# Lesson Video: Ohm’s Law Science

In this video, we will learn how to use Ohm’s Law to calculate the current through a component, the potential difference across a component, and the resistance of a component.

15:48

### Video Transcript

In this video, we will learn how to use Ohm’s law to calculate the current through a component, the potential difference across a component, and the resistance of a component.

Let’s start out by considering a simple circuit. In this circuit, we have a bulb. And this bulb is connected to a cell. We can recall that when we have a bulb connected in series with a cell like this, that bulb is going to light up. But why exactly is this? Well, the cell provides a potential difference across the bulb. We can recall that potential difference has units of volts, which has the symbol capital V. This potential difference causes charges to flow around the circuit. So these charges are going to flow through the bulb.

We might recall that the flow of electric charge is a current. So we can say that the potential difference provided by the cell causes a current through the bulb. Electric current has units of amperes, which has the symbol capital A. This current through the bulb is what causes it to light up. The greater the current, the brighter the bulb will be.

But then the next question we might ask is, what is it that determines the value of this current? It turns out that there are two things that are going to affect the current through the bulb. The first of these is the potential difference across the bulb. This potential difference is provided by the cell in the circuit. So if we change the potential difference of the cell that we use, then we change the potential difference across the bulb. So swapping out the cell that we use for one that provides a different potential difference is one way that we can affect the current through the bulb.

The second thing that affects the current is the bulb’s resistance. In the context of electricity, resistance is the opposition to the flow of charge. The greater the resistance of a component, the harder it is to make charge flow through it. Resistance is measured in units of ohms. The symbol for ohms is the Greek letter capital Ω.

So we’ve said that, in this circuit here, the current through the bulb is determined by both the potential difference across the bulb and the bulb’s resistance. In fact, it turns out that there is a simple relationship between the resistance of a component, the potential difference across that component, and the current through the component. This relationship was first discovered back in 1827 by a German physicist called George Ohm. As such, it is known as Ohm’s law.

Ohm’s law states that the potential difference across a component is equal to the current through that component multiplied by the component’s resistance. Typically, we label the potential difference as 𝑉, the current as 𝐼, and the resistance as 𝑅. Let’s add these labels to our circuit diagram to help us remember what they mean. 𝑉 is the potential difference across the bulb. And we know that this is provided by the cell. So let’s label the cell with the potential difference 𝑉. 𝐼 is the current through the bulb. If we look at the circuit diagram, we see that this circuit consists of a single loop of wire. This means that the current has to be the same at all points in the circuit. So let’s label our current 𝐼 at this point here. Lastly, we’ll label the bulb in this circuit with its resistance 𝑅.

Then we can rewrite Ohm’s law in terms of the symbols 𝑉, 𝐼, and 𝑅. In terms of these symbols, Ohm’s law says that 𝑉 is equal to 𝐼 multiplied by 𝑅. Generally, when we’re working with simple circuits like this one, we can assume that the resistance of components doesn’t change. In other words, in our Ohm’s law equation, we can assume that the value of 𝑅 remains constant. So then Ohm’s law is telling us that the potential difference 𝑉 is proportional to the current 𝐼.

So if we had a graph of current against potential difference, then that graph would be a straight line like this. For the case of the bulb in this circuit here, then this graph shows us visually that if we increase the potential difference 𝑉 across the bulb, then we also increase the current 𝐼 through the bulb. This is because as the value of 𝑉 increases, then the value of 𝐼 also increases in proportion to the increase in 𝑉. So by increasing the potential difference, we also increase the current. And we know that the greater the current, the brighter the bulb will be.

What this means in practice is that if we use a cell that provides a greater potential difference, then the bulb will be brighter. So we have seen that for any component with a fixed value of resistance, Ohm’s law tells us that the potential difference 𝑉 across that component is directly proportional to the current 𝐼 through it. But Ohm’s law is actually a lot more powerful than this. It provides us with a mathematical relationship between 𝑉, 𝐼, and 𝑅, which allows us to calculate the potential difference 𝑉 if we know the current 𝐼 and the resistance 𝑅.

Let’s clear some space so we can see how this works. We’ll consider the same circuit consisting of a bulb connected to a cell. Let’s assume that the bulb has a resistance of 0.5 ohms and that the current through the bulb is six amperes. Then the question is, what is the potential difference 𝑉 provided by the cell? Our Ohm’s law equation tells us that 𝑉 is equal to 𝐼 multiplied by 𝑅. In this case, we know the values of both 𝐼 and 𝑅. So we can go ahead and substitute these values in to this Ohm’s law equation.

Doing this, we find that 𝑉, the potential difference across the bulb, is equal to six amperes, that’s our value of the current 𝐼, multiplied by 0.5 ohms. That’s the bulb’s resistance 𝑅. Evaluating this expression, we find that 𝑉 is equal to three volts. It’s worth pointing out that whenever we have a current in units of amperes and a resistance in units of ohms, then we’re going to calculate a potential difference in units of volts. Volts, amperes, and ohms are the SI base units of potential difference, current, and resistance, respectively.

Now, the potential difference 𝑉 that we have calculated is the potential difference across the bulb in this circuit. That is, if we connected a voltmeter across our bulb like this, then that voltmeter would have a reading on it of three volts. It’s important to remember that whenever we use a voltmeter to measure the potential difference across a component, we need to connect the voltmeter in parallel with that component. This means that we connect it so that it forms a second branch in the circuit alongside the branch that the component is on.

So we have used Ohm’s law to find that a voltmeter connected in parallel with the bulb in a circuit will measure a potential difference of three volts. Since the bulb is the only component in the circuit besides the cell, then we know that the potential difference across the bulb must be equal to the potential difference provided by the cell. So we know that the cell provides a potential difference of three volts.

We have seen how we can use Ohm’s law to calculate a potential difference given values for current and resistance. We can also take this Ohm’s law equation and rearrange it to make either 𝐼 or 𝑅 the subject. Let’s clear some space so we can see how this works.

First, we’ll see how we can make the current 𝐼 the subject of the equation. We’ll start with the equation as we have seen it so far. That is, 𝑉 is equal to 𝐼 multiplied by 𝑅. Now if we want to make 𝐼 the subject, that means that we want to get an equation that says 𝐼 equals something. We can do this by dividing by 𝑅. And whenever we are rearranging an equation, it’s important to remember that whatever we do to one side of the equation, we also have to do the same thing to the other side.

So let’s divide both sides of this equation by the resistance 𝑅. On the right-hand side of the equation, we have an 𝑅 in the numerator of the fraction and an 𝑅 in the denominator. This right-hand side is 𝐼 multiplied by 𝑅 divided by 𝑅. And so the 𝑅 in the numerator cancels with 𝑅 in the denominator. This leaves us with an equation that says 𝑉 divided by 𝑅 is equal to 𝐼. And we can also flip this around to say that 𝐼 is equal to 𝑉 divided by 𝑅. That is, we now have an equation that tells us how to calculate the current 𝐼 through a component if we know the potential difference 𝑉 across that component and the component’s resistance 𝑅.

Let’s use this equation for a specific example. We’ll consider a circuit with a cell of potential difference six volts connected to a resistor with a resistance of two ohms. We want to work out the current 𝐼 through that resistor. So let’s take these values of 𝑉 is equal to six volts and 𝑅 is equal to two ohms and substitute them into this equation to calculate the current 𝐼. Doing this, we find that 𝐼 is equal to six volts divided by two ohms. Then, evaluating this expression, we calculate that the current 𝐼 through the resistor is equal to three amperes.

Now let’s go back to this original Ohm’s law equation and rearrange it to make the resistance 𝑅 the subject. As before, we’re going to start from 𝑉 is equal to 𝐼 multiplied by 𝑅. And this time, we want to get an equation that says 𝑅 equals something. To do this, we can divide both sides of the equation by the current 𝐼. Then on the right-hand side of this equation, we have an 𝐼 in the numerator of the fraction, which cancels with the 𝐼 in the denominator. Notice that we’re doing this working out side by side with the method that we used to get an expression for the current 𝐼. In this way, we can highlight the similarities in the methods.

Canceling the 𝐼’s in the right-hand side of this expression, we get an equation that says 𝑉 divided by 𝐼 is equal to 𝑅. As we did before, we can flip this expression around to say that 𝑅 is equal to 𝑉 divided by 𝐼. And so we have an equation that tells us how to calculate the resistance 𝑅 of a component if we know the potential difference 𝑉 across it and the current 𝐼 through it.

Now let’s have a look at an example question.

The diagram shows a circuit consisting of a cell and a bulb. The bulb has a resistance of 20 ohms, and the current in the circuit is 1.5 amperes. What is the potential difference provided by the cell?

Okay, so in this question, we have a bulb that we’re told has a resistance of 20 ohms. We’ll label this resistance as 𝑅 so that we have 𝑅 is equal to 20 ohms. This bulb is connected to a cell. And we’re told that there is a current in the circuit of 1.5 amperes. Labeling this current as 𝐼, we have that 𝐼 is equal to 1.5 amperes. We are asked to work out the potential difference provided by the cell. And we’ll label this potential difference as 𝑉.

We can recall that these three quantities are linked by an equation known as Ohm’s law. Specifically, Ohm’s law tells us that potential difference 𝑉 is equal to current 𝐼 multiplied by resistance 𝑅. So now we can go ahead and substitute these values for 𝐼 and 𝑅 into this Ohm’s law equation. When we do this, we find that 𝑉 is equal to 1.5 amperes, so that’s our value for 𝐼, multiplied by 20 ohms. That’s our value for 𝑅.

We have a current in units of amperes and a resistance in units of ohms. So that means that we’re going to get a potential difference in units of volts. When we evaluate this expression, we find that 𝑉 is equal to 30 volts. This value of 𝑉 is the potential difference across the bulb in our circuit, which must be the same as the potential difference provided by the cell. And so our answer to the question is that the potential difference provided by the cell is 30 volts.

Let’s finish up by summarizing what we have learned in this video. We have seen that Ohm’s law tells us that the current through a component in a circuit is directly proportional to the potential difference across it. This means that if we plot a graph of current against potential difference, it will be a straight line like this. We then saw that we could write Ohm’s law mathematically as an equation that says potential difference 𝑉 is equal to current 𝐼 multiplied by resistance 𝑅. This equation means that we can calculate the potential difference across a component if we know the current through the component and the component’s resistance.

Finally, we saw how we could rearrange the Ohm’s law equation to make either the current 𝐼 or the resistance 𝑅 the subject. Making 𝐼 the subject, we got an equation that said 𝐼 is equal to 𝑉 divided by 𝑅. Making 𝑅 the subject, we got that 𝑅 is equal to 𝑉 divided by 𝐼. These three equations are all equivalent statements of Ohm’s law.