Transformer: Principle, Working, Formulae, Losses and Applications

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Transformer: Principle, Working, Formulae, Losses and Applications

Complete premium Physics notes with exam-ready diagrams, derivations, losses, transmission logic, solved numericals, case studies and practice questions.

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Section 1: Transformer Diagram

This diagram shows the soft iron core, laminated core indication, primary winding N1, secondary winding N2, AC input V1, output V2, currents I1 and I2, load resistance RL and magnetic flux Φ.

Section 2: Definition

A transformer is an electrical device which changes the magnitude of alternating voltage or alternating current without changing the frequency, using the principle of mutual induction.

Works only with AC: It needs changing current and changing magnetic flux.

Does not work with steady DC: steady DC produces constant flux after switching on.

Frequency unchanged: A transformer changes voltage/current magnitude, not frequency.

Section 3: Principle

A transformer works on mutual induction. When alternating current flows through the primary coil, changing magnetic flux is produced in the soft iron core. This changing flux links with the secondary coil and induces alternating emf in the secondary coil.

Changing current is required.
Changing magnetic flux is required.
Therefore AC works, but steady DC does not work.

Section 4: Uses of Transformer

Step-up transformer increases AC voltage.
Step-down transformer decreases AC voltage.
Used in long-distance power transmission.
Used in adapters and chargers.

Used in substations and local distribution.
Used in electrical appliances.
Used for isolation and voltage matching.
It can step up or step down only AC voltage, not steady DC voltage.

Section 5: Types of Transformer

          Step-up Transformer
          N2 > N1
V2 > V1
I2 < I1
Used for long-distance power transmission.

        

          Step-down Transformer
          N2 < N1
V2 < V1
I2 > I1
Used near homes and in local distribution.

        

Point	Step-up Transformer	Step-down Transformer
Turns ratio	N2 > N1	N2 < N1
Voltage	V2 > V1	V2 < V1
Current	I2 < I1	I2 > I1
Main use	Transmission at high voltage	Safe distribution to homes

Section 6: Components of Transformer

Component	Function
Primary winding	Receives AC input voltage and produces changing magnetic flux.
Secondary winding	Receives changing flux and provides induced output voltage.
Soft iron core	Provides low-reluctance path for magnetic flux.
Laminated core	Reduces eddy current loss by increasing resistance to circulating currents.
Insulation	Prevents short circuit between turns and between windings/core.
Load resistance	Receives output electrical power from secondary coil.
AC source	Supplies alternating current required for transformer action.

Section 7: Working of Transformer

AC voltage V1 is applied to the primary coil.
Alternating current I1 flows in the primary coil.
Changing magnetic flux is produced in the soft iron core.
This changing flux links with the secondary coil.
Alternating emf V2 is induced in the secondary coil.
If load is connected, current I2 flows through the load.

Section 8: Mathematical Derivation

Let primary turns = N1, secondary turns = N2, primary voltage = V1 and secondary voltage = V2.

From Faraday's law: V1 = -N1 dΦ/dt

V2 = -N2 dΦ/dt

Dividing the two equations: V2/V1 = N2/N1

or N2/N1 = V2/V1

For ideal transformer: Input power = Output power

V1 I1 = V2 I2

Therefore V2/V1 = I1/I2

Hence N2/N1 = V2/V1 = I1/I2

Voltage ratio is directly proportional to turns ratio. Current ratio is inversely proportional to turns ratio.

Section 9: Efficiency of Transformer

η = Output power / Input power η = (V2 I2)/(V1 I1) η% = (Output power / Input power) × 100

For an ideal transformer, η = 100%, so V1 I1 = V2 I2.

Example 1: Efficiency

Given: Input power = 500 W, output power = 450 W.

Formula: η% = Output/Input × 100

Solution: η% = 450/500 × 100 = 90%.

Final Answer: Efficiency = 90%.

Example 2: Output Power

Given: η = 80%, input power = 1000 W.

Formula: Output = η × Input

Solution: Output = 0.80 × 1000 = 800 W.

Final Answer: Output power = 800 W.

Section 10: Transformer Losses

Loss	Cause	Reduction Method
Copper loss	Current flowing through windings produces heat. Copper loss = I²R.	Use thick copper wires and low-resistance windings.
Hysteresis loss	Repeated magnetisation and demagnetisation of the core.	Use soft iron or high permeability material with narrow hysteresis loop.
Eddy current loss	Changing magnetic flux induces circulating currents inside the core.	Use laminated core. Laminations increase resistance to eddy current paths.
Flux leakage	Some flux produced by primary coil does not link with secondary coil.	Use closed soft iron core and proper winding design.
Heating and insulation loss	Heat weakens insulation and energy is wasted in practical operation.	Use proper cooling, insulation and rated operation.

Section 11: Why High Voltage Is Used For Long Distance Transmission

Power transmitted is P = VI. For the same power, I = P/V. If voltage is increased, current decreases. Power loss in transmission wires is P_loss = I²R. Therefore, at high voltage, current becomes small, so I²R loss becomes very small.

Step-up transformer is used at the generating station. Step-down transformer is used near the locality before supplying homes.

Section 12: Power Distribution Diagram

Section 13: Why Transformer Does Not Work On DC

A transformer requires changing magnetic flux. In steady DC, current becomes constant after switching on, magnetic flux becomes constant and dΦ/dt = 0. Therefore no emf is induced in the secondary coil.

At the instant of switching DC on or off, momentary induced emf may appear, but continuous transformer action is not possible.

Section 14: Formula Sheet

V2/V1 = N2/N1 N2/N1 = V2/V1 = I1/I2 V1 I1 = V2 I2 η = Output power / Input power η% = (V2 I2 / V1 I1) × 100 Copper loss = I²R Power loss in wires = I²R P = VI I = P/V

Section 15: Numerical Problems

Problem 1: Turns Ratio

Given: N1 = 500, N2 = 2000, V1 = 220 V.

Formula: V2/V1 = N2/N1

Solution: V2 = 220 × 2000/500 = 880 V.

Final Answer: V2 = 880 V, step-up transformer.

Common mistake: Reversing N1 and N2.

Exam tip: If N2 > N1, voltage increases.

Problem 2: Step-down Voltage

Given: V1 = 11000 V, N1 = 5000, N2 = 100.

Formula: V2 = V1 × N2/N1

Solution: V2 = 11000 × 100/5000 = 220 V.

Final Answer: V2 = 220 V.

Common mistake: Forgetting that domestic supply needs step-down operation.

Exam tip: Near homes, transformers are step-down.

Problem 3: Current Ratio

Given: V1 = 200 V, V2 = 1000 V, I1 = 5 A.

Formula: V1I1 = V2I2

Solution: I2 = 200 × 5 / 1000 = 1 A.

Final Answer: I2 = 1 A.

Common mistake: Thinking current also increases in a step-up transformer.

Exam tip: Voltage increases but current decreases for ideal step-up transformer.

Problem 4: Efficiency

Given: V1 = 220 V, I1 = 4 A, V2 = 400 V, I2 = 2 A.

Formula: η% = V2I2/V1I1 × 100

Solution: η% = 400 × 2 /(220 × 4) × 100 = 90.9%.

Final Answer: η = 90.9%.

Common mistake: Using only voltage ratio for efficiency.

Exam tip: Efficiency is power ratio, not voltage ratio.

Problem 5: Copper Loss

Given: Current = 8 A, winding resistance = 0.5 Ω.

Formula: Copper loss = I²R

Solution: Loss = 8² × 0.5 = 32 W.

Final Answer: Copper loss = 32 W.

Common mistake: Using IR instead of I²R.

Exam tip: Heating loss always depends on square of current.

Problem 6: Long-distance Transmission

Given: Power = 100 kW, voltage = 10 kV.

Formula: I = P/V

Solution: I = 100000/10000 = 10 A.

Final Answer: Transmission current = 10 A.

Common mistake: Not converting kW and kV.

Exam tip: High voltage keeps current small.

Problem 7: Loss Reduction

Given: Current is reduced to one-fourth at same wire resistance.

Formula: P_loss = I²R

Solution: New loss = (I/4)²R = I²R/16.

Final Answer: Power loss becomes 1/16 of original.

Common mistake: Saying loss becomes one-fourth.

Exam tip: Square-law reduction is very important.

Problem 8: DC Supply

Given: A transformer is connected to steady DC.

Formula: Induced emf = -N dΦ/dt

Solution: For steady DC, flux is constant, so dΦ/dt = 0. Hence no secondary emf is produced continuously.

Final Answer: Transformer does not work on steady DC.

Common mistake: Forgetting momentary switching emf.

Exam tip: Mention continuous transformer action is impossible on steady DC.

Section 16: Exam Practice Questions

CBSE Class 12

Define transformer and state its principle.
Why does a transformer not work on DC?
Derive V2/V1 = N2/N1.
Explain copper, hysteresis and eddy current losses.
Numerical: N1 = 1000, N2 = 50, V1 = 2200 V. Find V2.

NEET

MCQ: In a step-up transformer, current in secondary is greater or smaller?
Find current ratio for turns ratio 1:5.
Which material reduces hysteresis loss?
Why is core laminated?
Power loss in wires becomes what if voltage is doubled?

JEE Main

Calculate efficiency from V1, I1, V2 and I2.
Assertion-reason: Transformer changes voltage but not frequency.
Compare step-up and step-down transformer.
Find copper loss for given winding resistance.
Explain high voltage transmission mathematically.

JEE Advanced

Discuss flux leakage and non-ideal transformer behavior.
Power loss reduces by factor 100. Find current reduction factor.
Analyze transformer with efficiency 85% and load current.
Explain why current ratio is inverse of voltage ratio.
Case question based on transmission grid.

IB Physics

Explain energy conservation in an ideal transformer.
Draw labelled transformer diagram.
Describe the role of the iron core.
State one environmental benefit of high voltage transmission.
Numerical based on efficiency.

IGCSE / GCSE Physics

State whether transformer works with AC or DC.
Name the input and output coils.
Explain step-up transformer use.
Calculate secondary voltage from turns ratio.
Why are power lines at high voltage?

ICSE / ISC Physics

State Faraday's law relation used in transformer.
Why is soft iron used in the core?
What is eddy current loss?
Define efficiency.
Find output power for given efficiency.

A-Level / British Curriculum

Derive transformer equation from Faraday's law.
Evaluate losses in a loaded transformer.
Explain ideal versus practical transformer.
Calculate transmission loss at two different voltages.
Explain magnetic flux linkage.

Section 17: Case Study Questions

Case Study 1: Transformer in Power Transmission

Passage: A power station uses a step-up transformer before sending electrical power through long transmission lines. Near the city, a step-down transformer supplies safer voltage to homes.

Why is voltage increased before transmission?
Which loss is reduced?
Which transformer is used near homes?
Does frequency change?

Answers: To reduce current; I²R loss; step-down transformer; no, frequency remains unchanged.

Explanation: For same power, I = P/V. High voltage means small current and small I²R loss.

Case Study 2: Step-up and Step-down Transformer

Passage: A transformer has more turns in secondary than primary in one setup, and fewer turns in secondary in another setup.

Which setup is step-up?
What happens to secondary current in step-up?
Which setup is used in chargers?
Write voltage ratio formula.

Answers: N2 > N1; current decreases; step-down; V2/V1 = N2/N1.

Explanation: Voltage follows turns ratio, while current follows inverse ratio.

Case Study 3: Why Transformer Works Only On AC

Passage: A student connects a transformer to a battery and observes no continuous output from the secondary coil.

What type of current does a battery provide?
Why is no continuous emf induced?
What is dΦ/dt for steady DC?
Can a momentary emf appear?

Answers: DC; flux becomes constant; zero; yes, during switching on/off.

Explanation: Transformer action requires changing magnetic flux, which steady DC cannot provide continuously.

Case Study 4: Transformer Losses

Passage: A practical transformer becomes warm during operation and its output power is less than input power.

Name the winding heat loss.
How is copper loss reduced?
Which loss is reduced by laminations?
Which core material reduces hysteresis loss?

Answers: Copper loss; thick low-resistance copper wires; eddy current loss; soft iron.

Explanation: Practical transformers lose energy as heat, magnetic hysteresis, eddy currents and flux leakage.

Case Study 5: Efficiency of Transformer

Passage: A transformer takes 1000 W input power and gives 920 W output power to a load.

Find efficiency.
Is it ideal?
What is lost power?
Give one cause of loss.

Answers: 92%; no; 80 W; copper loss, hysteresis loss, eddy current loss or flux leakage.

Explanation: Efficiency = output/input × 100 = 920/1000 × 100 = 92%.

Section 18: Kumar Sir Exam Tips

Transformer works on mutual induction.
Transformer works only on AC, not steady DC.
Frequency does not change in a transformer.
Step-up transformer increases voltage but decreases current.
Step-down transformer decreases voltage but increases current.
For ideal transformer: N2/N1 = V2/V1 = I1/I2.
High voltage transmission reduces I²R loss.
Laminated core reduces eddy current loss.
Soft iron core reduces hysteresis loss.
Copper loss is reduced by using low-resistance thick copper wires.

If Transformer, AC circuits or Electromagnetic Induction is not clear, contact Kumar Sir for one-to-one online Physics classes.

Contact: +91-9958461445

Email: kumarsirphysics@gmail.com

Website: https://kumarphysicsclasses.com