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
We can call the voltage in the primary winding, 110 volts, 𝑉 sub one and the voltage
output from the secondary winding, 9.0 volts, as 𝑉 two. We’re told that the primary winding has 500 turns in it. We’ll call that number 𝑁 sub one. We’re told that the final current in the system is 500 milliamps. We’ll call that 𝐼 sub two.
In part one, we want to solve for the number of turns that the secondary winding has. We’ll call that number 𝑁 sub two. And second, we want to solve for the current that runs through the primary
winding. We’ll call that 𝐼 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 we’re working with has both a
primary side and a secondary side. Current 𝐼 one flows into the primary side and moves through a number of loops 𝑁 sub
one. This circuit is powered by a potential difference of 𝑉 one.
On the secondary side, a current 𝐼 sub two moves through the loops of which there
are 𝑁 sub two. This circuit delivers potential difference of 𝑉 two to an appliance. Transformers like this follow particular mathematical equations for the ratios of
voltage, current, and the number of turns 𝑁. In particular, 𝑁 one over 𝑁 two is equal to 𝑉 one over 𝑉 two, which equals 𝐼 two
over 𝐼 one.
In part one of our exercise, we want to solve for 𝑁 two. Rearranging this expression, we find that 𝑁 two is equal to 𝑉 two over 𝑉 one times
𝑁 one. In the exercise statement, we’re told 𝑉 one, 𝑉 two, and 𝑁 one. So we can now plug in in this equation and solve for 𝑁 two.
When we enter these values on our calculator, we find that, to two significant
figures, 𝑁 two is 41. That’s the number of turns in the secondary coil. Now we move on to solving for 𝐼 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 𝐼 one is equal to 𝑉 two over 𝑉 one
times 𝐼 two. We know all three of these values and can plug them in to solve for 𝐼 one.
When we do, making sure to use a value in units of amperes for our current 𝐼 two,
when we plug these terms in on our calculator, we find that 𝐼 one, to two
significant figures, is 41 milliamps. That’s the current that runs through the primary coil.