💡 Getting Started: The Magic of Transformers
Hello! Welcome to the essential topic of transformers. These devices are arguably the most important invention enabling our modern electrical grid. Why? Because they allow us to efficiently change AC voltages—stepping them up for long-distance transmission and stepping them down for safe use in our homes.
In this chapter, we will connect the ideas you’ve learned about magnetic fields and electromagnetic induction to understand exactly how a transformer works, why it's so important for power transmission, and why they only operate using AC.
1. The Operating Principle: Electromagnetic Induction
A transformer is a passive electrical device that transfers electrical energy from one electrical circuit to another, or multiple circuits. It achieves this using two fundamental physics concepts: magnetic fields and induction.
A. The Basic Structure
A typical transformer consists of three main parts:
- Primary Coil (\(N_p\)): The coil connected to the input AC power supply.
- Secondary Coil (\(N_s\)): The coil connected to the output load (the device receiving the power).
- Soft Iron Core: A magnetic material (like soft iron) that links the magnetic flux perfectly between the primary and secondary coils.
B. Why AC is Essential
Transformers only work with Alternating Current (AC). This is a key concept!
Remember Faraday's Law of Induction (from Section 3.10.4): An electromotive force (e.m.f.) is induced in a coil when there is a change in magnetic flux linkage.
Here is the step-by-step process of induction in a transformer:
- AC Input: The primary coil is connected to an AC supply. AC current is constantly changing direction and magnitude.
- Changing Flux: This changing current creates a constantly changing magnetic field (and therefore changing magnetic flux, \(\Phi\)) inside the soft iron core.
- Flux Linkage: The soft iron core guides this changing flux so that it passes through the turns of the secondary coil.
- Induced EMF: Because the secondary coil experiences a changing magnetic flux linkage (\(\Delta \Phi / \Delta t\)), an e.m.f. (\(V_s\)) is induced across the secondary coil, providing AC power to the load.
If you used a DC (Direct Current) supply, the current would be steady. The magnetic flux would be constant, \(\Delta \Phi / \Delta t\) would be zero, and no e.m.f. would be induced in the secondary coil! The transformer would not work.
Key Takeaway (Section 1)
The operation of a transformer relies entirely on electromagnetic induction, which requires a changing magnetic flux. Therefore, transformers must use AC.
2. The Ideal Transformer: Equations and Ratios
An ideal transformer is one that loses no energy, meaning the input power equals the output power. While no real transformer is ideal, we use this model to derive the essential relationships between voltage, current, and the number of turns.
A. The Transformer Equation (Voltage and Turns)
The magnitude of the induced e.m.f. in a coil is proportional to the number of turns in that coil. This gives us the crucial transformer equation relating the secondary voltage \(V_s\) and the primary voltage \(V_p\) to the number of turns \(N_s\) and \(N_p\):
Transformer Equation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
This equation defines the two types of transformers:
- Step-Up Transformer: \(N_s > N_p\). The voltage is increased (\(V_s > V_p\)).
- Step-Down Transformer: \(N_s < N_p\). The voltage is decreased (\(V_s < V_p\)).
Memory Aid: If you want to Step Up the voltage, you need More Turns in the Secondary coil. (SU-M-TS).
B. Power Conservation (Current Ratio)
In an ideal transformer, the power input to the primary coil equals the power output from the secondary coil.
\( P_{\text{in}} = P_{\text{out}} \implies V_p I_p = V_s I_s \)
By rearranging this, we can find the relationship between the current and the turns ratio:
Current Ratio: \( \frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
Important Concept: Notice the relationship for current is inverse to the voltage relationship.
If you step up the voltage (V increases), you must step down the current (I decreases) by the same factor, to ensure power is conserved. This is vital for safety and efficiency.
Quick Review: Ideal Transformer Rules
- Voltage and turns are directly proportional: \( V \propto N \)
- Voltage and current are inversely proportional: \( V \propto 1/I \)
3. Real Transformers and Efficiency
In reality, all transformers waste some energy, primarily as heat. We define the effectiveness of a real transformer using efficiency.
A. Defining Transformer Efficiency
Efficiency is the ratio of the useful power output to the total power input, usually expressed as a percentage.
\( \text{Efficiency} = \frac{\text{Useful Output Power}}{\text{Input Power}} \)
Since \( P = IV \), the formula required by the syllabus is:
\( \text{Efficiency} = \frac{I_s V_s}{I_p V_p} \)
Modern power transformers can achieve efficiencies of over 99%, but it is important to understand the sources of the small losses.
B. Causes of Inefficiencies (Power Losses)
1. Resistance Loss (Joule Heating)
The copper wires used for the primary and secondary coils have resistance (\(R\)). As current flows, energy is lost as heat, calculated by \(P = I^2 R\). This is often called copper loss.
- Mitigation: Use thicker wires (lower resistance) for the coils, especially on the high-current side (the low-voltage side).
2. Eddy Current Loss
The changing magnetic flux passing through the iron core not only induces an e.m.f. in the secondary coil but also induces unwanted currents within the core itself. These are called eddy currents. These currents flow in small loops, generating heat (\(I^2 R\)) in the core, wasting energy.
- Mitigation: The core is not made of one solid block of iron, but rather a stack of thin sheets (called laminations) separated by insulating material. This insulation dramatically increases the resistance along the path of the eddy currents, reducing their magnitude and minimizing loss.
3. Magnetic Flux Leakage
Ideally, all the magnetic field lines produced by the primary coil should pass through the secondary coil. In practice, some lines escape ("leak"), meaning the flux linkage in the secondary coil is less than the flux produced by the primary coil.
- Mitigation: Using a continuous soft iron ring core (a closed loop) forces nearly all the magnetic field lines to stay confined within the core material, minimizing leakage.
4. Application: Electrical Power Transmission
This is the most critical real-world application of transformers and the core reason we use AC for the national grid.
A. Why We Transmit Power at High Voltage
Imagine a power station generating 10 MW of power that needs to be sent to a city far away via transmission lines (pylons).
The power delivered (\(P_{delivered}\)) is related to the current (\(I\)) and the voltage (\(V\)) by \(P_{delivered} = IV\).
The power wasted (lost) in the transmission cables due to their resistance (\(R_{\text{line}}\)) is given by:
\( P_{\text{loss}} = I^2 R_{\text{line}} \)
To minimize \(P_{\text{loss}}\), we must minimize the current \(I\) flowing through the lines.
Step-by-Step Logic:
- Since \(P_{delivered} = IV\) must remain constant (we need to deliver the same amount of power).
- If we use a step-up transformer near the power station to increase the voltage (V) massively (e.g., from 25 kV to 400 kV).
- The required current (I) drops dramatically.
- Since power loss is proportional to \(I^2\), even a small reduction in current leads to a massive reduction in wasted energy.
Example: Doubling the voltage halves the current, but reduces the power loss by a factor of four (\( (1/2)^2 = 1/4 \)).
B. The Role of Step-Down Transformers
Once the power reaches the local substations, step-down transformers are used sequentially to reduce the dangerously high voltage back to safe levels (e.g., 230 V or 110 V) for domestic and industrial consumption.
C. Calculation of Voltage and Power Losses
The syllabus requires calculations involving power and voltage losses in transmission lines.
If you are asked to calculate the voltage drop across the transmission line, use Ohm's law:
\( V_{\text{drop}} = I_{\text{line}} \times R_{\text{line}} \)
The total power loss is calculated using:
\( P_{\text{loss}} = I^2 R_{\text{line}} \)
Common Mistake to Avoid: When calculating power loss, make sure you use the current in the transmission line (\(I_{\text{line}}\)), not the input or output current of the transformer itself. \(I_{\text{line}}\) is the current determined by the high transmission voltage \(V\).
Key Takeaway (Section 4)
High-voltage transmission minimizes current (\(I\)), which minimizes power loss (\(P \propto I^2\)) across the resistive transmission cables, drastically improving the efficiency of the national grid.