### Video Transcript

The circuit shown in the diagram
consists of three resistors connected in series with a cell. The first resistor has a resistance
of π
, the second resistor has a resistance of two π
, and the third resistor has a
resistance of three π
. The cell provides a potential
difference of 24 volts. The current through the circuit is
0.1 amperes. What is the value of π
?

Taking a look at our circuit, we
see that indeed there are these three individual resistors of values π
, two π
, and
three π
in series with our 24-volt cell. Weβre told a bit more about the
circuit. Weβre also told that it carries a
current of 0.1 amperes. We can call that current πΌ. Knowing all this, we want to solve
for the value of π
, which we see all three of our resistors are expressed in terms
of.

As we get to work, letβs clear some
space on screen. Okay, to briefly recap, we were
told that the current running through the circuit is 0.1 amps. And overall, we want to solve for
the value of π
. To do that, we would like to relate
the two known quantities in our circuit, which are the voltage and the current, to
the unknown quantity, which is the resistance. Whenever we name these three
quantities, voltage, current, and resistance, that can remind us of a law for
electrical circuits β Ohmβs law.

This law tells us that given a
resistor of constant value, if we multiply the value of that resistor by the current
running through it, the net product is equal to the potential difference across the
resistor. Looking at our circuit, we see that
we donβt have one resistor, but we have three. But really, we would like to have
one. We would like to be able to express
these three resistors as one equivalent resistance.

Thankfully, weβre able to do that
because these resistors are arranged in series and we can recall the addition rule
for resistors arranged this way. We can remember that for π series
resistors β that is π resistors arranged in series with one another β the total
resistance of all those resistors equals the value of the first resistor plus the
value of the second resistor plus dot dot dot all the way up to the value of the
πth resistor.

In our case, in our circuit, π is
equal to three. Thatβs how many series resistors we
have. So we can say that the total
resistance in our particular circuit is equal to the value of the first resistor π
plus the value of the second resistor two π
plus the value of the third resistor
three π
. Looking at the right-hand side of
this expression, we see we can add all these terms together: one π
plus two π
plus
three π
is six π
. Thatβs the total resistance in this
circuit.

Itβs wonderful that weβve solved
for that because now we can return to Ohmβs law, where in this case, our resistor
value is π
sub π‘, the value weβve just solved for. When we apply Ohmβs law to our
equivalent or total resistance in our circuit, we write that six π
which is our
total equivalent resistance multiplied by the current in the circuit β weβve called
it πΌ β is equal to the potential difference across the entire circuit. Letβs call it π.

As we look though, we see that we
know the value for πΌ; itβs 0.1 amperes. And we also know the value for π;
itβs given as 24 volts. This means we can substitute those
values in. And now, we want to arrange this
equation so that we solve for π
. Dividing both sides by 0.1 amps
times six, both those terms cancel out on the right-hand side. And we see that π
is equal to 24
volts divided by 0.1 amps times six. When we calculate this fraction, we
find a result of 40 ohms. Thatβs the resistance value of
π
.