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Transformers and generators are some of the most practical topics in Class 12 physics, and often straightforward marks, as long as you keep three ideas straight: what is at its peak and when, which way the turns ratio goes, and why stepping up voltage does not create energy. This post works through each with numbers and gives you a short practice plan.
Where are these topics in the syllabus?
The JEE (Main) 2026 syllabus ends its electromagnetic induction and alternating current unit with "AC generator and transformer". The JEE (Advanced) 2026 brochure lists Faraday's and Lenz's laws and LCR circuits with a.c. sources, but does not name generators or transformers, so check the current brochure if you are preparing for Advanced. NCERT covers the generator in Class 12 Chapter 6, section 6.8 and the transformer in Chapter 7, section 7.8.
How does an AC generator produce an alternating emf?
A coil of N turns and area A spins at angular speed ω in a uniform field B. The flux through it is NBA cos ωt, so by Faraday's law the emf is ε = NBAω sin ωt. The peak emf NBAω occurs when the coil's plane is parallel to the field, because that is when the flux is changing fastest, even though the flux itself is zero then. NCERT also notes that the frequency of rotation is 50 Hz in India.
Worked example. A 100-turn coil of area 0.050 m² turns at 50 revolutions per second in a 0.20 T field.
ω = 2π × 50 ≈ 314 rad/s.
Peak emf ε₀ = NBAω = 100 × 0.20 × 0.050 × 314 ≈ 314 V.
The rms value is ε₀/√2 ≈ 222 V, close to household mains.
Students have trouble with the ideas underneath this. A teaching study by Zuza and colleagues (2014) notes that physics education research has shown the basic concepts of electromagnetic induction are barely understood by students, and that teaching often rushes the concepts to spend time on rote problem solving. Our post on Lenz's law and induced current covers the direction questions.
How does the turns ratio work?
For an ideal transformer, V_s/V_p = N_s/N_p, and since input power equals output power, I_s/I_p = N_p/N_s. NCERT's own example: 100 turns in the primary and 200 in the secondary step 220 V at 10 A up to 440 V at 5.0 A. Voltage doubles; current halves; power stays at 2200 W.
Worked example. A phone charger steps 240 V down to 12 V.
Turns ratio N_s/N_p = 12/240 = 1/20. With 2000 primary turns you need 100 secondary turns.
The phone draws 24 W at 12 V, so I_s = 2.0 A. For an ideal transformer I_p = 24/240 = 0.10 A.
If the transformer is 90% efficient, input power is 24/0.90 ≈ 26.7 W, the primary current is about 0.111 A, and about 2.7 W is lost as heat.
Where does a real transformer lose energy?
NCERT section 7.8 says a well designed transformer may have an efficiency of more than 95%, and lists four losses:
Flux leakage: not all the primary's flux links the secondary. Winding the coils one over the other reduces it.
Winding resistance: I²R heating in the wires, cut by using thick wire for high-current windings.
Eddy currents: currents induced in the iron core, reduced by a laminated core.
Hysteresis: energy lost each time the core's magnetisation reverses, kept low by choosing a suitable core material.
Why do power lines use high voltage?
NCERT explains that generator output is stepped up so the current, and so the I²R loss, is lower, then stepped down in stages to the 240 V supply in homes. A simple model shows the size of the effect. Send 100 kW through lines of total resistance 2 Ω:
At 1 kV, I = 100 A and the line loss is 100² × 2 = 20 kW, a fifth of the power.
At 10 kV, I = 10 A and the loss is 10² × 2 = 200 W.
Ten times the voltage gives one hundredth of the loss. (Real grids are more complex, but the scaling is the point.)
Common mistakes
Transformer on DC. Steady current gives steady flux, so no emf is induced in the secondary.
Turns ratio upside down for current. Current goes in the inverse ratio of turns.
Peak versus rms. Mains "240 V" is the rms value; its peak is about 340 V.
Peak emf at maximum flux. The emf peaks when the flux is zero and changing fastest.
A practice routine
Five generator problems: find peak emf, rms emf and the coil angle at which each occurs.
Five transformer problems with an efficiency given, finding both currents and the loss.
Two transmission problems comparing line losses at two voltages.
If the link between changing flux and emf still feels abstract, Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Next, rework the 240 V to 12 V charger example with 80% efficiency and compare the primary current.
