Question Video: Finding the Output Current of a Transformer Physics • Third Year of Secondary School

Video Transcript

A 100 percent efficient transformer has five times as many turns on its secondary coil as it does on its primary. If the current through the primary coil is 20 amperes, what is the current through the secondary coil?

In this question, we want to find the current through the secondary coil of a 100 percent efficient transformer. We are told that the transformer has five times as many turns on its secondary coil as it does on its primary coil and that the current through the primary coil is 20 amperes.

Before we tackle this question, let’s clear up some terminology. A transformer has two coils: a primary coil and a secondary coil. The current and potential difference associated with the primary coil are called the input current and the input potential difference. The current and potential difference associated with the secondary coil are called the output current and the output potential difference.

Recall that for a transformer, the ratio of the number of turns 𝑁 in the input and output coil is the same as the ratio of the potential difference 𝑉 across these coils. We can write this as 𝑁 input over 𝑁 output equals 𝑉 input over 𝑉 output. We are told that the secondary coil has five times as many turns compared to the primary coil. So 𝑁 output is equal to five times 𝑁 input.

We can then rearrange this equation to get 𝑁 input over 𝑁 output equals one-fifth, which is helpful because now we can directly substitute this value into our previous equation. This substitution shows us that the ratio of potential differences across these coils must also equal one-fifth. We can then rewrite this here as 𝑉 input over 𝑉 output equals one-fifth. This is a helpful proportion to know, but we still need to find the current through the secondary coil, 𝐼 output. In order to relate potential difference to current, we can use power.

Now, let’s recall that the power 𝑃 in each coil is given by 𝑃 equals 𝐼𝑉, where 𝐼 is the current in that particular coil and 𝑉 is the potential difference of that particular coil. Since we have a 100 percent efficient transformer, the power in each coil will be equal: 𝑃 input equals 𝑃 output. This means that we will have 𝐼 input times 𝑉 input equals 𝐼 output times 𝑉 output. We are looking to isolate 𝐼 output, which we can do by dividing both sides by 𝑉 output, causing the 𝑉 output terms on the right side to cancel. Looking now at what remains in the equation, we can spot a familiar proportion: 𝑉 input over 𝑉 output.

Now, we have already found earlier that 𝑉 input over 𝑉 output is equal to one-fifth. So, substituting this value into our equation, we find that 𝐼 input divided by five equals 𝐼 output. The value of current through the primary coil that we are given is 20 amperes. So we can substitute this value into the equation. After doing so, we find that 𝐼 output equals 20 amperes divided by five, which simplifies to just four amperes.

Therefore, the current going through the secondary coil, 𝐼 output, is four amperes. This is the correct answer to this question.