Electromagnetic Induction in Alternating Current Systems
3. Transformers
Learning outcomes
- I can explain how transformers operate using mutual induction.
- I can distinguish between step-up and step-down transformers.
- I can calculate transformer voltage ratios.
- I can explain energy conservation in ideal transformers.
- I can describe practical applications of transformers.
Transformers
A transformer is an electrical device that transfers electrical energy from one circuit to another using mutual induction.
Transformers can change alternating voltage without requiring a direct electrical connection between the input and output circuits.
The basic process is:
AC in primary coil → changing magnetic field → changing magnetic flux in core → induced AC voltage in secondary coil
Transformers are essential in electrical power systems because they allow voltages to be increased or decreased efficiently.
Main Parts of a Transformer
A basic transformer contains:
- primary coil
- secondary coil
- magnetic core
The primary coil is connected to the input AC supply.
The secondary coil provides the output voltage.
The magnetic core helps guide magnetic flux from the primary coil through the secondary coil.
The coils are normally electrically separate but magnetically coupled.
The Primary Coil
When alternating current flows through the primary coil:
- current continuously changes
- the magnetic field changes
- the direction of the magnetic field periodically reverses
- magnetic flux through the core changes
This changing magnetic flux is necessary for transformer operation.
A steady magnetic field would not continuously induce a voltage in the secondary.
The Magnetic Core
The transformer core provides a low-reluctance path for magnetic flux.
It helps concentrate the changing magnetic field so that a large fraction of the magnetic flux produced by the primary links the secondary.
This improves magnetic coupling between the coils.
Common transformer core materials are selected to have useful magnetic properties while minimizing energy losses.
The Secondary Coil
The changing magnetic flux from the primary passes through the secondary coil.
According to Faraday's law:
ε = −N ΔΦ/Δt
Therefore, changing flux through the secondary produces an induced EMF.
Because the magnetic flux repeatedly changes direction and magnitude, the induced voltage in the secondary is also alternating.
Mutual Induction
Transformer operation is an application of mutual induction.
Mutual induction occurs when:
changing current in one circuit produces an induced EMF in another circuit
The primary current creates the changing magnetic field.
The secondary experiences the resulting changing magnetic flux.
Therefore:
primary circuit → magnetic coupling → secondary circuit
No direct conducting connection between the two windings is required in an ordinary transformer.
Why Transformers Require Changing Flux
Suppose steady DC is connected to a transformer.
Immediately after switching:
- current changes
- magnetic field changes
- induction can occur briefly
Once the current becomes steady:
ΔI/Δt = 0
The magnetic field becomes approximately constant.
Therefore:
ΔΦ/Δt = 0
and continuous induced secondary voltage stops.
Transformers therefore require changing magnetic flux.
AC naturally provides this.
AC Transformer Operation
During one part of the AC cycle:
- primary current increases in one direction
- magnetic flux increases
- voltage is induced in the secondary
Later:
- primary current decreases
- magnetic flux changes
- the secondary voltage changes
During the opposite half-cycle:
- primary current reverses
- magnetic field reverses
- induced secondary voltage reverses
The output is therefore also AC.
Transformer Turns Ratio
The voltage produced in each winding depends on the number of turns.
For an ideal transformer:
V_s/V_p = N_s/N_p
where:
- V_s = secondary voltage
- V_p = primary voltage
- N_s = number of turns in secondary
- N_p = number of turns in primary
The ratio:
N_s/N_p
is called the turns ratio.
Understanding the Turns Ratio
If:
N_s > N_p
then:
V_s > V_p
The transformer increases voltage.
If:
N_s < N_p
then:
V_s < V_p
The transformer decreases voltage.
If:
N_s = N_p
then ideally:
V_s = V_p
The voltage is unchanged.
Step-Up Transformers
A step-up transformer increases voltage.
Therefore:
V_s > V_p
and:
N_s > N_p
The secondary has more turns than the primary.
Worked Example: Step-Up Transformer
A transformer has:
N_p = 200 turns
N_s = 1000 turns
V_p = 24 V
Use:
V_s/V_p = N_s/N_p
V_s/24 = 1000/200
V_s/24 = 5
V_s = 120 V
V_s = 120 V
The transformer increases the voltage from 24 V to 120 V.
Step-Down Transformers
A step-down transformer decreases voltage.
Therefore:
V_s < V_p
and:
N_s < N_p
The secondary has fewer turns than the primary.
Worked Example: Step-Down Transformer
A transformer has:
N_p = 1200 turns
N_s = 200 turns
V_p = 240 V
Use:
V_s/V_p = N_s/N_p
V_s/240 = 200/1200
V_s/240 = 1/6
V_s = 40 V
V_s = 40 V
The transformer reduces the voltage from 240 V to 40 V.
Finding the Number of Turns
The transformer equation can be rearranged to find an unknown number of turns.
Suppose:
V_p = 120 V
V_s = 12 V
N_p = 600 turns
Find N_s.
Use:
V_s/V_p = N_s/N_p
12/120 = N_s/600
0.10 = N_s/600
N_s = 60
N_s = 60 turns
Finding Primary Turns
Suppose:
V_p = 50 V
V_s = 200 V
N_s = 800 turns
Use:
V_s/V_p = N_s/N_p
200/50 = 800/N_p
4 = 800/N_p
N_p = 200
N_p = 200 turns
This is a step-up transformer.
Identifying the Transformer Type
A quick way to identify transformer type is to compare the number of turns.
More secondary turns than primary turns:
N_s > N_p
→ step-up
Fewer secondary turns than primary turns:
N_s < N_p
→ step-down
You can also compare the voltages.
V_s > V_p → step-up
V_s < V_p → step-down
Voltage Ratio and Turns Ratio
Suppose:
N_s/N_p = 4
Then:
V_s/V_p = 4
The secondary voltage is four times the primary voltage.
If:
N_s/N_p = 0.25
then:
V_s/V_p = 0.25
The secondary voltage is one-quarter of the primary voltage.
The voltage changes by the same factor as the number of turns.
Energy Conservation in an Ideal Transformer
An ideal transformer has no energy losses.
Therefore:
input power = output power
Electrical power is:
P = VI
So for an ideal transformer:
V_pI_p = V_sI_s
where:
- V_p = primary voltage
- I_p = primary current
- V_s = secondary voltage
- I_s = secondary current
This relationship follows from conservation of energy.
Voltage and Current Trade-Off
A transformer does not create additional electrical power.
If voltage increases, current must decrease.
If voltage decreases, current can increase.
Therefore:
step-up voltage → step-down current
and:
step-down voltage → step-up current
for an ideal transformer supplying a load.
Current Ratio
From:
V_pI_p = V_sI_s
and:
V_s/V_p = N_s/N_p
we obtain:
I_s/I_p = N_p/N_s
Notice that the current ratio is the inverse of the turns ratio.
This is important.
Step-Up Transformer Current
For a step-up transformer:
V_s > V_p
but:
I_s < I_p
For example, if voltage increases by a factor of 10, the ideal current decreases by a factor of 10.
Power remains unchanged.
Step-Down Transformer Current
For a step-down transformer:
V_s < V_p
but:
I_s > I_p
For example, if voltage decreases by a factor of 5, the ideal current can increase by a factor of 5.
Again, the transformer does not create energy.
Worked Example: Current in an Ideal Transformer
A transformer has:
V_p = 120 V
I_p = 2.0 A
V_s = 24 V
Use:
V_pI_p = V_sI_s
(120)(2.0) = (24)I_s
240 = 24I_s
I_s = 10 A
I_s = 10 A
The voltage decreased by a factor of 5.
The current increased by a factor of 5.
Worked Example: Step-Up Voltage and Current
A transformer receives:
V_p = 50 V
I_p = 4.0 A
and produces:
V_s = 200 V
Input power:
P_p = V_pI_p
P_p = (50)(4.0)
P_p = 200 W
For an ideal transformer:
P_s = 200 W
Therefore:
I_s = P_s/V_s
I_s = 200/200
I_s = 1.0 A
Voltage increased by a factor of 4 while current decreased by a factor of 4.
Why Transformers Cannot Create Energy
Suppose a transformer increases voltage from:
100 V → 1000 V
This does not mean the transformer produces ten times as much energy.
If the ideal voltage increases by a factor of 10, current decreases by a factor of 10.
For example:
Primary:
100 V × 10 A = 1000 W
Secondary:
1000 V × 1 A = 1000 W
The power remains:
1000 W
Energy has been transferred and transformed, not created.
Ideal vs Real Transformers
An ideal transformer assumes:
- no electrical resistance
- no magnetic flux leakage
- no eddy-current losses
- no hysteresis losses
- perfect magnetic coupling
Therefore:
P_input = P_output
Real transformers are not perfect.
Therefore:
P_output < P_input
Transformer Efficiency
Efficiency can be calculated using:
efficiency = P_output/P_input × 100%
A well-designed transformer can have high efficiency, especially at its intended operating conditions.
Worked Example: Transformer Efficiency
A transformer receives:
P_input = 500 W
and supplies:
P_output = 450 W
Efficiency:
efficiency = 450/500 × 100%
efficiency = 90%
Power lost:
500 − 450
P_loss = 50 W
This lost energy is mostly transferred into thermal energy.
Copper Losses
Transformer windings are made from conducting wire.
Real wire has resistance.
Therefore, some electrical energy becomes thermal energy.
The power loss in a winding can be described by:
P_loss = I²R
These losses are often called copper losses.
Reducing winding resistance can improve efficiency.
Eddy-Current Losses
The changing magnetic flux in the core can induce circulating currents inside conducting core material.
These are eddy currents.
Eddy currents produce unwanted heating.
One way to reduce them is to construct the core from thin insulated layers called laminations.
Why Laminations Reduce Eddy Currents
A solid conducting core can allow relatively large circulating current loops.
Laminating the core:
- divides it into thin sections
- interrupts large current paths
- increases resistance to circulating currents
- reduces eddy-current heating
This improves transformer efficiency.
Hysteresis Losses
The transformer core is repeatedly magnetized in opposite directions as AC changes direction.
The magnetic domains inside the material repeatedly rearrange.
Some energy is lost during this process.
This is called hysteresis loss.
Suitable magnetic materials are chosen to reduce this loss.
Flux Leakage
Ideally, all magnetic flux produced by the primary would link the secondary.
In reality, some magnetic flux does not pass through both windings.
This is called leakage flux.
Flux leakage reduces effective magnetic coupling.
Transformer designs attempt to keep the primary and secondary strongly magnetically coupled.
Heating
Copper losses, eddy currents, and hysteresis all contribute to heating.
Large transformers therefore require effective cooling.
Cooling methods may include:
- natural air cooling
- forced-air cooling
- insulating cooling fluids
- radiators
- heat exchangers
Managing heat is an important part of transformer design.
Why Transformers Are Important for Power Transmission
Transformers are extremely important in electrical power grids.
Electrical power is:
P = VI
For a fixed amount of transmitted power:
higher V → lower I
Transmission-line heating is approximately:
P_loss = I²R
Therefore, reducing current greatly reduces energy loss in transmission wires.
Example: Why High Voltage Helps
Suppose:
P = 100,000 W
If electricity is transmitted at:
V = 1000 V
then:
I = P/V
I = 100,000/1000
I = 100 A
Suppose instead it is transmitted at:
V = 10,000 V
I = 100,000/10,000
I = 10 A
Current has decreased by a factor of 10.
Since:
P_loss = I²R
the resistive loss for the same wire resistance decreases by a factor of:
10² = 100
This demonstrates why high-voltage transmission is so useful.
Transformers in the Electrical Grid
Electrical power systems use multiple transformers.
A simplified sequence is:
generator → step-up transformer → high-voltage transmission → step-down transformers → consumers
Near the generator, voltage is increased.
Near users, voltage is reduced to levels appropriate for distribution and equipment.
Step-Up Transformers in Power Stations
Generators produce electrical power at a particular voltage.
A step-up transformer increases this voltage before long-distance transmission.
This allows the same power to be transmitted with lower current.
Lower current reduces:
I²R heating losses
in transmission lines.
Step-Down Transformers
High transmission voltages are unsuitable for direct use by most equipment.
Step-down transformers progressively reduce voltage during electrical distribution.
This makes the electrical supply appropriate for:
- homes
- businesses
- schools
- factories
- electronic devices
Different systems require different voltage levels.
Transformers in Chargers and Power Supplies
Many electrical devices require voltages different from the available supply.
Transformers can be part of systems that:
- reduce voltage
- provide electrical isolation
- transfer power
- prepare voltage for further conversion
Modern electronic chargers often use high-frequency switching circuits and relatively small transformers.
Why High-Frequency Transformers Can Be Small
The induced voltage depends on how rapidly magnetic flux changes.
At higher operating frequencies, the flux changes more rapidly.
This can allow transformers designed for high-frequency switching power supplies to be much smaller than traditional low-frequency transformers handling comparable power.
This is one reason modern chargers can be compact.
Electrical Isolation
A transformer can transfer energy without a direct electrical connection between the primary and secondary windings.
This can provide galvanic isolation.
Isolation can be useful for:
- electrical safety
- protecting sensitive electronics
- separating circuit grounds
- measurement equipment
- medical and industrial systems
Not every transformer configuration provides isolation; for example, an autotransformer uses a shared winding.
Isolation Transformers
An isolation transformer may have approximately equal primary and secondary voltages.
For example:
N_s ≈ N_p
so:
V_s ≈ V_p
Its main purpose is therefore not necessarily to change voltage.
Instead, it provides electrical separation between circuits.
This shows that transformers can have functions beyond simply stepping voltage up or down.
Audio and Signal Transformers
Transformers can also transfer changing electrical signals.
They may be used for:
- impedance matching
- signal isolation
- audio equipment
- communication circuits
These transformers operate using the same principle of mutual induction.
Instrument Transformers
High-power electrical systems can contain voltages and currents that are too large to measure directly with ordinary instruments.
Special transformers can produce smaller proportional quantities for:
- measurement
- monitoring
- protection systems
Examples include current transformers and voltage transformers.
Welding Transformers
Some welding equipment requires:
- relatively low voltage
- very high current
A step-down transformer can reduce voltage while increasing available current.
The large current produces intense heating at the welding point.
This is an excellent example of the voltage-current trade-off in transformer operation.
Doorbells and Low-Voltage Systems
Some low-voltage devices use transformers to reduce supply voltage.
Applications can include:
- doorbells
- control circuits
- lighting systems
- older low-voltage power adapters
The transformer allows equipment to operate at a more suitable voltage.
Microwave Oven Transformers
Traditional microwave ovens may use transformers to produce a much higher voltage required by the magnetron circuitry.
This is an example of a step-up application.
Such circuits can involve dangerously high voltages and are not suitable for experimentation.
Transformers and Wireless Charging
Traditional transformers use a magnetic core to strongly couple nearby windings.
Wireless charging uses a related principle with physically separated coils.
The transmitter coil produces changing magnetic flux.
The receiver coil experiences this changing flux and develops an induced EMF.
Both technologies rely on mutual induction.
Transformer Frequency
Transformers are designed for particular operating frequencies or frequency ranges.
Changing frequency affects:
- magnetic flux
- core behaviour
- losses
- heating
- efficiency
Using a transformer far outside its designed frequency can cause poor performance or damage.
Magnetic Saturation
Transformer cores can only support a limited magnetic flux density before approaching saturation.
When saturation occurs:
- the core no longer responds proportionally
- effective inductive behaviour decreases
- primary current can increase greatly
- overheating can occur
Transformer design therefore considers voltage, frequency, turns, and core properties together.
A Complete Transformer Problem
A transformer has:
V_p = 240 V
N_p = 1200 turns
N_s = 100 turns
The primary current is:
I_p = 0.50 A
First calculate secondary voltage:
V_s/V_p = N_s/N_p
V_s/240 = 100/1200
V_s = 20 V
V_s = 20 V
For an ideal transformer:
V_pI_p = V_sI_s
(240)(0.50) = (20)I_s
120 = 20I_s
I_s = 6.0 A
I_s = 6.0 A
The transformer reduces voltage by a factor of 12 and increases current by a factor of 12.
Input power:
P_p = (240)(0.50)
P_p = 120 W
Output power:
P_s = (20)(6.0)
P_s = 120 W
Energy is conserved in the ideal model.
A Real Transformer Problem
Suppose the transformer above actually produces:
V_s = 20 V
I_s = 5.4 A
Output power:
P_s = V_sI_s
P_s = (20)(5.4)
P_s = 108 W
Input power was:
120 W
Efficiency:
efficiency = 108/120 × 100%
efficiency = 90%
Power lost:
120 − 108
P_loss = 12 W
A Reliable Strategy for Transformer Problems
First identify:
- V_p
- V_s
- N_p
- N_s
- I_p
- I_s
For voltage and turns:
V_s/V_p = N_s/N_p
For an ideal transformer's power:
V_pI_p = V_sI_s
For current ratio:
I_s/I_p = N_p/N_s
For a real transformer's efficiency:
efficiency = P_output/P_input × 100%
Then determine whether the transformer is:
step-up: V_s > V_p and N_s > N_p
or:
step-down: V_s < V_p and N_s < N_p
Finally, check that the result makes physical sense.
If voltage increases in an ideal transformer, current should decrease.
Common Mistakes
Thinking transformers create energy
Transformers transfer energy. An ideal transformer conserves energy.
Confusing primary and secondary
The primary receives input. The secondary provides output.
Reversing the turns ratio
Use:
V_s/V_p = N_s/N_p
Keep secondary quantities together and primary quantities together.
Thinking step-up means more current
A step-up transformer increases voltage but decreases current in the ideal loaded case.
Thinking step-down means less power
An ideal step-down transformer has the same input and output power; voltage decreases while current increases.
Thinking transformers work continuously with steady DC
Continuous transformer action requires changing magnetic flux.
Assuming real transformers are 100% efficient
Real transformers lose energy through resistance, eddy currents, hysteresis, leakage, and other effects.
Thinking laminations increase eddy currents
Laminations are used to reduce unwanted eddy currents.
Ignoring units
Use volts for voltage, amperes for current, watts for power, and turns as a simple count.
Key Terms
Transformer — device that transfers electrical energy between circuits through mutual induction.
Primary coil — winding connected to the input electrical supply.
Secondary coil — winding in which the output voltage is induced.
Mutual induction — induction of an EMF in one circuit caused by changing current and magnetic flux from another circuit.
Magnetic core — material that provides a path for magnetic flux and improves coupling between windings.
Turns ratio — ratio of the number of turns in the secondary to the number in the primary.
Step-up transformer — transformer that produces a higher secondary voltage than primary voltage.
Step-down transformer — transformer that produces a lower secondary voltage than primary voltage.
Ideal transformer — theoretical transformer with no energy losses.
Efficiency — fraction or percentage of input power converted into useful output power.
Copper loss — energy loss caused by electrical resistance in transformer windings.
Eddy current — circulating current induced in conducting material by changing magnetic flux.
Lamination — thin insulated core layer used to reduce eddy currents.
Hysteresis loss — energy loss caused by repeated magnetization and demagnetization of a magnetic material.
Leakage flux — magnetic flux that does not effectively link both windings.
Galvanic isolation — electrical separation between circuits with no direct conductive path.
Magnetic saturation — condition where increasing magnetizing influence produces relatively little additional magnetization in a core.
Key Takeaways
- Transformers operate using mutual induction.
- Alternating primary current creates changing magnetic flux.
- Changing magnetic flux induces an alternating voltage in the secondary.
- A magnetic core helps couple flux between the windings.
- Transformer voltage depends on the turns ratio.
- The ideal transformer equation is V_s/V_p = N_s/N_p.
- A step-up transformer has more secondary turns and produces higher secondary voltage.
- A step-down transformer has fewer secondary turns and produces lower secondary voltage.
- An ideal transformer conserves power: V_pI_p = V_sI_s.
- Increasing voltage requires decreasing current in an ideal transformer.
- Decreasing voltage allows current to increase.
- Transformers do not create electrical energy.
- Real transformers lose some energy through resistance, eddy currents, hysteresis, leakage, and heating.
- Laminated cores help reduce eddy-current losses.
- Transformers make efficient high-voltage electrical transmission possible.
- High transmission voltage allows lower current for the same power.
- Lower current greatly reduces I²R transmission losses.
- Transformers are used in power grids, chargers, power supplies, welding equipment, measurement systems, isolation systems, and electronic equipment.
- High-frequency transformer operation allows compact transformers in many modern switching power supplies.
- Transformers are one of the most important practical applications of electromagnetic induction.
Check Your Understanding
- Explain how mutual induction allows a transformer to transfer electrical energy from the primary to the secondary.
- Why must the magnetic flux in a transformer change with time?
- Identify the primary coil, secondary coil, and magnetic core, and describe the function of each.
- A transformer has N_p = 500 turns and N_s = 2000 turns. Is it a step-up or step-down transformer? Explain.
- A transformer has V_p = 120 V, N_p = 600 turns, and N_s = 150 turns. Calculate V_s.
- A transformer changes 20 V to 100 V. If the primary contains 200 turns, calculate the required number of secondary turns.
- An ideal transformer receives 240 V at 2.0 A and produces 24 V. Calculate the secondary current.
- Explain why the secondary current of an ideal step-up transformer is lower than its primary current.
- A transformer receives 800 W and produces 720 W. Calculate its efficiency.
- Identify three major sources of energy loss in a real transformer and explain how at least one can be reduced.
- Explain why transformer cores are often laminated.
- A power line must transmit the same amount of power at either 10 kV or 100 kV. Compare the currents and explain why the higher transmission voltage reduces energy loss.
- Explain why connecting steady DC directly to a conventional transformer does not produce continuous secondary voltage and may cause problems.
- Compare a step-up and step-down transformer in terms of turns, voltage, and current.
- Describe three practical applications of transformers and explain why a transformer is useful in each application.