Lesson Explainer: Ohm’s Law Science

In this explainer, 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 begin by considering the circuit shown in the diagram below.

The circuit consists of a bulb connected to a cell.

The cell provides a potential difference across the bulb. Potential difference has units of volts, which have the symbol .

This potential difference creates a current through the bulb. Current has units of amperes, which have the symbol . The current through the bulb causes it to light up.

We may ask what determines the brightness of the bulb. The short answer to this question is that the brightness depends on the current through the bulb. The greater this current, the brighter the bulb. But, in order to answer this question more fully, we must also ask: What factors does the current depend on?

The current through the bulb depends on the potential difference across it and the bulb’s electrical resistance. Recall that electrical resistance is the opposition to the flow of charge. The greater the electrical resistance of a component, the harder it is to make charge flow through it. Resistance is measured in units of ohms, with the symbol .

There is a simple relationship between the current through a circuit component, the potential difference across it, and its resistance. This relationship was first discovered in 1827 by German physicist Georg Ohm, and it is known as Ohm’s law.

Let us again consider a circuit in which a bulb is connected to a cell. We will label the potential difference provided by the cell as , the resistance of the bulb as , and the current through the bulb as . We may indicate this on the circuit diagram as shown below.

We will assume that the resistance of the bulb, , has a fixed value. We will also suppose that we can choose the potential difference, , provided by the cell.

Ohm’s law tells us that if the resistance of a component has a fixed value, then the current through the component is directly proportional to the potential difference across it.

This means that the greater the potential difference across the component, the greater the current through the component. In the case of the bulb, this means that the greater the potential difference of the cell we use, the greater the current through the bulb, and therefore the brighter the bulb.

The graph below shows this relationship between potential difference and current.

This graph shows a straight line. This relationship means that if we made the potential difference bigger by a factor of 10, then the current through the resistor would also increase by a factor of 10.

Let’s have a look at an example question.

Ohm’s law enables us to go further than statements about proportionality. It provides a mathematical relationship between potential difference, ; current, ; and resistance, . This allows us to calculate the value of if we know the values of and .

Suppose that we know the values of the current through a component and the resistance of that component. Then, we can use Ohm’s law to calculate the value of the potential difference across the component.

Consider the circuit shown in the diagram below.

We have a circuit containing a resistor with a resistance of . We know that the current through that resistor is . Let’s see how we can use Ohm’s law to calculate the potential difference, , across the resistor.

Ohm’s law tells us that . We know the values of and for this circuit, so we can substitute these values into the right-hand side of the Ohm’s law equation. Doing this gives us

Doing the multiplication on the right-hand side, we can work out that the value of the potential difference is

Since the resistor is the only component in this circuit besides the cell, we know that the potential difference across the resistor must be equal to the potential difference provided by the cell. This means that we know the cell must provide a potential difference of 10 V.

It may be the case that we have a cell that provides a known potential difference and a resistor or other circuit component of known resistance and that we want to know what current we will have through the component.

In this case, we know the values of and and we want to work out the value of . We can do this using Ohm’s law. However, in this case we first need to rearrange the Ohm’s law equation in order to make the subject.

We can also use the Ohm’s law equation to work out the resistance of a component in a circuit if we know both the potential difference across the component and the current through it.

In order to do this, we need to rearrange the equation to make the resistance, , the subject.

Let us now have a look at a couple of example problems in which we will see how to rearrange Ohm’s law.

Example 2: Using Ohm’s Law to Calculate the Current through a Component

The diagram shows a circuit consisting of a cell and a resistor. The cell provides a potential difference of 6 volts, and the resistor has a resistance of 3 ohms. What is the current at point P in the circuit?

Answer

In this question, we are presented with a circuit diagram. We are asked to find the current at the point marked P in the circuit.

The circuit consists of a single loop. The current will be the same at all points around this loop.

The question tells us the resistance of the resistor in the circuit. We will label this , so we have that .

The other information given to us in the question is the potential difference provided by the cell. This will be equal to the potential difference across the resistor. We will label this , so we have .

We can recall that Ohm’s law relates the potential difference across a component, the current through the component, and the resistance of the component:

We can use Ohm’s law to work out the current, , through the component. To do this, we need to rearrange the equation to make the subject.

If we divide both sides of the equation by , we get

Then, the in the numerator of the fraction on the right-hand side cancels with the in the denominator of this fraction. This gives us which we can also write as

Now we just need to substitute in our values for and . If we substitute in that and , we get the following expression for the current through the resistor:

Working out the division on the right-hand side, we find that

Since we said that the current will be the same at every point in this circuit, we know that this current through the resistor will also be the current at point P.

So, our answer to the question is that the current at point P in the circuit is equal to 2 A.

Example 3: Using Ohm’s Law to Calculate the Resistance of a Component

The diagram shows a circuit consisting of a battery and a resistor. What is the resistance of the resistor?

Answer

This question gives us a diagram of a circuit containing a battery and a resistor. In this diagram, the battery is labeled as providing a potential difference of 24 V. The current through the circuit is labeled as 3 A.

We are asked to find the resistance of the resistor, which we will label .

The 24 V potential difference provided by the battery will be equal to the potential difference across the resistor. We will label the potential difference across the resistor as , so that we have .

The circuit is a single loop, so the current must be the same at all points in this circuit. We therefore know that the current through the resistor, which we will label , is given by .

So, we know the current, , through the resistor and we know the potential difference, , across it. We want to find its resistance, .

We can recall that Ohm’s law links these three quantities:

We are trying to work out the value of , so we need to rearrange the equation to make the subject.

Dividing both sides of the equation by , we get that

On the right-hand side, the in the numerator cancels with the in the denominator. This gives us which we can equally write as

Now that we have an equation for the resistance, , in terms of the potential difference, , and the current, , we are ready to substitute in the values for and . Substituting in that and , we have that

Doing the division on the right-hand side gives us our answer: the resistance of the resistor is

In all of the questions we have seen so far, we have started from a position of knowing either the potential difference across a component, the current through the component, or both of these quantities.

However, this will not always be the case in practice. Let us imagine that we are asked to build a circuit in order to work out the resistance of a bulb.

We will further suppose that we are given a battery to use in order to power this circuit but that we are not told what potential difference this battery provides.

We can connect the battery to the bulb to form a circuit as shown in the diagram below.

So far, we do not know the potential difference across the bulb or the current through it. We have no way of calculating the bulb’s resistance.

However, we may recall that measurement devices exist that can be used to measure the potential difference across and the current through a component. Let us suppose that we have a voltmeter and an ammeter available to use.

We may recall that if we connect a voltmeter in parallel with a component, the reading on the voltmeter will be the potential difference across that component. The correct way to connect a voltmeter to measure the potential difference across the bulb is shown in the diagram below.

To use an ammeter, we can recall that we need to connect it in series with the component that we wish to measure the current through. If we place the ammeter along the same path that the charges must be flowing to go through the bulb, all those same charges must flow through the ammeter. The ammeter will then give us a reading that tells us the current through the bulb. This is shown in the diagram below.

Now, we could make these measurements one after the other. In other words, we could first connect the voltmeter and measure the potential difference across the bulb. We could then disconnect the voltmeter and connect the ammeter to measure the current through the bulb.

However, we can also connect both the voltmeter and the ammeter at once so that we may measure the potential difference and the current at the same time. The complete circuit, containing both the voltmeter and the ammeter, is shown in the diagram below.

Using the readings from the voltmeter and the ammeter, we will know the potential difference across the bulb as well as the current through it. We may then use these measured values along with Ohm’s law to calculate the resistance of the bulb.

Let’s see how this works by looking at an example problem.

Example 4: Using Ohm’s Law along with Measurement Device Readings to Calculate the Resistance of a Component

The diagram shows a circuit consisting of a cell, a resistor, a voltmeter, and an ammeter. The reading on the voltmeter is 3 volts, and the reading on the ammeter is 0.1 amperes. What is the resistance of the resistor?

Answer

In this question, we are given a circuit diagram. We are asked to work out the resistance of the resistor in the diagram.

We are told that the voltmeter gives a reading of 3 volts. We can see from the diagram that the voltmeter is connected across the resistor. This means that the reading on the voltmeter gives us the potential difference across the resistor. We will call this potential difference , so we have that .

We are also told that the ammeter gives a reading of 0.1 amperes. We can see from the diagram that the ammeter is connected in series with the resistor, that is, along the same loop of wire that the resistor is on. This means that the ammeter reading gives us the current through the resistor. We will label this current , so we have that .

We are asked to find the resistance of the resistor, which we will label . We can recall that Ohm’s law relates the resistance of a component, the potential difference across the component, and the current through the component:

In our case, we know the values of , the potential difference across the resistor, and , the current through the resistor. We are trying to find its resistance, . Therefore, we need to rearrange the equation to make the subject.

To do this, we divide both sides of the equation by . This gives us

On the right-hand side, we can then cancel the in the numerator with the in the denominator. Then, we have that which we can also write as

Now we can substitute in that and to give us

Finally, we can do the division on the right-hand side to find that the resistance of the resistor is given by

Let us now summarize what has been learned in this explainer.