### Video Transcript

A 100 percent efficient step-down transformer has 100 turns on its primary coil and 20 turns on its secondary coil. The input potential difference is 250 volts and the input power is 7500 watts. What is the output current?

So let’s start by underlining all the important information in the question. Well, firstly, we know that we’ve got a transformer that’s 100 percent efficient. We also know that it’s a step-down transformer. But then we can see that from the number of turns on the primary and secondary coil. The primary has 100 turns and the secondary has 20 turns. Therefore, it must be a step-down transformer. It’s got more turns on the primary coil than it does on the secondary.

Also we know that the input potential difference is 250 volts and the input power is 7500 watts. What we’re asked to do is to find the output current. Now, we can draw a diagram here. So let’s do that.

And here it is. Now, a transformer usually consists of a primary coil, here drawn in orange, a secondary coil, drawn in pink, and the core, which is usually iron drawn in blue. Here, we’ve also labelled all the quantities we’ve been given in the question. For example, 𝑉 sub 𝑖 or the input potential difference has been labelled as 250 volts. Also the input power 𝑃 sub 𝑖 is 7500 watts. In the question, we’ve been asked to calculate 𝐼 sub 𝑜 or the output current.

We can also label the number of turns on the primary coil as 𝑁 sub 𝑖 because number sub input coil. And the secondary coil will be 𝑁 sub 𝑜, number sub output. Now, we know that this transformer is 100 percent efficient. We’ve been told that in the question. This means that it doesn’t lose any energy to its surroundings. All of the energy remains in the system.

Therefore, the power must not change from one side to the other as well. This is because power is the rate of transfer of energy. And if we’re not losing any energy to the surroundings, then the power which is just how much energy is being transferred per unit time must be the same on both sides. Using this logic, we know that the output power 𝑃 sub 𝑜 is also 7500 watts, same as on the primary side. And this is a property of ideal or 100 percent efficient transformers. They don’t lose power.

In real life of course, ideal transformers are difficult to manufacture because there will always be some form of losses such as resistive losses in the wire or eddy current being formed in the iron core, stuff like that. But in this question, we’re assuming that we’ve got an ideal 100 percent efficient transformer. So let’s run with it.

So we know what the power is on both sides of the transformer — the voltage on the primary side. And we want to find out what the current is on the secondary side. To do this, first, we can find out what the voltage is on the secondary side. This can be done by using the transformer equation.

The transformer equation basically describes the whole point of a transformer. So in this case, we’ve got a step-down transformer, which means that the voltage from one side to the other is stepped down. Specifically, what this means is that the ratio of voltages on both sides is the same as the ratio of the number of turns on both sides. Now, that’s a very long and complicated sentence. But it’s a lot easier to understand when we write it in symbols.

And that happens to be 𝑉 sub 𝑜 over 𝑉 sub 𝑖, the ratio of the output and input voltages, is equal to 𝑁 sub 𝑜 over 𝑁 sub 𝑖, the ratio of the number of turns in each coil. And that is a hope for another transformer. It transforms the voltage from one side to the other, depending on the ratio of the number of coils on each side.

So we can take this equation and rearrange it to find out what the voltage on the output side is. If we multiply both sides of the equation by the input voltage 𝑉 sub 𝑖, then we get 𝑉 sub 𝑜 is equal to 𝑉 sub 𝑖 multiplied by the ratio of the number of turns 𝑁 sub 𝑜 over 𝑁 sub 𝑖. We can then plug in our numbers 𝑉 sub 𝑖 being 250 volts, 𝑁 sub 𝑜 being 20, and 𝑁 sub 𝑖 being 100. And that gives us a value for the output voltage of 50 volts. So we can write down that useful piece of information on the secondary side in our diagram.

Now, we’ve got the voltage on the secondary side and we’ve got the power on the secondary side. We now need to work out the current on the secondary side. At this point, we can ignore the primary side. We’ve got all the information that we can from that side. So therefore, we only need to focus on the pink bit now. We need to find a relationship that links together the voltage, the current, and the power — in this case, on the secondary side.

We can recall that the power in a circuit is simply given by voltage multiplied by current. So we can take this relationship and rearrange it to find the current. If we divide both sides of the equation by the voltage 𝑉, then we get that the power divided by the voltage is equal to the current.

Now, in this case, we’re looking at the power, voltage, and current on the secondary side of the coil or the output side. So we can add the subscript 𝑜 for all of these values. Then, it’s just a matter of substituting in our values for the output power and the output voltage. And this leads us to our final answer, which is that the output current is 150 amps.