Transformers and Power Transfer
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Direct answer
A transformer changes alternating voltage according to the turns ratio: for an ideal transformer V(s)/V(p) = N(s)/N(p) = I(p)/I(s), stepping voltage up, current down, power unchanged. Real transformers lose energy through four channels — copper losses (I^2 R heating of windings), eddy-current losses in the core (suppressed by laminating the silicon-steel core into insulated sheets), hysteresis losses (area of the B–H loop per cycle) and flux leakage — and run at 95-99% efficiency at ratings. The transformer only works on AC because mutual induction needs changing flux, which is why grids transmit at extra-high voltage: at high V the current for a given power is small, and the I^2 R line loss shrinks as the square of the voltage step-up.
What you must remember
- Ideal transformer equations: V(s)/V(p) = N(s)/N(p) and I(s)/I(p) = N(p)/N(s); power in equals power out, so stepping voltage up steps current down by the same factor.
- Why AC only: mutual induction requires d phi/dt; a transformer fed steady DC produces no secondary emf (and burns its primary — the winding is a near-short at DC).
- Four losses: copper (I^2 R in windings), eddy currents (laminated core), hysteresis (soft magnetic material with a narrow loop), flux leakage (common core geometry); efficiency = P(out)/P(in), typically above 95%.
- Line-loss arithmetic: for a line of resistance R(carried) delivering P at voltage V, current I = P/V and loss = P^2 R / V^2 — raising V by a factor of 10 cuts line loss a hundredfold.
- Indian grid anchors: generation near 11 kV steps up to 220/400/765 kV for transmission, then steps down through 33 kV and 11 kV to the 230 V, 50 Hz domestic supply.
- Rating conventions: transformers are rated in kVA, not kW, because the load's power factor decides the real power the same kVA delivers.
- Load logic: a step-down transformer feeding more appliances draws more primary current — the secondary demand sets the primary current through the power balance.
- Pattern note: Main tests turns-ratio numericals and loss identification; Advanced tests line-loss minimisation and power-balance reasoning with efficiency included.
Following power from generator to plug
A 2 MW load at the end of a line of total resistance 20 ohms makes the voltage choice vivid. Transmit at 2000 V and the current is 1000 A — the loss is I^2 R = 1000^2 × 20 = 2 × 10^7 W: the entire payload burns in the wires. Step up to 200 kV and the current falls to 10 A, loss to 2000 W — one ten-thousandth — the entire argument for high-tension lines in one calculation. Repeat at 400 kV and the loss falls fourfold: the physics behind the 220/400/765 kV pylons and the 33 kV/11 kV/230 V step-down chain at the delivery end.
Turns-ratio arithmetic stays honest the same way. A 230 V to 11.5 V doorbell transformer has N(p)/N(s) = 20; if the bell draws 0.5 A, the secondary delivers 5.75 W and the primary draws 5.75/230 = 25 mA ideally — power, not voltage, is the conserved currency.
Where students slip
Applying the turns ratio to voltages but forgetting the current inversion is routine; the ideal transformer conserves power, so a step-up in V is exactly a step-down in I, and answers claiming both rise violate conservation. Second, the DC question: a transformer on steady DC delivers nothing on the secondary while the primary behaves nearly as a short circuit — "transformer works on DC" is a permanently wrong option. Third, line loss is I^2 R on the line, often confused with the load's power; the clean form loss = P^2 R/V^2 makes the V-squared advantage explicit, and candidates who write loss = V^2/R with V the transmission voltage have used the wrong voltage — R sees only the line's share. Fourth, when an efficiency is quoted, input = output + losses; kVA versus kW ratings is the professional nuance increasingly tested.
Frequently asked questions
What does the turns ratio of a transformer determine?
The voltage ratio V(s)/V(p) = N(s)/N(p), with the inverse current ratio I(s)/I(p) = N(p)/N(s) for an ideal transformer, keeping power conserved.
Why does a transformer not work on direct current?
Steady DC produces no changing flux, so d phi/dt = 0 and no secondary emf appears, while the low-resistance primary draws excessive current.
Why is electrical power transmitted at very high voltage?
At high voltage the current for a given power is small, and since line loss is I^2 R = P^2 R/V^2, raising transmission voltage slashes the loss as the square of the step-up factor.
How are eddy-current losses in the core reduced?
By laminating the core into thin sheets insulated from one another, which blocks the large circulating current loops while preserving the magnetic path.
What is the difference between a transformer's kVA rating and kW output?
The rating in kVA is apparent power; the real power in kW equals kVA multiplied by the load power factor, which the manufacturer cannot predict.