### Video Transcript

A transformer is used to step down 110 volts from a wall socket to 9.0 volts for a
radio. If the primary winding has 500 turns, how many turns does the secondary winding
have? If the radio operates at a current of 500 milliamps, what is the current through the
primary winding?

We can call the voltage in the primary winding, 110 volts, π sub one and the voltage
output from the secondary winding, 9.0 volts, as π two. Weβre told that the primary winding has 500 turns in it. Weβll call that number π sub one. Weβre told that the final current in the system is 500 milliamps. Weβll call that πΌ sub two.

In part one, we want to solve for the number of turns that the secondary winding has. Weβll call that number π sub two. And second, we want to solve for the current that runs through the primary
winding. Weβll call that πΌ sub one.

To start on our solution, we can recall the ratio relationships for a
transformer. If we draw a sketch of the transformer, the transformer weβre working with has both a
primary side and a secondary side. Current πΌ one flows into the primary side and moves through a number of loops π sub
one. This circuit is powered by a potential difference of π one.

On the secondary side, a current πΌ sub two moves through the loops of which there
are π sub two. This circuit delivers potential difference of π two to an appliance. Transformers like this follow particular mathematical equations for the ratios of
voltage, current, and the number of turns π. In particular, π one over π two is equal to π one over π two, which equals πΌ two
over πΌ one.

In part one of our exercise, we want to solve for π two. Rearranging this expression, we find that π two is equal to π two over π one times
π one. In the exercise statement, weβre told π one, π two, and π one. So we can now plug in in this equation and solve for π two.

When we enter these values on our calculator, we find that, to two significant
figures, π two is 41. Thatβs the number of turns in the secondary coil. Now we move on to solving for πΌ sub one, the current in the primary coil.

We can once again use our transformers equation. And when we rearrange it slightly, we find that πΌ one is equal to π two over π one
times πΌ two. We know all three of these values and can plug them in to solve for πΌ one.

When we do, making sure to use a value in units of amperes for our current πΌ two,
when we plug these terms in on our calculator, we find that πΌ one, to two
significant figures, is 41 milliamps. Thatβs the current that runs through the primary coil.