### Video Transcript

A transformer has 200 turns on its
primary coil and 50 turns on its secondary coil. If the input potential difference is 20
volts, what is the output potential difference?

Okay, so in this question, weβre dealing
with a transformer which weβve been told has 200 turns on its primary coil. So we can say that π π, the number of
turns on the primary coil, is 200. And we also know that it has 50 turns on
its secondary coil. In other words, the number of turns π
sub π on the secondary coil is 50.

As well as this, we know the input
potential difference or in other words the potential difference across the primary coil
which weβll call π sub π and we know that itβs 20 volts. This is because the primary coil
is on the input side of the transformer and the secondary coil is on the output side. Hence, the potential difference across
the primary coil is the same thing as the input potential difference. And in this question, we need to find the
output potential difference, which in other words is the potential difference across the
secondary coil, π sub π .

Now, to find π sub π , we can recall
that for a transformer the turns ratio π sub π over π sub π or in other words the number
of turns on the secondary coil divided by the number of turns on the primary coil is equal
to the voltage ratio of the same coils, in other words the output voltage divided by the
input voltage or the voltage across the secondary coil divided by the voltage across the
primary coil.

So we can use this equation to rearrange
and solve for the output potential difference. We can do this by multiplying both sides
of the equation by π π, the potential difference across the primary coil. This way, π π cancels on the right-hand
side. And what weβre left with is that π π
multiplied by π π over π π, the turns ratio, is equal to π π . Then we can plug in the values for π π,
π π , and π π. So we have 20 volts as π π multiplied
by 50, thatβs π π , divided by 200, thatβs π π. And then, when we evaluate the left-hand
side of the equation, we find that π sub π is equal to five volts.

In other words, the output potential
difference for this transformer is five volts.