Question Video: Solving for the Current in the Secondary Coil of a Transformer | Nagwa Question Video: Solving for the Current in the Secondary Coil of a Transformer | Nagwa

# Question Video: Solving for the Current in the Secondary Coil of a Transformer Physics • Third Year of Secondary School

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A transformer with an iron core has a primary coil with 45 turns and a secondary coil that also has 45 turns. The coils have a mutual inductance of 2.5 H. A current in the primary coil induces a potential difference of 2.2 V across the secondary coil in a time of 0.18 s. What is the current through the secondary coil after 0.18 s? Give your answer to two decimal places.

04:07

### Video Transcript

A transformer with an iron core has a primary coil with 45 turns and a secondary coil that also has 45 turns. The coils have a mutual inductance of 2.5 henries. A current in the primary coil induces a potential difference of 2.2 volts across the secondary coil in a time of 0.18 seconds. What is the current through the secondary coil after 0.18 seconds? Give your answer to two decimal places.

Let’s say that this is the iron core of our transformer. Here is the primary coil of the transformer. We’ve drawn it with just a few turns rather than 45. And here is the secondary coil with the same number of turns. We’re told that these two coils have a mutual inductance of 2.5 henries. In the process of mutual inductance, a change in current in one of the coils, here we’ll say it’s the change in the current in the primary coil Δ𝐼 sub one, induces a magnetic flux, we’ll call it ΔΦ sub 𝐵, in the iron core. This change in flux is experienced by each one of the turns in the secondary coil.

If we call the number of turns in that coil 𝑁 sub two, then this product is equal to the total change in magnetic flux through the secondary coil. Mathematically, this is equal to the mutual inductance 𝑀 between the coils multiplied by the change in current in the primary coil. Physically, when there’s a change in magnetic flux through the turns of a coil, that induces a potential difference across the coil. Here, we’ll represent this as Δ𝑉 sub two, since we’re considering the secondary coil. The potential difference across the secondary coil is equal to the mutual inductance between the coils times the change in current in the primary coil all divided by the amount of time it takes for this current to change.

What we’re seeing then is that a change in current in one of the coils, coupled with the fact that the coils are mutually inductive, induces a potential difference in the other coil. Along with this, we can note that this potential difference that’s induced induces a current in that other coil, in this case, the secondary one. In this question, we’re given Δ𝑉 sub two, that’s 2.2 volts; 𝑀, that’s 2.5 henries; and Δ𝑡, 0.18 seconds.

What we don’t know is Δ𝐼 sub one, the change in current through the primary coil. Notice this though. Our question statement asks us not about the current in the primary coil but that in the secondary coil. What we really want to know is a quantity we can call Δ𝐼 sub two. Similarly to the equation right above it, if we multiply Δ𝐼 sub two by the mutual inductance between the coils, this equals the number of turns in the primary coil, we’ll call this 𝑁 sub one, multiplied by the change in magnetic flux through the iron core ΔΦ sub 𝐵.

In our problem statement, we’re told that both the primary and the secondary coils have 45 turns. The important bit here is not what that particular number is — that it’s 45 — but rather that 𝑁 one is equal to 𝑁 two. Looking at these two equations, we see that 𝑀 is the same in both cases, ΔΦ sub 𝐵 is the same in both cases, and we now know that 𝑁 two equals 𝑁 one. For these two equations to hold true then, it must also be the case that Δ𝐼 sub one equals Δ𝐼 sub two. In other words, because the two coils have the same number of turns, the current change in both of them must also be the same.

This fact means that we can return to our equation for Δ𝑉 sub two and replace Δ𝐼 sub one with Δ𝐼 sub two since these two quantities are equal. Now we have an equation we can work with to solve for the current in the secondary coil Δ𝐼 sub two. If we multiply both sides of this equation by the change in time Δ𝑡 divided by the mutual inductance 𝑀, then on the right-hand side the mutual inductance 𝑀 and Δ𝑡 both cancel out. We find then that Δ𝑡 times Δ𝑉 sub two divided by 𝑀 equals Δ𝐼 sub two.

Clearing a bit of space to work, we can reverse the sides of this equation and then substitute in 0.18 seconds for Δ𝑡, 2.2 volts for Δ𝑉 sub two, and 2.5 henries for 𝑀. Calculating this value, to two decimal places, it’s 0.16 amperes. This is the current through the secondary coil after 0.18 seconds.

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