### Video Transcript

Malala creates the potential divider shown in the diagram. π
one has a value of 30 ohms and π
two has a value of 90 ohms. What ratio of output voltage to input voltage does this produce?

Okay, so, in this diagram, we can see whatβs known as a potential divider. What this consists of is usually two resistors, π
one and π
two in this case. And because the input voltage is across both the resistors that are placed in series, and the output voltage is only across one of the resistors, this device can essentially be used to produce an output voltage, which weβll call π sub out in this case, thatβs a certain fraction of the input voltage, which we will call π sub in.

Now what weβve been told, first of all, is that the value of π
one is 30 ohms. And the value of π
two is 90 ohms. These are the resistances of these resistors here. What weβve been asked to do in the question is to find out the ratio of output voltage to input voltage produced by this potential divider. In other words, weβve been asked to find the ratio π out divided by π in. So, in order to work out this ratio, letβs start by finding an expression for π out.

Well, π out is the potential difference across this resistor π
two. Now because we know that the right-hand end of this circuit is at zero volts, this is going to work towards making life easier for us. Because the potential difference across the resistor π
two is simply the voltage at this point minus the voltage at this point. Or, in other words, thatβs π out minus zero volts, or just π out. In other words then, the value of π out that weβre trying to find out in our ratio is simply the potential difference across resistor π
two.

Now we can work this out using a law known as Ohmβs law. Ohmβs 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. So, in this case, we can say that π out is equal to, well, firstly, the current in the resistor, which we donβt know but weβll just call πΌ for now, multiplied by the resistance of the resistor, which is π
two. And so, at this point, we have an expression for π out, which is the top of our fraction here. Letβs then try and find an expression for π in.

Now π in is simply the potential difference across both the resistors π
two and π
one. So, if we want to apply Ohmβs law once again, then, first of all, we need to know the current through both of the resistors. And secondly, we need to know the resistance of both resistors combined. Now letβs start by discussing the current. This oneβs the simpler of the two.

Because the resistors π
two and π
one are in series, we know that the current flowing through both of them must be the same. In other words, there must be a current πΌ flowing through both the resistors. Because that was the current we decided was flowing through resistor π
two. And once again, because theyβre in series, thatβs the same current thatβs flowing through π
one. So, we found the current through both of the resistors, but what is the combined resistance of π
one and π
two?

Well, we can recall that the total resistance of resistors when connected in series is given by simply adding up all the resistances of all of the components until weβve added the resistance of the final component in the circuit. Now in this case, weβve only got two components, π
one and π
two. So, the total resistance of the two resistors together, which weβll call π
sub tot, is simply equal to π
one plus π
two. And now armed with this knowledge, we can finally work out an expression for π sub in.

Because π sub in, thatβs the total potential difference across both resistors, is equal to the current through both resistors, thatβs πΌ, multiplied by the resistance of both resistors combined, thatβs π
sub tot. But weβve just worked out an expression for π
sub tot. We know that π
sub tot is equal to π
one plus π
two. And now we have an expression for π out and an expression for π in. So, letβs divide π out by π in to give us this ratio.

We can say that π out divided by π in is equal to πΌπ
two, thatβs π out, divided by πΌ brackets π
one plus π
two, thatβs π in. And then, we can see that the factor of πΌ in the numerator and the denominator cancels. Hence, weβre simply left with the right-hand side π
two divided by π
one plus π
two. So, thatβs essentially the ratio of output voltage to input voltage. But this is only an algebraic expression. We can go further than this because weβve actually been told the values of π
one and π
two.

π
one has a value of 30 ohms and π
two has a value of 90 ohms. We can plug in the values of π
one and π
two to give us an expression for π out over π in, as being equal to 90 ohms, thatβs π
two, divided by 30 ohms, thatβs π
one, plus 90 ohms, thatβs π
two. The denominator simplifies to 120 ohms. At which point, the unit of ohms in the numerator and the denominator cancels. So, weβre just left with 90 divided by 120. And this simplifies down to 0.75. Hence, we can now say that the ratio of output voltage to input voltage produced by this potential divider is 0.75.