the resistance of the resistor. The more light that falls on this resistor, the smaller the resistance of the resistor. And conversely, the less light that falls on the resistor, the larger the resistance of the resistor.

Therefore, this potential divider circuit can be used as a sensing circuit, specifically a light-sensing circuit. In other words, we can use it as an automatic light switch because we can place this second resistor, this LDR, in the region that we want to light automatically, for example, a room or on the street, where this will control the street lighting. This way, once the room or the street gets very dark, the resistance of the LDR increases. And as we can see from this fraction, the larger the value of 𝑅 two, the larger the ratio of 𝑉 out divided by 𝑉 in, or in other words the larger the value of 𝑉 out.

We can then connect this part of the circuit to something known as a transistor, which can behave as a switch. Because as soon as this value of 𝑉 out exceeds a certain threshold voltage, the transistor will switch into the on state. And that can allow a current to flow through this part of the circuit, thus switching on the lights that are used to light up the street or the room. And so by placing an LDR into the position of the resistor 𝑅 two, we’ve created a light-sensing circuit using a potential divider. So now that we’ve seen what a potential divider is and what it can be used for, let’s take a look at an example question.

A potential divider has an input voltage of 48 volts. The resistance of the second resistor, 𝑅 two, is 100 kiloohms. The output voltage is drawn across the second resistor, 𝑅 two. What resistance must the first resistor, 𝑅 one, have in order to produce an output voltage of 32 volts?

Okay, so in this question, we’re dealing with a potential divider, which we can recall looks something like this. Firstly, we’ve got this wire in the circuit, and this wire is grounded or earthed. Therefore, it’s at a potential of zero volts relative to the earth. And then we’ve got a potential difference source across this part of the circuit. And what that potential different source is is not really relevant. But what is relevant is that that potential difference source is creating an input voltage.

We can say that this input voltage is 𝑉 subscript in. And from the question, we know that 𝑉 subscript in is equal to 48 volts. Then we see that we’ve got two different resistors which are placed in series. So we can say that the first resistor has a resistance 𝑅 one. And the second resistor has a resistance 𝑅 two. And we’ve been told that the output voltage is drawn across the resistor 𝑅 two. Therefore, we can say that the voltage at this wire is 𝑉 subscript out, the output voltage.

Now other information we’ve been given in the question is that the resistance of the second resistor, 𝑅 two, is 100 kiloohms, 100000 ohms. And we need to find the value of the resistance 𝑅 one in order to have an output voltage from this circuit, 𝑉 subscript out, of 32 volts. Now the way to go about doing this is to recall that we’ve got two resistors in series in this circuit. And because this part of the circuit is not yet connected to anything, it acts as an open switch. And so all of the current flowing through the resistor 𝑅 one is going to be the same as the current flowing through the resistor 𝑅 two.

We can say that this current in resistor 𝑅 one and 𝑅 two is 𝐼. And then we can recall that Ohm’s law tells us that the potential difference across a component in a circuit is equal to the current through that component multiplied by the resistance of that component. Therefore, firstly, focusing on the resistance 𝑅 two, we can say that the potential difference across it, which we know is 𝑉 out, is equal to the current through that resistor, which we’ve said is 𝐼, multiplied by the resistance of the resistor, which is 𝑅 two.

Similarly, for the whole circuit, we know that the potential difference across the entire circuit is 𝑉 in. And so we can say that 𝑉 in is equal to the current through that circuit, which we know is 𝐼. And we multiply this by the total resistance of the circuit. But because we’ve got two resistors in series, we can recall that the total resistance of this circuit is simply going to be the two resistances 𝑅 one and 𝑅 two added together.

Now at this point, we’ve got two different equations. And we can divide one by the other. Hence, we can say that 𝑉 out two divided by 𝑉 in is equal to 𝐼 multiplied by 𝑅 two divided by 𝐼 multiplied by 𝑅 one plus 𝑅 two. So that’s this equation divided by this equation. Then we see on the right-hand side that the current in the numerator cancels with the current at the denominator. And so what we’re left with is that 𝑉 out divided by 𝑉 in is equal to 𝑅 two divided by 𝑅 one plus 𝑅 two.

Now in this question, we’ve been given the value of 𝑉 out and 𝑉 in. And we also know the value of 𝑅 two. We’re trying to find the value 𝑅 one. So we need to rearrange this equation. We can do this by first multiplying both sides of the equation by 𝑅 one plus 𝑅 two. That’s the denominator on the right-hand side. This way, that denominator cancels with the numerator. And so what we’re left with is that 𝑅 one plus 𝑅 two multiplied by 𝑉 out over 𝑉 in is equal to 𝑅 two.

Then we can expand the parentheses. We multiply 𝑉 out over 𝑉 in by 𝑅 one and by 𝑅 two. That gives us 𝑅 one multiplied by 𝑉 out over 𝑉 in plus 𝑅 two multiplied by 𝑉 out over 𝑉 in is equal to 𝑅 two. Then we can subtract this term from both sides of the equation, which leaves us with 𝑅 two minus 𝑅 two 𝑉 out over 𝑉 in on the right-hand side. At this point, we can factorize 𝑅 two. And we’re left with one minus 𝑉 out over 𝑉 in in the parentheses.

Then all that’s left to do is to get rid of this fraction on the left-hand side. And to do this, we multiply by its reciprocal on both sides of the equation, because this way the fraction cancels on the left-hand side. And on the right-hand side, we’re left with 𝑉 in over 𝑉 out multiplied by 𝑅 two multiplied by one minus 𝑅 out over 𝑅 in. Then we can multiply 𝑉 in over 𝑉 out by each term in the parentheses. And what that leaves us with is that 𝑅 one is equal to 𝑅 two multiplied by 𝑉 in over 𝑉 out minus one.

Finally, we can substitute in the values we’ve been given here. We see that 𝑅 one is equal to 100 kiloohms — that’s 𝑅 two — multiplied by 48 volts over 32 volts — that’s 𝑉 in over 𝑉 out — minus one. Now we can see that, in the fraction, the unit of volts cancels. And 48 divided by 32 is the same as 1.5. Therefore, 1.5 minus one becomes 0.5. So what we’re left with in the parentheses is 0.5. In other words, 𝑅 one is equal to 100 kiloohms multiplied by 0.5 or half. And so, at this point, we have our final answer. The resistance of the first resistor, 𝑅 one, is equal to 50 kiloohms or 50000 ohms.

Okay, so now that we’ve had a look at an example question, let’s summarize what we’ve talked about in this lesson. Firstly, we saw that potential dividers exploit the sharing of potential difference between components in series. In other words, a potential divider can be used to create an output potential difference that is less than the input potential difference despite only having a voltage source that can produce the input potential difference.

We also saw that the ratio 𝑉 out divided by 𝑉 in, so the ratio of the output potential difference to the input potential difference, depends in the following way on the resistances 𝑅 one and 𝑅 two. We also saw that we can tune the upper potential difference by replacing the second resistor in this diagram, 𝑅 two, with a variable resistor.

And finally, we also saw that a light-sensing circuit can be created by replacing the resistance 𝑅 two with a light-dependent resistor, otherwise known as an LDR. So that is an overview of potential dividers and their potential uses.