Transformers

What is a transformer?

In plain words: the core carries the same changing flux through every turn of both coils, so each turn gets the same emf. More turns on the secondary means more emf in total: the voltage is stepped up. Energy is not created, so the current is stepped down in the same ratio.

Use the ratios carefully: V and I denote rms magnitudes. For an ideal transformer V_ₚI_ₚ = V_ₛI_ₛ equates apparent powers (VA). Real average power is VI cos φ; the ideal transformer preserves load power factor and real power. Writing power = VI in watts requires unity power factor. Loaded primary current balances the secondary ampere-turns, in addition to the small magnetising current of a real core.

Step-up and step-down transformers

• The two coils are wound on a soft-iron core, either one on top of the other or on separate limbs of the core (NCERT page 194, Figure 7.16). Soft iron is used because it is easily magnetised, so it carries nearly all the flux from one coil to the other, and it has a low hysteresis loss (reasoned from NCERT pages 194–196).
• In an ideal transformer both windings have zero resistance, all core flux links both windings, core losses vanish and magnetising current is negligible. With φ the flux in each turn, the emfs are (NCERT Eqs. (7.31)–(7.32), page 194) ε_s = −N_s dφ/dt in the secondary, by mutual induction, and the back emf ε_p = −N_p dφ/dt in the primary, by self-induction. Both follow from Faraday’s law.
• The primary back emf opposes the applied voltage: with a consistent loop direction, v_p + ε_p = 0. Their magnitudes are equal. With negligible winding drops, the secondary terminal-voltage magnitude equals its induced-emf magnitude. Choosing corresponding winding polarities gives the following voltage ratio; reversing one winding reverses its phase. Dividing (NCERT Eq. (7.33), page 195):

• NCERT page 195 motivates an approximation with small currents and negligible resistance. More precisely, winding voltage drops and leakage must be negligible; ideal-transformer ratios do not require the load current itself to be small.
• If no energy is lost, the power in equals the power out, i_pv_p = i_sv_s (NCERT Eq. (7.34), page 195), so i_p/i_s = v_s/v_p = N_s/N_p (NCERT Eq. (7.35), page 195). A well-designed transformer can be more than 95% efficient, so this is a good approximation (NCERT page 195).
• The same ratios hold for amplitudes and rms values (NCERT Eq. (7.36), page 195):

• A transformer changes the voltage and the current but not the frequency: both sides oscillate at the source frequency (NCERT page 195).
• Step-up: N_s > N_p, so V_s > V_p and I_s < I_p. Step-down: N_s < N_p, so V_s < V_p and I_s > I_p (NCERT page 195).
• Stepping up the voltage does not break the conservation of energy: the current is reduced in the same proportion (NCERT Points to Ponder 11, page 199).
• A transformer needs changing flux. An ordinary ac transformer cannot continuously transform steady dc: after switching transients it has no secondary emf. Pulsed supplies can drive suitably designed transformers during changing flux, with core reset between pulses (reasoned from Faraday’s law).

Energy losses in a transformer

NCERT lists four departures from ideal transformer operation (NCERT pages 195–196):

• Flux leakage: not all the primary’s flux passes through the secondary, because of poor core design or air gaps in the core. Winding the two coils one over the other reduces it. Leakage flux itself stores and returns magnetic energy rather than necessarily dissipating it; it reduces coupling and causes reactive voltage drops. Additional stray-field heating can occur in real construction.
• Resistance of the windings: the wires heat up (I²R). In high-current, low-voltage windings this is kept down by using thick wire.
• Eddy currents: the alternating flux induces eddy currents in the iron core, heating it. A laminated core reduces them.
• Hysteresis: the core’s magnetisation is reversed again and again by the alternating field, and the energy spent appears as heat. It is kept small by using a core material with a low hysteresis loss.

Long-distance transmission

• Electrical energy is transmitted and distributed over long distances using transformers. The generator’s voltage is stepped up, so that the current, and with it the I²R loss in the lines, is reduced (NCERT page 196). The linked page on electrical energy and power works out with numbers how a higher voltage cuts the loss.
• Near the consumers, the voltage is stepped down at an area sub-station, then again at distributing sub-stations and on utility poles, before it reaches homes (NCERT page 196).
• NCERT inconsistency: page 196 says a supply of 240 V reaches our homes, and NCERT Points to Ponder 1 (page 198) also calls the outlet voltage 240 V, but page 180 calls the household line voltage 220 V, the value the chapter’s examples and exercises use. In a problem, use the voltage the question gives.

Solved examples

Common mistakes to avoid

Quick revision summary

• A transformer uses mutual induction: an alternating flux in a soft-iron core links a primary of N_p turns and a secondary of N_s turns.
• Ideal transformer: V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s, so V_pI_p = V_sI_s.
• Step-up: N_s > N_p, more voltage, less current. Step-down: the reverse.
• The output has the same frequency as the input.
• Nonideal effects: flux leakage reduces coupling; winding resistance, eddy currents (reduced by lamination) and hysteresis cause heating; a well-designed transformer may exceed 95% efficiency.
• Power is transmitted at high voltage and low current to cut I²R losses, then stepped down near consumers.

Questions students ask

Why is a transformer rated in kVA and not in kW?

A transformer is rated in kVA because its limit is the heating from its losses, which depend on its current and voltage, whatever the load’s power factor: winding loss grows as I^²R and core loss depends on the voltage and frequency (beyond the NCERT text: US DOE Fundamentals Handbook: Electrical Science, 1992, Module 13, where the rating is V_sI_s in kVA). Not knowing the load, the maker states the largest VI it can carry safely. The real power delivered is that times the load’s power factor: at full current a 2.0 kVA transformer gives 2.0 kW to a heater but 1.7 kW to a load of power factor 0.85.

What are copper loss and iron loss in a transformer, and which one stays constant?

Copper loss and iron (core) loss are engineering names for the heating among the nonideal effects NCERT lists: copper loss is the I^²R heating of the windings, iron loss the heating of the core by hysteresis and eddy currents (NCERT pages 195–196; beyond the NCERT text: US DOE Fundamentals Handbook: Electrical Science, 1992, Module 13). Iron loss depends on the core flux, set by the supply voltage and frequency, so it hardly changes with the load, while copper loss grows as the square of the load current. At fixed supply voltage, frequency and load power factor, with approximately constant core loss, winding resistance and negligible other losses, efficiency peaks where copper and iron losses are equal: with 80 W of iron loss and 180 W of copper loss at full load, at two-thirds of full load (reasoned).

Where does the transformer emf equation E = 4.44 fNΦ_m come from?

The transformer emf equation E = 4.44 fNΦ_m is NCERT’s ε = −N dφ/dt (Eq. (7.31), page 194) for a core flux φ = Φ_m sin ωt: the emf of N turns then has the amplitude NωΦ_m = 2πfNΦ_m, and dividing by √2 for the rms value gives about 4.44 fNΦ_m (beyond the NCERT text, reasoned from NCERT page 180). The emf per turn, 4.44 fΦ_m, is the same in both windings, so it fixes the number of turns: a flux peaking at 2.25 mWb at 50 Hz gives about 0.50 V per turn, so a 230 V winding needs about 460 turns and a 9.0 V winding 18.

What is the difference between a transformer and a generator?

A generator converts mechanical energy into electrical energy by turning a coil in a magnetic field or a magnet inside coils, while a transformer changes an ac voltage to a larger or smaller one with coils and a core that do not move: its flux varies because of electrical excitation, the primary’s alternating current setting up a changing flux that also threads the secondary (NCERT pages 171–172 and 194). Both work by Faraday’s law, but a transformer only passes on the electrical energy it is given, in ordinary sinusoidal operation gives out the frequency it takes in, while a generator’s frequency is set by how fast it turns (NCERT pages 171 and 195). An AC generator makes the voltage; transformers step it up or down.

Is the transformer in the JEE Main and NEET syllabus, or has it been deleted?

The transformer has not been deleted: both syllabuses name it in Unit 14, Electromagnetic Induction and Alternating Currents, the JEE Main syllabus ending with “wattless current, AC generator and transformer.” and the NEET (UG) 2026 syllabus with “AC generator and transformer.” In NCERT it is Section 7.8 of Chapter 7, Alternating Current, in the 2026–27 Class 12 Physics Part I (pages 194–196): the principle of mutual induction, V_s/V_p = N_s/N_p, step-up and step-down transformers, the four nonideal effects and long-distance transmission. Neither syllabus names copper and iron losses or kVA ratings, which are beyond the NCERT text.

Is the turns ratio of a transformer N_s/N_p or N_p/N_s?

The turns ratio of a transformer is written both ways, so check which one a question uses: NCERT writes the ratios with N_s/N_p, greater than 1 for a step-up transformer (NCERT Eqs. (7.33) and (7.36), page 195), while the US DOE handbook defines the turns ratio as N_p/N_s, greater than 1 for step-down (beyond the NCERT text: US DOE Fundamentals Handbook: Electrical Science, 1992, Module 13, Equation (13-1)). The physics is the same whichever is used, V_s/V_p = N_s/N_p. So a 230 V to 9.2 V transformer has N_s/N_p = 0.04 but a turns ratio of 25, written 25 : 1.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. On what principle does a transformer work?

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   Mutual induction (NCERT page 194).

2. Q2. A transformer has 500 primary turns and 50 secondary turns. With 230 V on the primary, what is the secondary voltage?

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   V_s = 230 × 50/500 = 23 V: a step-down transformer (NCERT Eq. (7.36), page 195).

3. Q3. Why is the core of a transformer laminated?

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   To reduce the eddy currents induced in it by the alternating flux, and so the heating (NCERT page 196).