Transformer (Turns Ratio)

Use the turns ratio: U₂ = U₁·n₂/n₁. The voltages behave like the numbers of turns, because every turn encloses the same alternating magnetic flux and receives the same induced voltage. Example doorbell transformer: 230 V on the primary with 1,000 turns, secondary with 50 turns: U₂ = 230 × 50/1,000 = 11.5 V. If the secondary has more turns than the primary, the voltage is stepped up; with fewer turns it is stepped down. The calculation applies to the unloaded ideal transformer; under load and with real losses the actual secondary voltage is somewhat lower.

Energy conservation enforces this. An ideal transformer passes on the absorbed power completely: P₁ = P₂, hence U₁·I₁ = U₂·I₂. If the voltage is stepped down to one twentieth, the current must rise twentyfold, otherwise power would vanish or appear from nowhere. It follows that I₂/I₁ = U₁/U₂ = n₁/n₂. A welding transformer therefore delivers huge currents of several hundred amperes at a small voltage. Real transformers reach efficiencies above 95 %; the losses (winding resistance, eddy currents, core remagnetisation) reduce the secondary current only slightly. Mnemonic: voltage follows the turns, current runs opposite.

Induction needs a changing magnetic flux. After switch-on, a constant DC voltage drives a constant current through the primary coil, the flux in the iron core is then static, and nothing is induced in the secondary: U₂ = 0. Only at the instant of switching on or off does a brief voltage pulse appear. Worse still: without the inductive AC impedance, only the small ohmic winding resistance limits the primary current, so the coil overheats and burns out. That is why power supplies for DC devices either use AC ahead of the transformer or chop the DC electronically into a high-frequency AC voltage (switch-mode power supply).

The power lost in a line depends on the current: P_V = I²·R. Stepping the voltage up by a factor of 100 cuts the current to one hundredth for the same transmitted power, and the line losses fall to one ten-thousandth. Example: 100 MW over a line of 10 Ω. At 10 kV the current is 10,000 A and the loss I²·R = 1,000 MW, ten times the useful power, completely unusable. At 380 kV only 263 A flow, a loss of about 0.69 MW or 0.7 %. This is exactly why the AC grid was built: transformers step the voltage up with little loss and safely back down before the consumer.

The ideal model makes three assumptions: no winding losses (resistance-free coils), no core losses (no eddy currents, no remagnetisation work) and complete magnetic coupling (the whole flux passes through both coils). Only then do U₁/U₂ = n₁/n₂ and P₁ = P₂ hold exactly. Real grid transformers come remarkably close: large machines reach 98 to 99 % efficiency. The residual losses are called copper losses (I²·R of the windings) and iron losses; against eddy currents the core is laminated from mutually insulated sheets. In exams you almost always use the ideal model and describe the deviations only qualitatively.