Ideal Transformer Calculations

Hello! Welcome to your next lesson in the UK electrical grids course.

In our last session, we explored the fundamental principle of electromagnetic induction and how it allows transformers to function. We learned that transformers are essential for changing AC voltage levels, which is critical for transmitting power efficiently across the grid.

Today, we will build directly on that conceptual foundation. This lesson will focus on the mathematics behind transformers. By the end of this 60-minute session, you will be able to calculate voltage, current, and turns ratios in ideal transformers. This will give you the practical skills to determine exactly how a transformer alters voltage and current.

1. The Ideal Transformer and the Turns Ratio

In the real world, all devices have some energy losses, usually as heat. However, to understand the basic principles of transformer calculations, we start with a simplified model called an ideal transformer. In an ideal transformer, we assume it is 100% efficient, meaning no energy is lost.

This leads to a fundamental rule: Power In = Power Out.

Transformer Basics - Introduction to Ratios and Calculations

Let's start with a short video that introduces the concept of an ideal transformer and its power rating.

Watch the section from 01:15 to 03:09. The key takeaway is the assumption we make for basic calculations: the power going into the transformer is the same as the power coming out.

As we discussed in the last lesson, the "magic" of a transformer comes from the number of wire turns in its primary and secondary coils. The relationship between these turns is called the turns ratio.

The turns ratio directly determines the voltage change. The relationship is proportional: if the secondary coil has half the turns of the primary, it will produce half the voltage.

The formula is:
$\frac{V _{P}}{V _{S}} = \frac{N _{P}}{N _{S}}$
Where:

$V_{P}$ = Primary Voltage (input)
$V_{S}$ = Secondary Voltage (output)
$N_{P}$ = Number of turns on the Primary coil
$N_{S}$ = Number of turns on the Secondary coil

This ratio $\frac{N _{P}}{N _{S}}$ is the turns ratio.

If $N_{P} > N_{S}$ , the turns ratio is greater than 1, and it's a step-down transformer (voltage decreases).
If $N_{S} > N_{P}$ , the turns ratio is less than 1, and it's a step-up transformer (voltage increases).

Transformer Basics and Transformer Principles

This article provides a clear, text-based explanation of the turns ratio and its relationship to voltage.

Please read the sections titled 'A Transformers Turns Ratio' and the two examples that follow. This will solidify your understanding of the formula and how to apply it for step-down transformers.

2. The Current Ratio: An Inverse Relationship

We started with the principle that for an ideal transformer, Power In = Power Out. We also know from Module 1 that Power is Voltage multiplied by Current ( $P = V \times I$ ).

This means:
$V_{P} \times I_{P} = V_{S} \times I_{S}$

This simple equation reveals a crucial insight: if voltage is stepped up, current must be stepped down by the same factor to keep the power equal on both sides. This is an inverse relationship.

Therefore, the formula for current is flipped compared to the voltage formula:
$\frac{I _{S}}{I _{P}} = \frac{N _{P}}{N _{S}}$
Where:

$I_{P}$ = Primary Current (input)
$I_{S}$ = Secondary Current (output)

Notice how the secondary current ( $I_{S}$ ) is in the numerator on the left, while the primary turns ( $N_{P}$ ) is in the numerator on the right.

We can combine these relationships into one master equation for ideal transformers:
$\frac{V _{P}}{V _{S}} = \frac{N _{P}}{N _{S}} = \frac{I _{S}}{I _{P}}$

7 Key Facts: What Is the Ideal Transformer Turn Ratio?

This article explains the inverse relationship between turns and current very well, using the principle of energy conservation.

Read the subsection titled 'The Inverse Relationship: Turns and Current'. Focus on understanding why stepping voltage down causes current to be stepped up.

3. Putting It All Together: Worked Examples

Now that we have the formulas, let's see them in action. The best way to learn is by watching someone solve a problem and then trying it yourself.

Transformers Physics Problems - Voltage, Current & Power Calculations - Electromagnetic Induction

This video from The Organic Chemistry Tutor is excellent for demonstrating how to solve transformer problems. It clearly explains the formulas and then applies them step-by-step.

First, watch the summary of the equations from 02:22 to 04:06. Then, carefully follow the worked example in 'Problem 1' from 04:06 to 07:57. Pay attention to how the main formula is used to find both the output voltage and the input current.

4. Your Turn: Practice Problem

Now it's time for you to solve a typical transformer problem.

Problem:
A transformer is used to power a 12V garden lighting system from the 230V UK mains supply. The primary coil has 4600 turns.

Is this a step-up or a step-down transformer?
How many turns must the secondary coil have?
If the lighting system draws a current of 4A from the secondary coil, what is the current drawn from the mains supply by the primary coil (assuming an ideal transformer)?

Take your time to work through the calculations using the formulas we've discussed.

Click here to see the solution

1. Step-up or Step-down?
The transformer reduces the voltage from 230V to 12V, so it is a step-down transformer.

2. Secondary Turns ( $N_{S}$ ):
We use the voltage and turns ratio formula:
$\frac{V _{P}}{V _{S}} = \frac{N _{P}}{N _{S}}$
We need to rearrange it to solve for $N_{S}$ :
$N_{S} = N_{P} \times \frac{V _{S}}{V _{P}}$
Now, plug in the values:
$N_{S} = 4600 \times \frac{12 V}{230 V} \approx 240 turns$
The secondary coil must have 240 turns.

3. Primary Current ( $I_{P}$ ):
We use the current and turns ratio formula:
$\frac{I _{S}}{I _{P}} = \frac{N _{P}}{N _{S}}$
Rearrange to solve for $I_{P}$ :
$I_{P} = I_{S} \times \frac{N _{S}}{N _{P}}$
Plug in the values:
$I_{P} = 4 A \times \frac{240}{4600} \approx 0.209 A$
The primary coil draws approximately 0.21 Amps from the mains.

Check (Optional): Does Power In = Power Out?

Power In = $V_{P} \times I_{P} = 230 V \times 0.209 A = 48.07 W$
Power Out = $V_{S} \times I_{S} = 12 V \times 4 A = 48 W$
The values are almost identical (the small difference is due to rounding), confirming our calculations are correct.

Conclusion

Great work today! You've moved from the concept of a transformer to the practical calculations that govern its operation.

Here are the key takeaways from this lesson:

For an ideal transformer, we assume it is 100% efficient, so Power In = Power Out.
The turns ratio ( $N_{P} / N_{S}$ ) is the key to all calculations.
Voltage is directly proportional to the turns ratio ( $V_{P} / V_{S} = N_{P} / N_{S}$ ).
Current is inversely proportional to the turns ratio ( $I_{S} / I_{P} = N_{P} / N_{S}$ ).
The combined formula $\frac{V _{P}}{V _{S}} = \frac{N _{P}}{N _{S}} = \frac{I _{S}}{I _{P}}$ is your tool for solving transformer problems.

Preview of the next lesson:
Having covered the basics of transformers, we will now shift our focus to where the electricity comes from in the first place. In the next lesson, we will begin looking at generation by describing the shared energy conversion process in thermal power stations using the Rankine cycle.

Recommended resources

1. The Ideal Transformer and the Turns Ratio

2. The Current Ratio: An Inverse Relationship

3. Putting It All Together: Worked Examples

4. Your Turn: Practice Problem

Conclusion