Simple Iron-Core Transformer: Principles, Turns Ratio, Energy Losses & Examples Explained

22.6 TRANSFORMER

22.6.1 Applications of a Simple Transformer

1 A transformer is an electrical device which has an input and an output. Its main application is to transfer electrical energy from an alternating voltage supply applied to the input to an electrical load connected to the output by changing the input voltage or current.

2 The applied (input) voltage is said to have been
(a) stepped up if the output voltage is larger than the input voltage.
(b) stepped down if the output voltage is lesser than the input voltage.

3 A transformer is also used to
(a) match impedance of one device with that of another
(b) isolate one part of a circuit from another

22.6.2 Structure of a Simple Iron-core Transformer

Figure 22.69 The figure shows a transformer consisting of a primary coil, a secondary coil, and a laminated soft iron core. The core links the coils and increases the magnetic flux.

Figure 22.69 shows the structure of a simple transformer.
A transformer has the following components:

(a) A primary coil
A transformer has one coil which is connected to the ac power supply. The coil is known as the primary coil. It acts as an inductor and has self inductance.

(b) A secondary coil
A transformer has a second coil, which is connected to an electrical load. This coil is known as the secondary coil. It acts as an inductor and has self inductance.

(c) A laminated soft iron core
The primary and secondary coils are wound over a laminated soft iron core. The purpose of the iron core is to
(i) act as a magnetic ‘circuit’ for the magnetic field lines to flow, hence magnetically linking the primary and the secondary coils
(ii) increase tremendously the magnetic flux through each coil as iron has very large relative permeability.

22.6.3 Principle of a Simple Unloaded Iron-core Transformer

Figure 22.70 The figure shows an alternating voltage applied to the primary coil, producing an alternating current that generates a magnetic field in the iron core. The magnetic field links the primary and secondary coils through the core.

1. An alternating voltage E_p applied is applied across the primary coil so that an alternating current I_p is produced in the coil.

2. The current flowing through the primary coil produces a magnetic field in the iron core. The magnetic field lines travel round the iron core in closed loops, hence magnetically linking the secondary coil to the primary coil, as shown in Figure 22.70.

3. Since the current changes with time, the magnetic flux that passes through the primary coil also changes with time. Hence, self induction occurs within the primary coil. According to Faraday’s law of electromagnetic induction, the time varying magnetic flux has to produce a self-induced emf E_p across the primary coil. According to Lenz’s law, this induced emf opposes the applied voltage.

4. The time varying magnetic flux, which is produced by the current in the primary coil, flows through the secondary coil and brings about mutual induction in the secondary coil. Hence, a mutually-induced emf E_s is produced across the secondary coil.

22.6.4 Turns Ratio for a Simple Ideal Iron-core Transformer

Refer to the diagram shown above. Suppose that there is no electrical load connected to the secondary coil. The alternating voltage applied to the primary coil produces a current in the coil. The alternating current produces a self-induced emf E_p across the primary coil. According to Faraday’s law of electromagnetic induction, this emf is given by

$E_p = −N_p \frac{dΦ_p}{dt}$

22.6.2 Structure of a Simple Iron-core Transformer

Figure 22.69 shows the structure of a simple transformer.
A transformer has the following components:

(a) A primary coil
A transformer has one coil which is connected to the ac power supply. The coil is known as the primary coil. It acts as an inductor and has self inductance.

(b) A secondary coil
A transformer has a second coil, which is connected to an electrical load. This coil is known as the secondary coil. It acts as an inductor and has self inductance.

(c) A laminated soft iron core
The primary and secondary coils are wound over a laminated soft iron core. The purpose of the iron core is to
(i) act as a magnetic ‘circuit’ for the magnetic field lines to flow, hence magnetically linking the primary and the secondary coils
(ii) increase tremendously the magnetic flux through each coil as iron has very large relative permeability.

22.6.3 Principle of a Simple Unloaded Iron-core Transformer

1. An alternating voltage E_p applied is applied across the primary coil so that an alternating current I_p is produced in the coil.

2. The current flowing through the primary coil produces a magnetic field in the iron core. The magnetic field lines travel round the iron core in closed loops, hence magnetically linking the secondary coil to the primary coil, as shown in Figure 22.70.

3. Since the current changes with time, the magnetic flux that passes through the primary coil also changes with time. Hence, self induction occurs within the primary coil. According to Faraday’s law of electromagnetic induction, the time varying magnetic flux has to produce a self-induced emf E_p across the primary coil. According to Lenz’s law, this induced emf opposes the applied voltage.

4. The time varying magnetic flux, which is produced by the current in the primary coil, flows through the secondary coil and brings about mutual induction in the secondary coil. Hence, a mutually-induced emf E_s is produced across the secondary coil.

22.6.4 Turns Ratio for a Simple Ideal Iron-core Transformer

Refer to the diagram shown above. Suppose that there is no electrical load connected to the secondary coil. The alternating voltage applied to the primary coil produces a current in the coil. The alternating current produces a self-induced emf E_p across the primary coil. According to Faraday’s law of electromagnetic induction, this emf is given by

$E_p = −N_p \frac{dΦ_p}{dt}$

where N_p is the number of turns of the primary coil and dΦ_p/dt is the rate of change of magnetic flux through the primary coil.

The alternating magnetic flux through the secondary coil produces a mutually-induced emf E_s across the coil which is given by

$E_s = −N_s \frac{dΦ_s}{dt}$

where N_s is the number of turns of the secondary coil and $\frac{dΦ_s}{dt}$ is the rate of change of magnetic flux through the secondary coil.

For an ideal transformer, we make the following assumption: there is no leakage of magnetic field lines out of the iron core.
In other words, we assume

$Φ_p = Φ_s$    (assumption)

$\frac{dΦ_p}{dt}=\frac{dΦ_s}{dt}$

Hence, we have

Comparing the two expressions above, we get the following expression for a simple unloaded ideal transformer:

$\frac{E_p}{E_s}=\frac{N_p}{N_s}$

Let us make another assumption: the resistance R_p of the primary coil is negligible.

$R_p ≅ 0$    (assumption)

Then we have

$E_p ≅ E_{applied}$

Hence, we have

$\frac{E_{applied}}{E_s} = \frac{N_p}{N_s}$

Notice that we may have the following two cases:

(a) If N_s > N_p, then E_s > E_p. The transformer is a step-up transformer because the output voltage (E_s) is larger than the input voltage (E_applied).

(b) If N_s < N_p, then E_s < E_p. The transformer is a step-down transformer because the output voltage is less than the input voltage.

Note: Although the resistance of the primary coil is very small, it does not mean that the current in the primary coil is infinitely large. There exists some other form of opposition, other than resistance, which limits the current. This opposition originates from the self inductance of the primary coil.

22.6.5 Principle of a Simple Ideal Loaded Iron-core Transformer

Figure 22.71 The figure shows magnetic flux flowing through the secondary coil, producing a mutually induced emf due to mutual induction.

1. The alternating voltage applied to the primary coil produces an alternating current I_p. The current in turn produces an alternating magnetic field in the iron core.

2. The field flows through the primary coil, causing self-induction to occur in the coil. The field produces a self-induced emf E_p across the primary coil. For an ideal transformer, we have the ‘equilibrium’ condition where $E_p = E_{applied}$.

3. The magnetic field lines travel round the iron core in closed loops. Hence, the magnetic flux also flows through the secondary coil. The alternating magnetic flux brings about mutual induction, thus producing a mutually-induced emf E_s across the secondary coil, as shown in Figure 22.71.

4. A resistive load is connected across the secondary coil, thus forming a closed circuit. At the instant when the secondary coil circuit is closed, an alternating current I_s flows in the circuit. The current flows through the load and so electrical power is dissipated.

5. At the instant when current I_s starts to flow, an unbalanced condition is created. This is due to the fact that the current I_s produces a magnetic field, which opposes the field produced by current I_p. Imagine for the time being that the field produced by current I_p does not exist. Then the alternating magnetic flux produced by I_s travels through the primary coil and brings about mutual induction in the primary coil.

6. A mutually-induced emf E'_p is produced by the field of I_s. This emf opposes the emf E_p already existing across the primary coil. Consequently, E_p is reduced at the moment when the current I_s starts to flow in the secondary coil so that now $E_{applied} > E_p$.

7. The reduction of E_p causes the effective p.d. across the primary coil to increase, thus causing the current I_p in the primary coil to increase too.

8. As I_p increases, the magnetic flux through the primary coil also increases. This in turn causes E_p to increase. In the end, a time will come when we again obtain the ‘equilibrium’ condition where $E_p = E_{applied}$. But now the current I_p under loaded condition is higher than before.

9. The increase in I_p causes additional electrical energy to be drawn out of the power supply.

10. The end result is that additional electrical energy is drawn out of the power supply connected to the primary coil at the very moment that a current flows through the resistive load. This additional electrical energy is dissipated by the load as heat.

11. The transformer helps to transfer electrical energy from the power supply to the load.

22.6.6 Energy Transfer by a Simple Loaded Ideal Iron-core Transformer

1. Suppose a resistive load of resistance R is connected across the secondary coil, as shown in Figure 22.71. Now the secondary coil is a closed circuit. Since an emf exists across the coil, a current I_s is produced, which flows through the resistor. Heat is produced and so electrical power is dissipated by the resistor.

2. The transformer helps to transfer electrical power from the power supply connected to the primary coil (input) to the resistive load (output).

3. If the transformer system is ideal, there would be no loss of energy by the coils and transformer to the surroundings. The load would completely receive the electrical power supplied by the power source. In other words,

input power = output power (no losses)

or,

$I_p E_p = I_s E_s$

22.6.7 Energy Losses to Surroundings

In practice, a transformer is not ideal. There are several ways through which some of the input electrical energy is dissipated by the transformer system and lost to the surroundings. Some of the ways are as follows:

(a) Loss due to resistance
The primary and secondary coils have resistance. Hence, heat is produced when current flows through each coil. The heat produced is lost to the surroundings.

Figure 22.72 The figure shows a laminated iron core made of thin insulated iron plates, which reduces eddy current losses by increasing resistance.

(b) Eddy current loss
The current flowing in the primary coil is an alternating current. Hence, the strength of the magnetic field found in the iron core varies with time. This time varying field produces eddy current which flows within the iron core. Since the core has resistance, heat is produced by the current and lost to the surroundings.
To minimise energy lost due to eddy current, we use a laminated iron core. A laminated core is made of many identical thin iron plates. These plates are electrically insulated from one another so that the resistance experienced by the eddy current is very large. Such large resistance would greatly reduce the eddy current flowing in each plate (Figure 22.72).

(c) Magnetic flux leakage
Some magnetic field lines can leak out of the iron core. As a result, the magnetic flux linkage through the secondary coil will be less than that through the primary coil. This means that the electrical power received by the load will be less than that provided by the power supply.

(d) Hysteresis loss
Iron is a magnetic substance. As the strength of the magnetic field in the iron core varies with time, the core undergoes continuously cycles of magnetisation and demagnetisation. Energy is lost as the core undergoes each cycle, which is known as the hysteresis cycle. The energy lost is converted to heat.

EXAMPLE 22.40

The primary coil of an ideal iron-core transformer has 1 200 turns and the secondary coil has 60 turns. The primary coil is connected to an alternating voltage supply of 240 V. A 3.0 ohm resistor is connected to the secondary coil. Determine
(a) the voltage across the secondary coil
(b) the current in the secondary coil.

Answer

(a) Since the transformer is ideal, we have

$\frac{E_p}{E_s}= \frac{N_p}{N_s}$

$\frac{240}{E_s} = \frac{1200}{60}$

$E_s = 12 \ V$

(b)

$E_s = I_s R$

$12 = I_s (3)$

$I_s = 4.0 \ A$

EXAMPLE 22.41

The primary and secondary coils of an ideal transformer have 100 turns and 200 turns respectively. The self inductance of the primary coil is 250 mH. An alternating voltage of 20 V is connected to the primary coil.

(a) Determine the emf across the secondary coil, which is not connected to a load.

(b) At a particular instant the current flowing through the primary coil is increasing at the rate of 40 A s⁻¹. Determine
(i) the induced emf across the primary coil at that instant
(ii) the rate of change of magnetic flux through the primary coil at that instant
(iii) the mutual inductance of the coils.

Answer

(a)

$\frac{E_p}{E_s}= \frac{N_p}{N_s}$

$\frac{20}{E_s} = \frac{100}{200}$

$E_s = 40 \ V$

(b) (i) Since the transformer is unloaded, only self induction occurs in the primary coil. The self-induced emf across the primary coil is given by

$E_p = −L_p \left(\frac{dI_p}{dt}\right)$

$= −(0.250)(+40)$

$= −10 \ V$

(ii)

$E_p = N_p \left(\frac{dΦ_p}{dt}\right)$

$10 = 100 \left(\frac{dΦ_p}{dt}\right)$

$\frac{dΦ_p}{dt} = 0.10 \ Wb \ s^{-1}$

(iii) Mutual induction occurs in the secondary coil. We have

$|E_s| = M \left(\frac{dI_p}{dt}\right)$

At the instant when the current flowing through the primary coil is increasing at the rate of 40 A s⁻¹ we have

$E_s = \left(\frac{N_s}{N_p}\right)E_p$

$= \left(\frac{200}{100}\right)(10)$

$= 20 \ V$

Hence,

$20 = M (40)$

$M = 0.50 \ H$

⚡ Applications of Transformer

🔌 Power Transmission

Transformers are used in power stations to step up voltage for long-distance transmission and step down voltage before it reaches homes, reducing energy loss.

📱 Phone Chargers

Chargers use transformers to reduce high AC voltage (220V) into low voltage suitable for electronic devices like smartphones and tablets.

💻 Electronic Devices

TVs, laptops, and computers use transformers to convert electrical voltage into levels required by internal circuits.

🏥 Medical Equipment

Transformers provide isolation and stable voltage supply in sensitive medical equipment for safety and accuracy.

🏭 Industrial Machines

Factories use transformers to supply appropriate voltage levels for heavy machinery and industrial operations.

🔊 Audio Systems

Transformers are used in amplifiers and speakers to match impedance and improve sound quality.

🚆 Electric Trains

Electric trains use transformers to convert high voltage from power lines into usable levels for operation.

🔋 Power Adapters

Adapters convert voltage levels for devices like routers, gaming consoles, and other household electronics.

Conclusion: Understanding the Importance of Transformers

A transformer is a crucial electrical device used to transfer electrical energy efficiently between circuits using electromagnetic induction. It is widely applied in modern electrical systems to ensure safe and effective power distribution.

• Transformers can increase (step-up) or decrease (step-down) voltage depending on the turns ratio.
• Energy transfer occurs through magnetic flux without direct electrical connection.
• In an ideal transformer, input power equals output power (no energy loss).
• In real applications, energy losses occur due to resistance, eddy currents, hysteresis, and flux leakage.
• Transformers are essential in power transmission, electronics, and daily life applications.

In conclusion, transformers play a vital role in delivering electrical energy efficiently and safely, making them indispensable in both industrial systems and everyday technology.

Frequently Asked Questions (FAQ) - Transformer

1. What is a transformer?

A transformer is an electrical device used to transfer electrical energy between circuits using electromagnetic induction.

2. How does a transformer work?

It works based on mutual induction where a changing current in the primary coil produces a magnetic flux that induces voltage in the secondary coil.

3. What is a step-up transformer?

A step-up transformer increases voltage when the number of turns in the secondary coil is greater than in the primary coil.

4. What is a step-down transformer?

A step-down transformer decreases voltage when the number of turns in the secondary coil is less than in the primary coil.

5. What is the turns ratio?

The turns ratio is the ratio of the number of turns in the primary coil to the secondary coil, which determines voltage transformation.

6. What is an ideal transformer?

An ideal transformer has no energy loss, meaning input power equals output power.

7. What causes energy loss in transformers?

Energy losses occur due to resistance, eddy currents, hysteresis, and magnetic flux leakage.

8. Why is a laminated core used?

A laminated core reduces eddy current losses by increasing resistance within the core.

9. Where are transformers used in daily life?

Transformers are used in power transmission, phone chargers, electronic devices, medical equipment, and industrial machines.

10. Why are transformers important?

Transformers are important because they allow efficient transmission and safe use of electrical energy.

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