### 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 letβs 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. Weβll say that these resistors have
a resistance of π
one and π
two, respectively. Then we connect wires across one of
these resistors. So letβs say we connect these wires
across π
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 weβll call π, across this circuit. And because the resistors π
one
and π
two are in series, that potential difference gets shared between those two
resistors. In other words then, we can say
that thereβs a potential difference, π one, across resistor one and a potential
difference, π two, across resistor two, such that the total potential difference
across this circuit, π, is equal to π one plus π two.

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

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

Before we consider that in more
detail though, itβs 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 thereβs 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. Letβs call this voltage π
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,
thereβs 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 itβs 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 weβll call π
subscript out. And so now we know that the lower
wire is at zero volts. This wire up here is at π
subscript in. And this wire is at π subscript
out. So we know the voltage of each one
of these wires. And whatβs more is that the
potential difference between this top wire and this bottom wire is π in minus zero
volts, which is simply π in. And similarly, the potential
difference in this region is π out minus zero volts, which is simply π out.

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

However, itβs 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 π out. And if we say that the top resistor
has a resistance of π
one and the bottom resistor has a resistance of π
two, then
we can actually calculate this value of π out in terms of π
one, π
two, and the
input voltage π in.

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

And then we can recall Ohmβs 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 π one and the potential
difference across the second resistor was equal to π two. Which by the way is the same thing
as π out. Then we can use Ohmβs law on the
second resistor say. We can say that π two is equal to
the current for the resistor, which is πΌ, multiplied by the resistance of the
resistor, which is π
two.

And then we can use Ohmβs law on
the entire circuit. We can say that the total potential
difference across this circuit, which we know is π in, is equal to the current
through the circuit, which since weβve got a series circuit is πΌ 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 weβve got is π in
is equal to πΌ multiplied by π
one plus π
two. And then we can substitute π two
for π out because, as we said earlier, theyβre the same thing. After which we can divide this
equation by this equation.

So weβve got the left-hand sides on
this side. Weβve got π out divided by π
in. And the right-hand sides are πΌπ
two divided by πΌπ
one plus π
two. So now that weβve 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 weβre left with is that
the ratio π out divided by π in is equal to the ratio π
two β thatβs the
resistance of this resistor β divided by π
one plus π
two β thatβs the total
resistance of the circuit.

And so based on this, we can modify
the values of either π
one or π
two or both π
one and π
two to give us any
desired value of π out between zero volts and π in. So itβs important to know this
equation but also know how to derive this equation. And in fact, tuning the value of π
out becomes really easy when we change the second resistor into a variable resistor,
because this way we can change the value of π
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 π
one is plus
zero. And the whole fraction on the
right-hand side therefore becomes zero because weβve got zero in the numerator. And so we find that π out divided
by π in is equal to zero. In other words, whatever the value
of π in, π 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, π
one plus π
two, we see that π
two is so much larger than π
one that
we can basically ignore π
one. And this becomes more true the
larger the value of π
two becomes.

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

There is another one that we can
look at very quickly. We can take the resistance π
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 π
two, the larger the ratio of π out divided by
π in, or in other words the larger the value of π 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 π
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 π
two, weβve created a light-sensing circuit using a
potential divider. So now that weβve seen what a
potential divider is and what it can be used for, letβs take a look at an example
question.

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

Okay, so in this question, weβre
dealing with a potential divider, which we can recall looks something like this. Firstly, weβve got this wire in the
circuit, and this wire is grounded or earthed. Therefore, itβs at a potential of
zero volts relative to the earth. And then weβve 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 π subscript in. And from the question, we know that
π subscript in is equal to 48 volts. Then we see that weβve got two
different resistors which are placed in series. So we can say that the first
resistor has a resistance π
one. And the second resistor has a
resistance π
two. And weβve been told that the output
voltage is drawn across the resistor π
two. Therefore, we can say that the
voltage at this wire is π subscript out, the output voltage.

Now other information weβve been
given in the question is that the resistance of the second resistor, π
two, is 100
kiloohms, 100000 ohms. And we need to find the value of
the resistance π
one in order to have an output voltage from this circuit, π
subscript out, of 32 volts. Now the way to go about doing this
is to recall that weβve 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 π
one is going to be the same as the current flowing through
the resistor π
two.

We can say that this current in
resistor π
one and π
two is πΌ. And then we can recall that 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. Therefore, firstly, focusing on the
resistance π
two, we can say that the potential difference across it, which we know
is π out, is equal to the current through that resistor, which weβve said is πΌ,
multiplied by the resistance of the resistor, which is π
two.

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

Now at this point, weβve got two
different equations. And we can divide one by the
other. Hence, we can say that π out two
divided by π in is equal to πΌ multiplied by π
two divided by πΌ multiplied by π
one plus π
two. So thatβs 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 weβre left with is that
π out divided by π in is equal to π
two divided by π
one plus π
two.

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

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

Then all thatβs 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, weβre
left with π in over π out multiplied by π
two multiplied by one minus π
out over
π
in. Then we can multiply π in over π
out by each term in the parentheses. And what that leaves us with is
that π
one is equal to π
two multiplied by π in over π out minus one.

Finally, we can substitute in the
values weβve been given here. We see that π
one is equal to 100
kiloohms β thatβs π
two β multiplied by 48 volts over 32 volts β thatβs π in over
π out β 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 weβre left with in the
parentheses is 0.5. In other words, π
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, π
one, is equal to 50 kiloohms or 50000 ohms.

Okay, so now that weβve had a look
at an example question, letβs summarize what weβve 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 π out
divided by π in, so the ratio of the output potential difference to the input
potential difference, depends in the following way on the resistances π
one and π
two. We also saw that we can tune the
upper potential difference by replacing the second resistor in this diagram, π
two,
with a variable resistor.

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