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.
