The operation of a transformer (A-level only)

What you'll learn

How transformers use electromagnetic induction to step alternating voltages up or down.
How to use the transformer turns ratio equation and calculate transformer efficiency.
Why real transformers aren't completely efficient (including the production of eddy currents).
Why the National Grid relies on high-voltage transformers to minimise power loss over long distances.

How a transformer works

A transformer is a device that changes the amplitude of an alternating voltage. If you've ever charged a laptop from the mains or looked at the massive pylons connecting the National Grid, you are relying on transformers.

Transformers operate entirely on the principle of Faraday's Law of electromagnetic induction. A basic transformer consists of two coils of wire—the primary coil and the secondary coil—wound around a single continuous ring of magnetic material, usually a soft iron core.

Transformer diagram

Here is the step-by-step physical process of how they work:

An alternating voltage is applied across the primary coil.
This drives an alternating current (AC) through the primary coil.
The alternating current creates a continuously changing magnetic field around the primary coil.
The soft iron core provides a highly permeable path, channeling this changing magnetic flux through to the secondary coil.
Because the secondary coil is sitting in a changing magnetic flux, an alternating electromotive force (emf) is induced across it (Faraday's Law).

Common Mistake

Why direct current (DC) fails

If you apply a steady direct current to the primary coil, a constant magnetic field is produced. Because the magnetic flux is constant (not changing), no emf will be induced in the secondary coil. Transformers only work with alternating current!

The Transformer Equation

The size of the voltage induced in the secondary coil depends entirely on the ratio of the number of turns of wire on each coil.

If we assume the transformer is "ideal" (meaning all the magnetic flux from the primary coil passes through the secondary coil with no losses), the relationship is given by the transformer equation:

NsNp=VsVp\frac{N_s}{N_p} = \frac{V_s}{V_p}

Where:

$N_s$ is the number of turns on the secondary coil.
$N_p$ is the number of turns on the primary coil.
$V_s$ is the voltage across the secondary coil (in volts, V).
$V_p$ is the voltage across the primary coil (in volts, V).

Definition

Step-up and Step-down transformers

Step-up transformer: Has more turns on the secondary coil ( $N_s > N_p$ ). It increases the alternating voltage ( $V_s > V_p$ ).
Step-down transformer: Has fewer turns on the secondary coil ( $N_s < N_p$ ). It decreases the alternating voltage ( $V_s < V_p$ ).

Example

Calculating secondary voltage

A step-up transformer has 400 turns on its primary coil and 2000 turns on its secondary coil. It is connected to a $\text{ V}$ mains supply. Calculate the output voltage of the transformer.

Identify the given values: $N_p = 400$ , $N_s = 2000$ , and $VV_p = 230 \text{ V}$ .
Rearrange the transformer equation to make $V_s$ the subject: $Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p}$
Substitute the values into the equation: $Vs=230×2000400V_s = 230 \times \frac{2000}{400}$
Calculate the final answer: $VV_s = 230 \times 5 = 1150 \text{ V}$

Transformer Efficiency

An ideal transformer is 100% efficient, meaning the electrical power entering the primary coil is completely transferred to the secondary coil.

Remembering that electrical power $P = I V$ , for an ideal transformer:

I_p V_p = I_s V_s

Key Idea

Voltage goes up, current goes down

If a step-up transformer increases the voltage by a factor of 10, it must simultaneously decrease the current by a factor of 10 to conserve energy. You cannot get more power out of a transformer than you put in!

In reality, transformers are very efficient (often above 98%), but never perfectly 100%. We calculate the efficiency using the ratio of useful power output to total power input:

Efficiency=IsVsIpVp\text{Efficiency} = \frac{I_s V_s}{I_p V_p}

(Note: To express this as a percentage, simply multiply the result by 100).

Example

Calculating efficiency

A step-down transformer is connected to a $\text{ V}$ supply and draws a primary current of $\text{ A}$ . The secondary coil supplies a current of $\text{ A}$ at a voltage of $\text{ V}$ . Calculate the efficiency of the transformer.

Calculate the input power ( $Pin=IpVpP_{\text{in}} = I_p V_p$ ): $WP_{\text{in}} = 2.5 \times 3300 = 8250 \text{ W}$
Calculate the output power ( $Pout=IsVsP_{\text{out}} = I_s V_s$ ): $WP_{\text{out}} = 24 \times 320 = 7680 \text{ W}$
Use the efficiency formula: $Efficiency=76808250\text{Efficiency} = \frac{7680}{8250}$
Convert to a percentage: $Efficiency=0.9309...≈93.1%\text{Efficiency} = 0.9309... \approx 93.1\%$

Causes of inefficiencies in a transformer

If power is lost, where does it go? Almost all lost energy in a transformer is dissipated as heat. You need to know the four main causes of inefficiency and, crucially, how engineers design transformers to minimise them.

1. Copper losses (Joule heating)

The copper wires forming the primary and secondary coils have resistance. As alternating current flows through them, they heat up, dissipating power equal to $I^2 R$ . How it is reduced: Use thick copper wire with a large cross-sectional area to keep the resistance $R$ as low as possible.

2. Eddy currents

The soft iron core is a metal, meaning it is an electrical conductor. Because it is sitting in a continuously changing magnetic field, Faraday's Law states that emfs will be induced inside the core itself.

Definition

Eddy currents

Small, circulating loops of induced alternating current entirely within the metal core of the transformer. They cause the core to heat up due to the metal's electrical resistance.

How it is reduced: The core is laminated. Instead of a solid block of iron, the core is built from many thin layers (laminae) of iron separated by thin layers of insulating material. This greatly increases the electrical resistance of the core in the direction the eddy currents want to flow, reducing their size and minimising the $I^2 R$ heat loss.

3. Hysteresis losses

The alternating current causes the magnetic field in the core to constantly flip back and forth (50 times a second for UK mains). Energy is required to constantly re-align the magnetic domains in the iron core. How it is reduced: The core is made of a "magnetically soft" material, like soft iron, which is easy to magnetise and demagnetise.

4. Flux leakage

Not all the magnetic flux generated by the primary coil manages to travel through the core to cut the secondary coil. How it is reduced: The primary and secondary coils are often wound as close to each other as possible, sometimes physically layered on top of one another around the same part of the core.

High-Voltage Power Transmission

The ultimate application of transformers is the National Grid. Power stations generate electricity, which then needs to be transmitted over hundreds of kilometres to towns and cities.

Transmission cables have an electrical resistance, $R$ . When a current $I$ flows through them, they act like giant heaters, wasting power into the atmosphere. The power lost as heat in the transmission lines is given by:

Ploss=I2RP_{\text{loss}} = I^2 R

Because the current $I$ is squared in this equation, even a small reduction in current leads to a massive reduction in wasted power.

To transmit a specific amount of power ( $P = I V$ ), we can either use a high current and a low voltage, or a low current and a high voltage. By using a step-up transformer at the power station, we drastically increase the transmission voltage $V$ , which proportionally shrinks the current $I$ .

Common Mistake

Using the wrong power equation

When calculating power lost in the cables, always use $Ploss=I2RP_{\text{loss}} = I^2 R$ . Do not use $Ploss=V2RP_{\text{loss}} = \frac{V^2}{R}$ . The $V$ in that formula refers to the potential difference across the ends of the cable, not the transmission voltage relative to the ground. Using it will give you the wrong answer and lose you marks!

Example

Calculating power loss in transmission

A power station generates $\text{ kW}$ of electrical power. This is transmitted through cables with a total resistance of $\Omega4.0 \text{ \Omega}$ . Calculate the power lost as heat in the cables if the electricity is transmitted at: (a) $\text{ V}$ (b) $\text{ V}$

Part (a) - Transmitting at $\text{ V}$ :

Calculate the current in the cables using $P = I V$ : $\frac{P}{V} = \frac{120,000}{2000} = 60 \text{ A}$
Calculate the power lost using $I^2 R$ : $kW)P_{\text{loss}} = 60^2 \times 4.0 = 3600 \times 4.0 = 14,400 \text{ W} \text{ (or } 14.4 \text{ kW)}$

Part (b) - Transmitting at $\text{ V}$ :

Calculate the new current: $\frac{P}{V} = \frac{120,000}{40,000} = 3.0 \text{ A}$
Calculate the new power lost: $WP_{\text{loss}} = 3.0^2 \times 4.0 = 9 \times 4.0 = 36 \text{ W}$

Notice the enormous difference: stepping the voltage up by a factor of 20 reduced the current by a factor of 20, which reduced the power loss by a factor of $20^2$ (400 times less power wasted).

Exam technique

In the exam

Be rigorous with your terms: If a question asks why laminated cores are used, clearly state that they increase electrical resistance, which reduces the size of the eddy currents, thereby reducing $I^2 R$ heat loss in the core.
Watch your prefixes: Voltages in power transmission are often given in $kV\text{kV}$ and power in $MW\text{MW}$ . Always convert these to $V\text{V}$ and $W\text{W}$ before using $P = I V$ or $P = I^2 R$ .
Remember the sequence: Step-up transformer at the power station (high $V$ , low $I$ for transmission) $→\rightarrow$ Step-down transformers near towns (to drop $V$ back to safe levels for homes, like $\text{ V}$ ).

Self review

Check yourself

Can you explain why a transformer will not work if connected to a DC battery?
Can you list the four main causes of energy loss in a transformer and state how each is minimised?
If a transformer is 100% efficient and steps up the voltage by a factor of 5, what happens to the current?
Why does doubling the transmission voltage of the National Grid cut the power lost in the cables by a quarter?

The operation of a transformer (A-level only)

What you'll learn

How transformers use electromagnetic induction to step alternating voltages up or down.
How to use the transformer turns ratio equation and calculate transformer efficiency.
Why real transformers aren't completely efficient (including the production of eddy currents).
Why the National Grid relies on high-voltage transformers to minimise power loss over long distances.

How a transformer works

Transformer diagram

Here is the step-by-step physical process of how they work:

An alternating voltage is applied across the primary coil.
This drives an alternating current (AC) through the primary coil.
The alternating current creates a continuously changing magnetic field around the primary coil.
The soft iron core provides a highly permeable path, channeling this changing magnetic flux through to the secondary coil.
Because the secondary coil is sitting in a changing magnetic flux, an alternating electromotive force (emf) is induced across it (Faraday's Law).

Common Mistake

Why direct current (DC) fails

The Transformer Equation

The size of the voltage induced in the secondary coil depends entirely on the ratio of the number of turns of wire on each coil.

NsNp=VsVp\frac{N_s}{N_p} = \frac{V_s}{V_p}

Where:

$N_s$ is the number of turns on the secondary coil.
$N_p$ is the number of turns on the primary coil.
$V_s$ is the voltage across the secondary coil (in volts, V).
$V_p$ is the voltage across the primary coil (in volts, V).

Definition

Step-up and Step-down transformers

Step-up transformer: Has more turns on the secondary coil ( $N_s > N_p$ ). It increases the alternating voltage ( $V_s > V_p$ ).
Step-down transformer: Has fewer turns on the secondary coil ( $N_s < N_p$ ). It decreases the alternating voltage ( $V_s < V_p$ ).

Example

Calculating secondary voltage

A step-up transformer has 400 turns on its primary coil and 2000 turns on its secondary coil. It is connected to a $\text{ V}$ mains supply. Calculate the output voltage of the transformer.

Identify the given values: $N_p = 400$ , $N_s = 2000$ , and $VV_p = 230 \text{ V}$ .
Rearrange the transformer equation to make $V_s$ the subject: $Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p}$
Substitute the values into the equation: $Vs=230×2000400V_s = 230 \times \frac{2000}{400}$
Calculate the final answer: $VV_s = 230 \times 5 = 1150 \text{ V}$

Transformer Efficiency

An ideal transformer is 100% efficient, meaning the electrical power entering the primary coil is completely transferred to the secondary coil.

Remembering that electrical power $P = I V$ , for an ideal transformer:

I_p V_p = I_s V_s

Key Idea

Voltage goes up, current goes down

In reality, transformers are very efficient (often above 98%), but never perfectly 100%. We calculate the efficiency using the ratio of useful power output to total power input:

Efficiency=IsVsIpVp\text{Efficiency} = \frac{I_s V_s}{I_p V_p}

(Note: To express this as a percentage, simply multiply the result by 100).

Example

Calculating efficiency

Calculate the input power ( $Pin=IpVpP_{\text{in}} = I_p V_p$ ): $WP_{\text{in}} = 2.5 \times 3300 = 8250 \text{ W}$
Calculate the output power ( $Pout=IsVsP_{\text{out}} = I_s V_s$ ): $WP_{\text{out}} = 24 \times 320 = 7680 \text{ W}$
Use the efficiency formula: $Efficiency=76808250\text{Efficiency} = \frac{7680}{8250}$
Convert to a percentage: $Efficiency=0.9309...≈93.1%\text{Efficiency} = 0.9309... \approx 93.1\%$

Causes of inefficiencies in a transformer

1. Copper losses (Joule heating)

2. Eddy currents

Definition

Eddy currents

Small, circulating loops of induced alternating current entirely within the metal core of the transformer. They cause the core to heat up due to the metal's electrical resistance.

3. Hysteresis losses

4. Flux leakage

High-Voltage Power Transmission

The ultimate application of transformers is the National Grid. Power stations generate electricity, which then needs to be transmitted over hundreds of kilometres to towns and cities.

Ploss=I2RP_{\text{loss}} = I^2 R

Because the current $I$ is squared in this equation, even a small reduction in current leads to a massive reduction in wasted power.

Common Mistake

Using the wrong power equation

Example

Calculating power loss in transmission

Part (a) - Transmitting at $\text{ V}$ :

Calculate the current in the cables using $P = I V$ : $\frac{P}{V} = \frac{120,000}{2000} = 60 \text{ A}$
Calculate the power lost using $I^2 R$ : $kW)P_{\text{loss}} = 60^2 \times 4.0 = 3600 \times 4.0 = 14,400 \text{ W} \text{ (or } 14.4 \text{ kW)}$

Part (b) - Transmitting at $\text{ V}$ :

Calculate the new current: $\frac{P}{V} = \frac{120,000}{40,000} = 3.0 \text{ A}$
Calculate the new power lost: $WP_{\text{loss}} = 3.0^2 \times 4.0 = 9 \times 4.0 = 36 \text{ W}$

Notice the enormous difference: stepping the voltage up by a factor of 20 reduced the current by a factor of 20, which reduced the power loss by a factor of $20^2$ (400 times less power wasted).

Exam technique

In the exam

Be rigorous with your terms: If a question asks why laminated cores are used, clearly state that they increase electrical resistance, which reduces the size of the eddy currents, thereby reducing $I^2 R$ heat loss in the core.
Watch your prefixes: Voltages in power transmission are often given in $kV\text{kV}$ and power in $MW\text{MW}$ . Always convert these to $V\text{V}$ and $W\text{W}$ before using $P = I V$ or $P = I^2 R$ .
Remember the sequence: Step-up transformer at the power station (high $V$ , low $I$ for transmission) $→\rightarrow$ Step-down transformers near towns (to drop $V$ back to safe levels for homes, like $\text{ V}$ ).

Self review

Check yourself

Can you explain why a transformer will not work if connected to a DC battery?
Can you list the four main causes of energy loss in a transformer and state how each is minimised?
If a transformer is 100% efficient and steps up the voltage by a factor of 5, what happens to the current?
Why does doubling the transmission voltage of the National Grid cut the power lost in the cables by a quarter?

The operation of a transformer (A-level only)

What you'll learn

How a transformer works

The Transformer Equation

Transformer Efficiency

Causes of inefficiencies in a transformer

1. Copper losses (Joule heating)

2. Eddy currents

3. Hysteresis losses

4. Flux leakage

High-Voltage Power Transmission

The operation of a transformer (A-level only) Revision Guide

The operation of a transformer (A-level only)

What you'll learn

How a transformer works

The Transformer Equation

Transformer Efficiency

Causes of inefficiencies in a transformer

1. Copper losses (Joule heating)

2. Eddy currents

3. Hysteresis losses

4. Flux leakage

High-Voltage Power Transmission

The operation of a transformer (A-level only) Revision Guide