# Question Video: Finding the Number of Windings on the Coils of a Step-Down Transformer

A transformer is used to step down 110 V from a wall socket to 9.0 V 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 mA, what is the current through the primary winding?

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### 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. Weβre told that the primary winding has 500 turns in it. Weβll call that number π sub one. Weβre told that the final current in the system is 500 milliamps. Weβll call that πΌ sub two.

In part one, we want to solve for the number of turns that the secondary winding has. Weβll call that number π sub two. And second, we want to solve for the current that runs through the primary winding. Weβll call that πΌ 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 weβre working with has both a primary side and a secondary side. Current πΌ 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 πΌ 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 πΌ two over πΌ 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, weβre 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. Thatβs the number of turns in the secondary coil. Now we move on to solving for πΌ 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 πΌ one is equal to π two over π one times πΌ two. We know all three of these values and can plug them in to solve for πΌ one.

When we do, making sure to use a value in units of amperes for our current πΌ two, when we plug these terms in on our calculator, we find that πΌ one, to two significant figures, is 41 milliamps. Thatβs the current that runs through the primary coil.