### Video Transcript

In this video, we will be
discussing a particular type of circuit known as a potential divider. We will see how this circuit is
built, why itβs called a potential divider circuit, and how it is used. So letβs start by looking at how we
would actually build this kind of circuit.

We start by connecting two
resistors in series. Letβs say that the resistance of
the first resistor is π
one and the resistance of the second resistor is π
two. These two resistors can then be
connected to, say, a power source, in this case weβve drawn a battery, or to some
other more complicated circuit. In fact, weβre not particularly
interested in what we find in this part of the circuit, which is why weβre going to
draw two terminals, one here and one here, to show that we can connect multiple
different things between these terminals. All we care about is that it needs
to be something that supplies a potential difference.

But what really turns the circuit
from just being two resistors in series to a potential divider is when we connect
some components in parallel with one of the resistors. In this case, weβre using π
two. For example, we could choose to
connect a light bulb here or for that matter any other components. But the important thing is that the
components here are connected in parallel with only one of the resistors, in this
case π
two.

But once again, weβll draw some
terminals here and here because we donβt really care whatβs connected across this
gap. What weβre interested in is the
effect that all of this has. So letβs imagine that we did
connect a battery in this position here, with the positive terminal over here and
the negative terminal over here. This means that the battery sets up
a potential difference across the rest of the circuit. Letβs call this potential
difference supplied by the battery π subscript in for input voltage. And the result of all of this is
that a current is set up in this circuit.

Specifically, if we think about
conventional current, which is a flow of positively charged particles, then we can
see that this current would be moving away from the positive terminal of our battery
and, therefore, be moving clockwise through our circuit until it arrived at the
negative terminal again. Letβs say that the magnitude or
size of the current set up in this circuit is πΌ. And to keep things simple, we will
consider only the circuit consisting of our battery, resistor π
one, and resistor
π
two.

In other words, for now, letβs
ignore everything that was connected in parallel with resistor π
two because this
way all weβve got is a series circuit. And we can recall that for a series
circuit, firstly, the current through all components is the same. In other words, if thereβs a
current of magnitude πΌ moving this way through our wire in the circuit, then this
means that the current through our resistor π
one is πΌ as well and the current
through the resistor π
two is also πΌ.

Secondly, we can consider the fact
that thereβs a potential difference of π in across our circuit. What this means is that the battery
that weβve drawn here as the placeholder for whatever is actually connected in this
part of the circuit is providing a potential difference of π in across both of the
components in our circuit, in this case the resistors, at which point we can erase
our placeholder battery. Because like we said earlier, weβre
not that interested in whatβs actually connected here, just the fact that something
should be connected here. And that will provide a potential
difference of π in across both resistors.

But letβs also recall that for a
series circuit, the potential difference across the entire circuit is shared across
the components. And what we mean by this is that
the total potential difference across, in this case, both our components π
one and
π
two is shared in such a way that if we say that the potential difference across
resistor one is π one and the potential difference across resistor two is π two,
then we know that π one plus π two must be equal to π in. Intuitively, this tells us that the
potential difference across the first resistor plus the potential difference across
the second resistor is equal to the potential difference across both resistors
combined, which sounds like a very common-sense thing to say. But remember that this is not true
for parallel circuits. And weβll have to worry about those
in a moment.

But for now, letβs also recall 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 its resistance. And we can apply this law to
components in our circuit, firstly the resistor π
one and secondly the resistor π
two.

For the resistor π
one, we can say
that the potential difference across that resistor, π one, is equal to the current
through that resistor, which we know is πΌ, multiplied by its resistance, which is
π
one. And similarly, for the resistor π
two, we know that the potential difference across that resistor is equal to the
current through that resistor multiplied by its resistance.

And lastly, we can apply Ohmβs law
to the entire circuit itself. In this case, we know that the
potential difference across the circuit is π in. So thatβs the total potential
difference across both resistors combined. And we know that the current in the
circuit is πΌ, because itβs a series circuit, which means the current is constant
throughout. And we can link this to the total
resistance of the circuit. In other words, we can say that the
total potential difference across the circuit, π in, is equal to the current
through the circuit, which is πΌ, multiplied by the total resistance of the circuit,
which we will have to call π
subscript tot.

Now, weβre gonna have to go about
finding what π
subscript tot actually is. And to do this, weβll have to
recall that for resistors in series, the total resistance of a set of resistors, π
subscript tot, is equal to the resistance of the first resistor connected in series
plus the resistance of the second resistor connected in series plus so on and so
forth, however many resistances there are connected in series.

In our scenario, weβve got two
resistors, π
one and π
two. And hence, we can say that the
total resistance of our circuit here, π
subscript tot, is equal to π
one plus π
two, which means that we have an expression for π
subscript tot in terms of π
one
and π
two. And we can substitute that
expression into this equation here. Specifically, instead of π
subscript tot, we can substitute in π
one plus π
two, which now tells us that the
total potential difference across the circuit is equal to the current through that
circuit multiplied by the sum of the resistances π
one and π
two.

Now, at this point, we might be
wondering, whatβs the point of all of this? Why are we writing down all of
these mathematical expressions? Well, the reason is because a
potential divider circuit does something very special. To see this, weβre going to have to
do one more step of mathematics. Specifically, weβre going to want
to find an expression for the ratio between the potential difference across the
second resistor and the potential difference across the whole circuit. We want to find π two divided by
π in.

But we know from this equation here
that π two is equal to πΌ multiplied by π
two. Thatβs the current through the
resistor multiplied by its resistance. And we know that π in is equal to
πΌ multiplied by π
one plus π
two. So thatβs the current through the
whole circuit multiplied by the sum of the two resistances, at which point we can
see that both in the numerator and the denominator weβve got the current πΌ. And πΌ divided by πΌ is equal to
one. And so the expression on the
right-hand side simplifies to π
two divided by π
one plus π
two.

And what this means is the
following. Despite the fact that we have a
fixed potential difference across our circuit, which is π in, that is provided by
whatever battery was connected here, which we said we donβt really care about, we
can still modify the potential difference across resistor π
two simply by changing
the values of resistances π
one or π
two. In other words, the potential
difference π two across resistor π
two is only dependent on the resistances π
one
and π
two, as we see in this expression.

And this is the point at which we
bring back in the parallel connection to resistor π
two. We connect some components over
here. Again, it doesnβt matter what those
components are. The point is that by simply
modifying the values of the resistances π
one and π
two, we can change the
potential difference that is across π
two and, therefore, change the potential
difference thatβs dropped across the connected components here. Because now we can recall that in a
parallel circuit, the potential difference is the same across parallel branches. In other words, if the potential
difference across resistor π
two is π two, then the potential difference across
whatever is connected here is also π two. And as weβve seen from this
equation here, the ratio between π two and π in is only dependent on the values of
the resistances π
one and π
two. Those are the only things in our
expression.

Now, if π in is fixed, as weβve
been assuming up until this point, π in being the potential difference provided by,
say, the battery, then this equation is most useful to us if we rearrange it
slightly. If we multiply both sides of the
equation by π in, then we find that the potential difference π two across the
resistor π
two is equal to this ratio, π
two divided by π
one plus π
two,
multiplied by what weβre saying is a constant potential difference provided by the
battery, π in. And so this is telling us that we
can tune the value of π two, the potential difference not just across resistor π
two but across whatever components are connected here, simply by changing this ratio
of resistances. And we can see that π two can take
any value between zero volts and π in.

So despite the fact that we only
have a battery that provides π in of potential difference, we can actually tune our
circuit so that we can choose any potential difference between zero and π in to be
across any components that weβre connecting over here. And we can see that the range of π
two is between zero and π in if we consider the two following scenarios.

Firstly, if we consider a scenario
where the value of π
two is zero, in other words, thereβs not actually a resistor
connected here, itβs just a piece of wire, or equally weβre thinking about it as
connecting a resistor with zero resistance in this position. In that case, our expression for π
two becomes π two is equal to zero, because thatβs what π
two is, divided by
whatever π
one is plus zero. And we multiply this by the
potential difference provided by the battery, which is π in.

But then because weβve got zero in
the numerator, this whole fraction becomes zero. And therefore, this whole
right-hand side expression becomes zero. In other words, we find in this
particular scenario, where π
two is equal to zero, the potential difference across
this resistor is also zero. And therefore, once again, the
potential difference across all components connected here is zero.

If we now instead consider a
scenario where π
one is equal to zero and π
two has some nonzero value, so in this
case thereβs no resistor connected in the π
one position or a resistor with zero
resistance is connected there, then our expression for π two becomes π
two divided
by zero plus π
two multiplied by π in. And this fraction simplifies to
just π
two divided by π
two, which is equal to one. And so we find that π two is equal
to one multiplied by π in or simply π in.

And so considering these two
extreme cases, weβve found the range of π two. π two can be anything between zero
and π in depending on the values that we choose for the resistances π
one and π
two. But the point is that this equation
is an important one to be able to use.

Now, weβve looked at the basics of
a potential divider circuit and specifically what it does. But why is it a useful circuit to
know? How can it be used practically? Well, one practical use of a
potential divider circuit can be found if instead of having a box standard constant
resistor here, we replace it with a different kind of resistor known as a
thermistor.

Thermistors are
temperature-dependent resistors. In other words, the value of the
resistance π
two will change as the temperature of the environment surrounding this
resistor changes. Now, there are different kinds of
thermistors. But a common type of thermistor
will behave according to this plot shown.

Specifically, weβve plotted the
temperature of the environment surrounding our thermistor on the horizontal
axis. And weβve plotted the value of the
resistance of the thermistor π
two on the vertical axis. And what we see here is that if the
value of the temperature is relatively low, then the resistance of the thermistor π
two is quite high, whereas if the value of the temperature is high, then the
resistance of the thermistor is quite low.

And we can summarize this in the
following way. If the environment around the
thermistor is cold, the value of π
two is large. If the temperature surrounding the
thermistor is hot, the value of π
two is small. But then based on this equation
over here, we see that for large values of π
two, the value of π two is large as
well. And for small values of π
two, the
value of π two is small as well.

And based on this circuit, what we
can do is to connect some sort of heating element in this part of the circuit
here. Because this way when the
temperature of the environment around our thermistor is low, when itβs cold around
our thermistor, this results in a large value of π two, a large potential
difference across a heating element, which will be connected here. And therefore, the heating element
will have a current through it and start heating up the environment, whereas if the
environment is hot around our thermistor, then the value of π two will be
small. And so our heating element will not
actually heat up the environment.

So a potential divider circuit is
essentially acting as a very primitive thermostat. It detects the temperature in its
environment by using a thermistor and switches on or off a heating element connected
here accordingly. Now, real thermostats are much more
complicated than this. But still, we can use a thermistor
in a temperature-dependent circuit. And in this case, itβs specifically
a potential divider circuit.

Another practical use for a
potential divider circuit is instead of using a thermistor here, we could use whatβs
known as a light-dependent resistor or LDR. Now, an LDR displays similar
behavior to the thermistor we discussed earlier, except that it depends on the
intensity of light falling on it rather than the temperature of the surrounding
environment. We can even plot a similar graph
showing how the resistance of the LDR changes. But on the horizontal axis, weβd
have to plot light intensity rather than temperature.

And so what we find here is that
when the light intensity falling on our LDR is low, in other words, when the
surroundings of the LDR are dark, the resistance of this particular LDR, π
two, is
high, which means that the potential difference π two across that part of the
circuit is also high, whereas when the light intensity falling on our LDR is high,
in other words, when itβs light surrounding our LDR, this results in a small value
of the resistance π
two and, therefore, a small value of π two. And so instead of a heating
element, this time we can connect some sort of lighting element, some sort of light
bulb, for example, which would then light up if the surroundings of the LDR are
dark. And it would switch off if the
surroundings are light.

And so at this point, weβve seen a
couple of practical uses of a potential divider circuit, which means we can 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 case, weβre
dealing with a potential divider circuit. So letβs first start by drawing our
potential divider circuit. Letβs first start by recalling that
a potential divider circuit consists of two resistors connected in series. Weβll call the resistance of our
first resistor π
one and the resistance of our second resistor π
two, as weβve
been told to in the question. We also recall that across this
part of the circuit, some sort of power source is connected, often a battery, but
not necessarily so. And this power source is what
provides whatβs known as the input voltage. Letβs call this input voltage π
subscript in. And in this case, in the question,
weβve been told that this input voltage is equal to 48 volts.

Now, additionally, weβve been told
that the resistance of the second resistor π
two is 100 kiloohms. Thatβs 100000 ohms because the
prefix kilo- means 1000. Weβve also been told that the
output voltage in our potential divider circuit is drawn across the second
resistor. What this essentially means is that
we connect a pair of wires here and here so that we can connect some components in
parallel with our resistor π
two. In this case, weβre not really
worried about which components are connected here, but just that the output voltage,
which we will call π subscript out, is drawn across our second resistor.

Now, the reason that we know that
this is the output voltage is because the voltage across our second resistor must be
the same as π out. The reason for this is that our
resistor π
two and whatever components are connected here are connected in
parallel. And components in parallel have the
same potential difference across them. Therefore, to reiterate, the
potential difference across resistor π
two is π subscript out. Thatβs the output voltage. And thatβs also the potential
difference across whatever components are connected here and is therefore the output
voltage.

Weβve been told in the question
that the output voltage must be equal to 32 volts. And weβve been asked to try and
work out what the resistance of the first resistor π
one must be in order for this
to be true. So in order to answer this
question, we will need to recall the potential divider equation. This equation tells us that the
potential difference across resistor π
two, which weβre calling π two, is equal to
the resistance π
two divided by the sum of the resistances π
one plus π
two all
multiplied by the input potential difference π subscript in.

Now remember, the potential
difference across our second resistor π two is the same as our output potential
difference because theyβre connected in parallel. So we can replace π two with π
out here, at which point we see weβve already got a value for π out, a value for π
two, and a value for π in. We just need to rearrange to solve
for π
one. We can do this by multiplying both
sides of the equation by π
one plus π
two and dividing both sides of the equation
by π out. Because this way on the left-hand
side weβve got π out over π out, which is equal to one. And on the right-hand side, weβve
got π
one plus π
two both in the numerator and denominator.

Then we simply subtract π
two from
both sides of the equation so that weβre left with π
one on the left, at which
point we simply substitute in the values on the right-hand side, taking care to
notice that in this fraction the unit of volts is both in the numerator and
denominator and that if weβre going to stick with kiloohms, then our final answer
for π
one is going to be in kiloohms as well. When we simplify all of the
right-hand side, we find that our final answer is π
one is equal to 50
kiloohms.

So letβs summarize what weβve
talked about in this lesson. Weβve seen that potential dividers
consist of resistors connected in series, with an output voltage drawn across one of
the resistors. We saw the relationship between the
output and input voltage and how it links to the resistances used in the
circuit. And finally, we saw that
thermistors and light-dependent resistors, or LDRs, can be used in these circuits
for practical applications.