PhysicsISC Class 12 · Electromagnetic Induction

The Transformer: Principle, Turns Ratio, Losses and Power Transmission

Updated 15 October 2026 · Fact-checked

A transformer changes an alternating voltage using mutual induction between two coils on a common soft iron core. For an ideal transformer $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$, because input power equals output power. Step-up transformers raise voltage and lower current; step-down transformers do the reverse.

Changing flux in the core links both coils; more secondary turns raise voltage but lower current, since Vp·Ip = Vs·Is.

Illustration drawn for QuestionBuddy.

Understand The Transformer and Power Transmission

Principle. A transformer works on mutual induction. An alternating current in the primary coil produces an alternating magnetic flux in the core; this changing flux links the secondary coil and induces an alternating emf in it. A steady d.c. current gives a constant flux and so no induced emf: a transformer works only on a.c.

Construction. Two insulated copper coils, the primary ( $N_{p}$ turns, connected to the a.c. supply) and the secondary ( $N_{s}$ turns, connected to the load), wound on a closed laminated soft iron core (thin sheets insulated from each other). Often the windings are placed one over the other on the same limb to reduce flux leakage.

Theory (ideal transformer). Assume no flux leakage, no resistance and no core losses. The same flux φ links every turn of both coils, so
$ε_{p} = - N_{p} \frac{d ϕ}{d t}, ε_{s} = - N_{s} \frac{d ϕ}{d t} \Rightarrow \frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = k,$
the turns ratio (transformation ratio). With no losses, input power equals output power: $V_{p} I_{p} = V_{s} I_{s}$ , so
$\frac{I _{s}}{I _{p}} = \frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}} .$

Step-up ( $N_{s} > N_{p}$ ): voltage increases, current decreases.
Step-down ( $N_{s} < N_{p}$ ): voltage decreases, current increases.

A transformer does not create energy: raising the voltage lowers the current in the same ratio. (The ISC 2025 examiners wanted exactly this argument, $V_{s} I_{s} = V_{p} I_{p}$ , not just "fewer turns in the secondary".)

Efficiency. $η = \frac{output power}{input power} \times 100% = \frac{V _{s} I _{s}}{V _{p} I _{p}} \times 100%$ . Large transformers reach 95–99%.

Energy losses and how they are reduced.

Loss	Cause	Remedy
Copper (I²R) loss	resistance of the windings	thick copper wire, especially for the high-current coil
Eddy current loss	currents induced in the core itself	laminated core, sheets insulated from each other
Hysteresis loss	repeated magnetisation of the core round the B-H loop	soft iron, which has a narrow loop (low coercivity)
Flux leakage	not all primary flux links the secondary	windings one over the other on a closed core
Humming (sound)	vibration of the laminations	tightly clamped core

Soft iron is chosen for the core because its high permeability gives a large flux for a small current and its narrow hysteresis loop wastes little energy each cycle (the 2024 ISC 2-mark question).

Power transmission. A power station sends power P through lines of resistance R. At voltage V the current is $I = \frac{P}{V}$ , and the power lost as heat in the lines is
$P_{l oss} = I^{2} R = \frac{P ^{2} R}{V ^{2}} .$
Raising V by a factor of 10 cuts the current by 10 and the loss by 100. So electricity is generated at a moderate voltage, stepped up to tens or hundreds of kilovolts for transmission, and stepped down in stages near towns and homes. This is possible only because a.c. can be transformed.

Uses. Power distribution, mobile and laptop chargers (step-down), welding (step-down for very large current), induction furnaces, neon sign supplies and X-ray tubes (step-up), voltage stabilisers.

The Transformer and Power Transmission: formulas and key facts

Turns ratio: $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$; Ideal transformer.
Current ratio: $\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$; From $V_{p} I_{p} = V_{s} I_{s}$ .
Efficiency: $η = \frac{V _{s} I _{s}}{V _{p} I _{p}} \times 100%$; Real transformers: < 100%.
Line current: $I = \frac{P}{V}$; Power P sent at voltage V.
Line loss: $P_{l oss} = I^{2} R = \frac{P ^{2} R}{V ^{2}}$; Falls as $\frac{1}{V ^{2}}$ .

How to solve The Transformer and Power Transmission questions

Use the turns ratio for voltages, power balance for currents, and $I^{2} R$ for line losses.

1Identify step-up or step-down from the turns or voltages.
2Voltage: $V_{s} = V_{p} \frac{N _{s}}{N _{p}}$ .
3Current (ideal): $I_{p} = I_{s} \frac{N _{s}}{N _{p}}$ or from $V_{p} I_{p} = V_{s} I_{s}$ .
4With efficiency η: $V_{p} I_{p} = \frac{V _{s} I _{s}}{η}$ .
5For transmission, find $I = \frac{P}{V}$ , then $I^{2} R$ ; compare as a percentage of P.
6State the reason or remedy in words when losses are asked about.

The Transformer and Power Transmission in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Step-down transformer in a charger

Suppose a laptop charger's transformer stage takes 230 V a.c. and delivers 19 V at 3.4 A to the rest of the circuit, and assume it is ideal.

Find the output power, the primary current and the turns ratio.

Try it first, then show the solutionSolution

$P = 19 \times 3.4 = 64.6$ W.
$I_{p} = \frac{64.6}{230} \approx 0.28$ A.
$\frac{N _{p}}{N _{s}} = \frac{230}{19} \approx 12.1$ .

Answer: 64.6 W; about 0.28 A; primary has about 12 times as many turns.

Memory hook: Volts down, amps up.

A welding transformer

Imagine a welding transformer that steps 230 V down to 23 V while the primary draws 10 A, and assume it is ideal.

What current flows in the secondary? Why is that useful for welding?

Try it first, then show the solutionSolution

$I_{s} = I_{p} \frac{V _{p}}{V _{s}} = 10 \times 10 = 100$ A.
The large current produces intense $I^{2} R$ heating at the joint, melting the metal.

Answer: 100 A; large current gives the heat needed to melt the metal.

Memory hook: Step-down = current step-up.

Tricks to remember The Transformer and Power Transmission

Short memory aids. Say each one aloud once, then use it on a question the same day.

Volts follow turns, amps oppose
$V \propto N$ ; $I \propto \frac{1}{N}$ .
Use it when: Ratios.
Power in = power out
For an ideal transformer, VI is the same on both sides.
Use it when: Explaining current changes.
Ten times volts, hundredth of loss
$P_{l oss} \propto \frac{1}{V ^{2}}$ .
Use it when: Transmission.
C-E-H-F
Copper, Eddy, Hysteresis, Flux leakage: thick wire, laminations, soft iron, close windings.
Use it when: Losses and remedies.

The Transformer and Power Transmission: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Why a.c. only · 1 mark · MCQBoard level

A transformer does not work on steady d.c. because:
(a) d.c. heats the core
(b) d.c. produces no changing flux, so no emf is induced in the secondary
(c) d.c. cannot flow in copper
(d) the core saturates instantly

Try it first, then show the solutionSolution

Mutual induction needs a changing flux.

Answer: (b).

Example 2: Voltage and current · 2 marks · Short answerBoard level

A transformer has 1200 turns in the primary and 60 in the secondary. The primary is connected to 240 V a.c. and the secondary supplies 3 A. Assuming it is ideal, find the secondary voltage and primary current. Is it step-up or step-down?

Try it first, then show the solutionSolution

$V_{s} = 240 \times \frac{60}{1200} = 12$ V: step-down. (1)
$I_{p} = 3 \times \frac{60}{1200} = 0.15$ A. (1)

Answer: 12 V; 0.15 A; step-down.

Example 3: Efficiency of a real transformer · 2 marks · Short answerHarder board level

A transformer draws 0.5 A from a 230 V supply and delivers 4.5 A at 22 V. Find its efficiency and the power lost.

Try it first, then show the solutionSolution

$P_{in} = 230 \times 0.5 = 115$ W; $P_{o u t} = 22 \times 4.5 = 99$ W. (1)
$η = \frac{99}{115} \approx 86%$ ; loss = 16 W. (1)

Answer: About 86%; 16 W.

Example 4: Losses and remedies · 4 marks · Long answerHarder board level

Name three causes of energy loss in a transformer and state how each is reduced. Why is soft iron, rather than steel, used for the core?

Try it first, then show the solutionSolution

Copper loss ( $I^{2} R$ in the windings): use thick copper wire. (1)
Eddy current loss in the core: laminate the core with insulated sheets. (1)
Hysteresis loss in the core: use soft iron with a narrow B-H loop. (Flux leakage: wind coils over each other.) (1)
Soft iron has high permeability and low coercivity, so it is magnetised easily and wastes little energy per cycle; steel's wide loop would heat the core. (1)

Answer: Copper, eddy and hysteresis losses (and flux leakage), reduced as described; soft iron for low hysteresis loss and high permeability.

Example 5: Why transmit at high voltage · 3 marks · Short answerHOTS / competency

A small generator must send 50 kW to a workshop through lines of total resistance 2 Ω. Find the power lost in the lines if it is sent at (a) 500 V, (b) 5000 V. Comment.

Try it first, then show the solutionSolution

(a) $I = \frac{50 000}{500} = 100$ A; loss $= 10 0^{2} \times 2 = 20$ kW (40% of the power). (1)
(b) $I = 10$ A; loss $= 100 \times 2 = 200$ W (0.4%). (1)
Ten times the voltage gives a hundredth of the loss; a step-up transformer at the generator and a step-down one at the workshop make this possible. (1)

Answer: 20 kW at 500 V; 200 W at 5000 V.

Example 6: A neon-sign transformer · 4 marks · Case-basedHOTS / competency

Suppose a neon sign needs 6900 V and is supplied from 230 V mains through a transformer with 400 primary turns. The tube draws 20 mA.
(i) Find the number of secondary turns. (1 mark)
(ii) Find the primary current if the transformer is ideal. (1 mark)
(iii) Find the primary current if its efficiency is 90%. (1 mark)
(iv) Why could this sign not be run from a 230 V battery bank through the same transformer? (1 mark)

Try it first, then show the solutionSolution

(i) $N_{s} = 400 \times \frac{6900}{230} = 12000$ .
(ii) $I_{p} = \frac{6900 \times 0.02}{230} = 0.6$ A.
(iii) $I_{p} = \frac{138}{0.9 \times 230} \approx 0.67$ A.
(iv) A battery gives d.c.; the flux would not change, so no emf would be induced in the secondary.

Answer: (i) 12 000; (ii) 0.6 A; (iii) ≈ 0.67 A; (iv) d.c. cannot be transformed.

Where students lose marks in The Transformer and Power Transmission

The slips examiners see most often, and what to write instead.

Mistake: Naming the wrong device
In ISC 2024 some candidates wrote "rectifier" or "step transformer" for a step-down transformer.
Do this: Name it fully: step-up or step-down transformer.
Usually costs: 1 mark.
Mistake: Conflicting reasons for a soft iron core
Also 2024: "greater coercivity" or only "heat loss" was written.
Wrong working
Soft iron has high coercivity.
Right working
Soft iron has low coercivity and high permeability, so little hysteresis loss.
Do this: Link the property to the loss it reduces.
Usually costs: 1 mark.
Mistake: Incomplete current explanation
In 2025 candidates only said "fewer turns in the secondary".
Do this: Use power balance: $V_{s} I_{s} = V_{p} I_{p}$ .
Usually costs: 1 mark.
Mistake: Using V²/R for line loss
V is the transmission voltage, not the p.d. across the line.
Do this: Use $I = \frac{P}{V}$ , then $I^{2} R$ .
Usually costs: 1 to 2 marks.

Exam tips

Draw a labelled diagram (core, primary, secondary, a.c. source, load) when asked about construction.
Say "ideal transformer" before using $V_{p} I_{p} = V_{s} I_{s}$ .
Give losses in pairs: cause and remedy.
Express transmission losses as a percentage to make the comparison clear.

Practice questions on The Transformer and Power Transmission

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The working principle of a transformer is:
(a) self-induction
(b) mutual induction
(c) the magnetic effect of current alone
(d) electrostatic induction
Hint
Two coils.
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
An ideal transformer steps 240 V down to 12 V. Explain why the secondary current is larger than the primary current, and find it when the primary current is 0.1 A. (A variation on ISC 2025.)
Hint
Power in = power out.
Answer
Output power equals input power, $V_{s} I_{s} = V_{p} I_{p}$ ; since $V_{s} < V_{p}$ , $I_{s} > I_{p}$ . $I_{s} = \frac{240 \times 0.1}{12} = 2$ A.
Q3 · 2 marks · Short answerHarder board level
Why is soft iron, rather than steel, used for the core of a transformer? Give two properties and the loss or leakage each one reduces. (A variation on ISC 2024.)
Hint
Think of permeability and the hysteresis loop.
Answer
High permeability: the core carries a large flux for a small magnetising current and links nearly all of it to the secondary (little flux leakage). Low coercivity with a narrow hysteresis loop: little energy is lost as heat in each cycle (small hysteresis loss). Steel has a wide loop and would waste far more energy.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a step-up transformer the secondary current is smaller than the primary current.
Reason (R): For an ideal transformer, $V_{p} I_{p} = V_{s} I_{s}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Higher V, lower I.
Answer
(a).
Q5 · 1 mark · Very short answerHarder board level
A transformer must give 9 V from 230 V mains. If its secondary has 45 turns, how many turns must the primary have?
Hint
$N_{p} = N_{s} \frac{V _{p}}{V _{s}}$ .
Answer
1150 turns.
Q6 · 3 marks · Short answerHOTS / competency
A transformer of efficiency 90% steps 230 V down to 115 V and delivers 2 kW. Find the secondary current, the primary current and the power wasted.
Hint
$P_{in} = \frac{P _{o u t}}{η}$ .
Answer
$I_{s} \approx 17.4$ A; $P_{in} \approx 2222$ W, $I_{p} \approx 9.66$ A; loss ≈ 222 W.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a town needs 1 MW from a power station through transmission lines of total resistance 10 Ω.
(i) Find the line current if power is sent at 11 kV. (1 mark)
(ii) Find the power lost and its percentage. (1 mark)
(iii) Repeat (ii) for 66 kV. (1 mark)
(iv) By what factor does the loss fall, and how is the voltage changed at each end? (1 mark)
Hint
$P_{l oss} = \frac{P ^{2} R}{V ^{2}}$ .
Answer
(i) ≈ 90.9 A. (ii) ≈ 82.6 kW, about 8.3%. (iii) ≈ 2.30 kW, about 0.23%. (iv) 36 times smaller ( $6^{2}$ ); step-up transformer at the station, step-down transformers near the town.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Principle of a transformer?Show answer
Mutual induction.
2.Turns ratio relation?Show answer
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ .
3.Why is the core laminated?Show answer
To reduce eddy current loss.
4.Np = 500, Ns = 2500, Vp = 220 V: Vs?Show answer
1100 V.
5.Why is power transmitted at high voltage?Show answer
The current is smaller, so $I^{2} R$ loss in the lines is much smaller.

The Transformer and Power Transmission: frequently asked questions

Does a step-up transformer increase power?

No. It raises voltage but lowers current; ideally the power stays the same, and in practice a little is lost.

Why does a transformer hum?

The alternating magnetisation makes the laminations vibrate slightly at twice the supply frequency (magnetostriction), producing a humming sound.

Why are the coils wound one over the other?

So that nearly all the flux of the primary also passes through the secondary, reducing flux leakage.

Where is the transformer in the ISC 2027 syllabus?

It is part of Electromagnetic Induction: principle, working and uses, step-up and step-down, efficiency, power transmission, energy losses and their reduction.

How was it tested recently?

In 2024 a 2-mark question asked for the type of transformer and why its core is soft iron; in 2025 a 1-mark item asked why the secondary current of a step-down transformer is larger.