# Transformers and Power Transfer > Transformers and power transfer for JEE Physics: turns ratio, losses, efficiency and why Indian grids transmit at extra-high voltage. - Canonical URL: https://prepelephant.com/topics/jee/physics/transformers-and-power-transfer - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Transformers and Power Transfer", PrepElephant, https://prepelephant.com/topics/jee/physics/transformers-and-power-transfer ## 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.