PhysicsCBSE Class 12 · Alternating Current

Transformer Class 12: Principle, Working and Energy Losses

Q: Does a step-up transformer increase power?

No. It raises voltage but lowers current in the same ratio, so the power out is at most equal to the power in. Real transformers lose a little energy as heat.

Q: Why is soft iron used for the core?

Soft iron has high permeability, so it guides the flux from primary to secondary with little leakage, and it is easy to magnetise and demagnetise, so little energy is wasted each cycle.

Q: What are the main energy losses in a transformer?

Flux leakage, heating of the windings (I²R), currents induced in the core, and the energy used in repeatedly magnetising the core. They are reduced by good design, thick wires, laminated cores and soft iron.

Q: Why does a transformer hum?

The alternating magnetic field makes the core laminations vibrate slightly at twice the supply frequency; this vibration is heard as a hum and wastes a little energy.

Q: What happens to the primary current when the secondary load increases?

The larger secondary current reduces the net flux slightly, so the back emf in the primary falls and more primary current flows. Power drawn from the mains rises to match the power delivered.

Updated 15 October 2026 · Fact-checked

A transformer changes an alternating voltage using mutual induction: a changing current in the primary coil produces a changing flux in a shared iron core, inducing an emf in the secondary. For an ideal transformer Vs/Vp = Ns/Np = Ip/Is. Step-up transformers raise voltage and lower current; real ones lose some energy as heat.

Changing flux in the shared iron core links both coils; more secondary turns give higher voltage but lower current (Vs/Vp = Ns/Np).

Illustration drawn for QuestionBuddy.

Understand Transformer

Electricity is generated at a few thousand volts, sent across the country at hundreds of thousands of volts and used at home at 230 V. The device that changes the voltage at each stage, with very little loss, is the transformer.

Principle. A transformer works on mutual induction. An alternating current in one coil sets up a changing magnetic flux; when this flux passes through a second coil, it induces an alternating emf there. A transformer therefore needs AC (or a changing current); a steady DC gives constant flux and no induced emf.

Construction. Two coils, the primary (connected to the input AC) and the secondary (connected to the load), are wound on a closed core of laminated soft iron. The iron has high permeability, so it guides almost all the flux of the primary through the secondary. The coils may be wound one over the other, or on separate limbs of the core.

Working (ideal transformer). Assume no flux leaks out of the core, the coils have negligible resistance and there are no core losses. The same flux Φ passes through every turn of both coils. Then:

Primary: the applied voltage balances the back emf, $v_{p} = N_{p} \frac{d Φ}{d t}$ (with negligible primary resistance).
Secondary: $ε_{s} = N_{s} \frac{d Φ}{d t}$ , and with a small load current $v_{s} \approx ε_{s}$ .
Dividing: $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ , the turns ratio.
Power in = power out: $V_{p} I_{p} = V_{s} I_{s}$ , so $\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$ .

A step-up transformer ( $N_{s} > N_{p}$ ) raises voltage and lowers current; a step-down transformer ( $N_{s} < N_{p}$ ) does the opposite. Voltage is gained at the cost of current, so energy is conserved.

Energy losses in real transformers.

Flux leakage: not all primary flux passes through the secondary. Reduced by winding the coils over each other on a well-designed core.
Resistance of the windings: current heats the copper coils ( $I^{2} R$ ). Reduced by using thick wire, especially for the high-current coil.
Currents in the core (eddy currents): the changing flux induces currents in the iron itself, which heat it. Reduced by building the core from thin, insulated laminations.
Magnetisation loss: energy is used up in repeatedly magnetising the core one way and then the other. Reduced by using a soft iron core that magnetises easily.
Large transformers also hum, losing a little energy as sound.

Well-made transformers have efficiencies above 90%: $η = \frac{P _{o u t}}{P _{in}} \times 100%$ .

Why transmit at high voltage? For a given power P = VI, raising V lowers I. The heat lost in the line, $I^{2} R$ , then falls as $\frac{1}{V ^{2}}$ . Raising the voltage 20 times cuts the line loss 400 times. So power is stepped up at the generating station, transmitted at high voltage, and stepped down in stages near the consumer.

Transformer: formulas and key facts

Turns ratio (ideal): $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$; N_s > N_p: step-up; N_s < N_p: step-down.
Current ratio (ideal): $\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$; From V_pI_p = V_sI_s.
Efficiency: $η = \frac{V _{s} I _{s}}{V _{p} I _{p}} \times 100%$; Less than 100% in practice.
Induced emfs: $ε_{p} = - N_{p} \frac{d Φ}{d t}, ε_{s} = - N_{s} \frac{d Φ}{d t}$; Same flux through every turn of both coils.
Transmission loss: $P_{l oss} = I^{2} R = (\frac{P}{V})^{2} R$; Raising V by a factor k cuts the loss by k².

How to solve Transformer questions

Use this method for transformer and power transmission numericals.

1Write the known turns and voltages; decide whether it is step-up or step-down.
2Use $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ to find the missing voltage or turns.
3For currents in an ideal transformer, use $V_{p} I_{p} = V_{s} I_{s}$ ; for a real one, $V_{s} I_{s} = η V_{p} I_{p}$ .
4For efficiency, compare output and input power; the difference is the power lost.
5For transmission, find the line current $I = \frac{P}{V}$ , then the loss $I^{2} R$ and the voltage drop IR.
6Check that a step-up transformer lowers the current (power cannot be created).

Transformer 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

Suppose a doorbell runs from a small transformer that steps 230 V mains down to 9 V. Its secondary has 45 turns and the bell draws 0.5 A when it rings. Treat the transformer as ideal.

(a) How many turns does the primary have? (b) Find the power delivered to the bell. (c) Find the current drawn from the mains.

Try it first, then show the solutionSolution

$N_{p} = N_{s} \frac{V _{p}}{V _{s}} = 45 \times \frac{230}{9} = 1150$ turns.
$P = V_{s} I_{s} = 9 \times 0.5 = 4.5$ W.
$I_{p} = \frac{P}{V _{p}} = \frac{4.5}{230} \approx 0.020$ A.

Answer: (a) 1150 turns; (b) 4.5 W; (c) about 0.020 A.

Memory hook: Fewer turns, fewer volts, more amps.

High-voltage transmission

Imagine a village is supplied with 100 kW through a line whose two wires have a total resistance of 4 Ω. Compare sending the power at 2 kV and at 20 kV.

(a) Find the line current in each case. (b) Find the power lost in the line in each case, as a percentage of 100 kW. (c) What equipment makes the high-voltage option possible?

Try it first, then show the solutionSolution

At 2 kV: $I = \frac{1 0 ^{5}}{2000} = 50$ A. At 20 kV: $I = 5$ A.
Loss at 2 kV: $5 0^{2} \times 4 = 10000$ W (10%). At 20 kV: $5^{2} \times 4 = 100$ W (0.1%).
A step-up transformer at the supply end and step-down transformers near the village.

Answer: (a) 50 A and 5 A; (b) 10 kW (10%) and 100 W (0.1%); (c) step-up and step-down transformers.

Memory hook: Ten times the voltage, a hundred times less loss.

Tricks to remember Transformer

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

Volts follow turns, amps go the other way
V ∝ N, I ∝ 1/N.
Use it when: Turns-ratio numericals.
Up the volts, down the amps
Step-up raises V and lowers I; the product VI stays (ideally) the same.
Use it when: Energy-conservation reasoning.
FRED for losses
Flux leakage, Resistance of windings, Eddy currents in the core, Demagnetising-remagnetising of the core.
Use it when: Listing the losses and their remedies.
AC only
No change in current, no change in flux, no induced emf: a transformer does nothing with steady DC.
Use it when: Why transformers need AC.

Transformer: solved examples

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

Example 1: Turns ratio and currents · 2 marks · Very short answerBoard level

An ideal transformer has 2000 turns in the primary and 100 turns in the secondary. The primary is connected to 220 V AC. Find (a) the secondary voltage, and (b) the primary current when the secondary delivers 4 A. Is it step-up or step-down?

Try it first, then show the solutionSolution

(a) $V_{s} = 220 \times \frac{100}{2000} = 11$ V: a step-down transformer. (1)
(b) $I_{p} = I_{s} \frac{N _{s}}{N _{p}} = 4 \times \frac{100}{2000} = 0.2$ A (check: 220 × 0.2 = 11 × 4 = 44 W). (1)

Answer: (a) 11 V, step-down; (b) 0.2 A.

Example 2: Principle, working and losses · 5 marks · Long answerBoard level

(a) State the principle of a transformer and describe its construction. (2)
(b) Show that for an ideal transformer $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ and $\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$ . (2)
(c) Name two sources of energy loss and how each is reduced. (1)

Try it first, then show the solutionSolution

(a) Principle: mutual induction; an alternating current in the primary produces a changing flux that induces an emf in the secondary. (1)
Construction: primary and secondary coils wound on a closed laminated soft-iron core that links the flux of one coil with the other. (1)
(b) Same flux in every turn: $ε_{p} = - N_{p} \frac{d Φ}{d t}$ and $ε_{s} = - N_{s} \frac{d Φ}{d t}$ . For an ideal transformer $v_{p} = ε_{p}$ in size and $v_{s} = ε_{s}$ , so $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ . (1)
With no losses, $V_{p} I_{p} = V_{s} I_{s}$ , so $\frac{I _{s}}{I _{p}} = \frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}$ . (1)
(c) Any two: copper (winding) resistance, reduced by thick wires; currents in the core, reduced by laminating it; flux leakage, reduced by winding coils over each other; magnetisation loss, reduced by using soft iron. (1)

Answer: Mutual induction; V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s for an ideal transformer; losses include winding resistance and core currents, reduced by thick wire and lamination.

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

A step-up transformer takes 2 A from a 220 V supply and delivers 0.36 A at 1100 V. Find (a) the input and output power, (b) the efficiency, and (c) the power lost.

Try it first, then show the solutionSolution

(a) $P_{in} = 220 \times 2 = 440$ W; $P_{o u t} = 1100 \times 0.36 = 396$ W. (1)
(b) $η = \frac{396}{440} \times 100% = 90%$ . (1)
(c) Loss = 44 W, mostly as heat in the windings and core. (1)

Answer: (a) 440 W and 396 W; (b) 90%; (c) 44 W.

Example 4: Transmission at two voltages · 3 marks · Short answerHarder board level

A power of 1 MW is sent through a line of total resistance 10 Ω. Find the line current and the power lost if it is sent at (a) 11 kV and (b) 220 kV. By what factor is the loss reduced?

Try it first, then show the solutionSolution

(a) $I = \frac{1 0 ^{6}}{11 \times 1 0 ^{3}} \approx 90.9$ A; loss $= I^{2} R \approx 8.26 \times 1 0^{4}$ W. (1)
(b) $I = \frac{1 0 ^{6}}{2.2 \times 1 0 ^{5}} \approx 4.55$ A; loss $\approx 207$ W. (1)
The voltage is 20 times higher, so the loss is $2 0^{2} = 400$ times smaller. (1)

Answer: (a) about 90.9 A and 82.6 kW; (b) about 4.55 A and 207 W; 400 times less.

Example 5: A transformer on DC · 5 marks · Long answerHOTS / competency

The primary of a transformer has a resistance of 2 Ω and is designed for 220 V AC, on which it draws only a small current.
(a) Explain why the primary current on AC is small. (2)
(b) What happens if the primary is connected to a 220 V DC supply? Calculate the current and the heating rate. (2)
(c) What is the secondary output in that case? (1)

Try it first, then show the solutionSolution

(a) On AC the changing flux induces a back emf in the primary that nearly balances the applied voltage. (1)
So only a small current is needed to keep the flux changing; the primary's large inductive reactance, not its 2 Ω resistance, limits the current. (1)
(b) On DC there is no changing flux and no back emf; only the resistance limits the current: $I = \frac{220}{2} = 110$ A. (1)
Heating rate $= I^{2} R = 11 0^{2} \times 2 = 24200$ W: the winding would overheat and burn out. (1)
(c) Zero (apart from a brief pulse at switch-on), since a steady flux induces no emf. (1)

Answer: (a) the back emf limits the AC current; (b) 110 A and 24.2 kW of heating, burning the coil; (c) no output.

Example 6: A phone charger transformer · 4 marks · Case-basedHarder board level

Read the passage and answer.

A charger uses a transformer to step 230 V AC down to 5 V AC (before converting it to DC). Its primary has 4600 turns and it supplies 2 A on the secondary side.
(a) Find the number of secondary turns. (1)
(b) Find the primary current if the transformer is ideal. (1)
(c) Find the primary current if the efficiency is 85%. (1)
(d) Which coil should be wound with thicker wire, and why? (1)

Try it first, then show the solutionSolution

(a) $N_{s} = 4600 \times \frac{5}{230} = 100$ turns. (1)
(b) $I_{p} = \frac{5 \times 2}{230} \approx 0.0435$ A. (1)
(c) $I_{p} = \frac{10}{0.85 \times 230} \approx 0.0512$ A. (1)
(d) The secondary, which carries the larger current (2 A); thicker wire has less resistance and so less $I^{2} R$ heating. (1)

Answer: (a) 100 turns; (b) about 0.0435 A; (c) about 0.0512 A; (d) the secondary, as it carries more current.

Where students lose marks in Transformer

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

Mistake: Inverting the current ratio.
Assuming current follows turns like voltage.
Wrong working
I_s/I_p = N_s/N_p
Right working
I_s/I_p = N_p/N_s
Do this: Derive from V_pI_p = V_sI_s each time.
Usually costs: 1 mark.
Mistake: Saying a step-up transformer increases power.
Confusing voltage with power.
Do this: Power out ≤ power in; voltage rises only as current falls.
Usually costs: 1 mark in reasoning.
Mistake: Listing losses without remedies.
Questions usually ask 'how are they reduced?'
Do this: Pair each loss with its remedy in one line.
Usually costs: ½ mark per loss.
Mistake: Computing transmission loss as V²/R with the supply voltage.
Using the wrong voltage in the formula.
Wrong working
Loss = (11 000)²/10
Right working
Loss = I²R with I = P/V = 90.9 A
Do this: Use the line current, not the supply voltage, to find line loss.
Usually costs: Most of the numerical.
Mistake: Missing the diagram in the 5-mark answer.
Saving time.
Do this: Draw the core with both coils, label primary, secondary, input and output.
Usually costs: 1 mark.

Exam tips

Start long answers with the principle (mutual induction) and a labelled diagram.
When deriving the ratios, state the ideal-transformer assumptions clearly.
Transformer working, losses and numericals carried 5-mark parts in 2024 (55/1/1 Q32(b) and 55/5/1 Q32(b)) and 2026 (55/1/1 Q33(b)).
Use the transmission example (loss ∝ 1/V²) whenever asked why power is sent at high voltage.
Check that power balances in ideal-transformer answers; it catches inverted ratios.

Practice questions on Transformer

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
A transformer works on the principle of
(a) self-induction (b) mutual induction (c) electrostatic induction (d) the heating effect of current
Hint
Two coils share flux.
Answer
(b) mutual induction.
Q2 · 1 mark · Very short answerBoard level
Why is the core of a transformer laminated?
Hint
Currents induced in the core itself.
Answer
The changing flux induces currents in the iron core that waste energy as heat. Thin insulated laminations break up the paths of these currents and reduce the loss.
Q3 · 2 marks · Very short answerHarder board level
An ideal transformer has 500 primary and 2500 secondary turns. The primary is connected to 120 V AC and the secondary supplies 0.5 A. Find the secondary voltage and the primary current.
Hint
Turns ratio 5.
Answer
$V_{s} = 120 \times 5 = 600$ V; $I_{p} = 0.5 \times 5 = 2.5$ A.
Q4 · 1 mark · Very short answerHarder board level
A step-down transformer converts 11 000 V to 220 V. Find the turns ratio Np : Ns.
Hint
Voltage ratio = turns ratio.
Answer
Np : Ns = 11 000 : 220 = 50 : 1.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A transformer cannot be used to step up a DC voltage.
Reason (R): A steady current produces a constant magnetic flux, which induces no emf in the secondary.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Faraday's law needs a change.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
A transformer of efficiency 80% draws 1 kW from the mains and gives an output at 200 V. Find the output current.
Hint
P_out = ηP_in.
Answer
$P_{o u t} = 0.8 \times 1000 = 800$ W, so $I_{s} = \frac{800}{200} = 4$ A.
Q7 · 2 marks · Very short answerHOTS / competency
A student says: 'A step-up transformer gives out more voltage than it takes in, so it produces energy.' Explain the error.
Hint
Compare VI on both sides.
Answer
The secondary current falls in the same ratio as the voltage rises (I_s/I_p = N_p/N_s), so V_sI_s is at most equal to V_pI_p. No energy is created; in practice some is lost as heat.

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.Voltage ratio of an ideal transformer?Show answer
V_s/V_p = N_s/N_p.
3.Current ratio of an ideal transformer?Show answer
I_s/I_p = N_p/N_s.
4.Why is the core laminated?Show answer
To reduce currents induced in the core and the heat they produce.
5.Why is power transmitted at high voltage?Show answer
Line current is smaller, so I²R loss is smaller.
6.Why can't a transformer work on DC?Show answer
Steady DC gives constant flux, so no emf is induced in the secondary.

Transformer: frequently asked questions

Does a step-up transformer increase power?

No. It raises voltage but lowers current in the same ratio, so the power out is at most equal to the power in. Real transformers lose a little energy as heat.

Why is soft iron used for the core?

Soft iron has high permeability, so it guides the flux from primary to secondary with little leakage, and it is easy to magnetise and demagnetise, so little energy is wasted each cycle.

What are the main energy losses in a transformer?

Flux leakage, heating of the windings (I²R), currents induced in the core, and the energy used in repeatedly magnetising the core. They are reduced by good design, thick wires, laminated cores and soft iron.

Why does a transformer hum?

The alternating magnetic field makes the core laminations vibrate slightly at twice the supply frequency; this vibration is heard as a hum and wastes a little energy.

What happens to the primary current when the secondary load increases?

The larger secondary current reduces the net flux slightly, so the back emf in the primary falls and more primary current flows. Power drawn from the mains rises to match the power delivered.