A transformer changes AC voltage without any moving parts: the ratio of the voltages is the ratio of the turns. That one rule sizes everything from phone chargers to grid substations. But a real transformer is not perfect — some power is always lost as heat in the windings and the core. This post works a complete example: a 230 V to 12 V transformer, its turns, its currents, and exactly where the lost power goes.

Figure 1: The 230 V to 12 V transformer (a), and how its efficiency changes with load (b). Efficiency is lowest at light load because the core loss is there whether the transformer is loaded or not. (Schematic, not to scale.)

The turns ratio

For an ideal transformer, the voltages follow the turns:

V₂ / V₁ = N₂ / N₁

and, because power in equals power out, the currents follow the inverse ratio: I₁ / I₂ = N₂ / N₁. Step the voltage down by 19 times and the current steps up by 19 times — that is how a small 12 V, 2 A load can draw only about 0.1 A from the 230 V mains.

Worked example — 230 V down to 12 V

Design a transformer with V₁ = 230 V, V₂ = 12 V, and a 2 A load on the secondary.

Step 1 — turns: the turns ratio is N₁ / N₂ = 230 / 12 = 19.17. If the primary has N₁ = 460 turns, the secondary needs N₂ = 460 / 19.17 = 24 turns. Check: 460 / 24 = 19.17, and 230 / 19.17 = 12.0 V. ✓

Step 2 — output power and ideal input current: P_out = 12 × 2 = 24 W. In an ideal transformer P_in = P_out, so I₁ = 24 / 230 = 104 mA.

Step 3 — the real transformer: no transformer is ideal. Suppose this one is 92% efficient at full load. Then P_in = 24 / 0.92 = 26.09 W, so 2.09 W is lost as heat, and the real primary current is I₁ = 26.09 / 230 = 113 mA — a little more than the ideal 104 mA.

Step 4 — where the 2.09 W goes: the losses split into two kinds. Copper loss is I²R heating in the windings — with the winding resistances of this example (8 Ω primary, 0.05 Ω secondary) that is 0.113² × 8 + 2² × 0.05 = 0.10 + 0.20 = 0.30 W. Core loss is the hysteresis and eddy-current loss in the iron, which is there whenever the transformer is energised: 2.09 − 0.30 = 1.79 W. Efficiency is η = P_out / P_in = 24 / 26.09 = 92%.

Efficiency against load

Core loss is nearly constant, but copper loss grows with the square of the load current — so efficiency rises with load:

25% load (6 W out): η = 6 / 7.81 = 76.8%  •  50% load (12 W out): η = 12 / 13.86 = 86.5%  •  75% load (18 W out): η = 18 / 19.96 = 90.2%  •  100% load (24 W out): η = 92.0%  •  125% load (30 W out): η = 93.0%

At quarter load the fixed 1.79 W core loss eats a big share of a small output; at full load it is a small share of a large output. That is why a transformer left energised with no load still gets warm — and why oversizing a transformer hurts efficiency: a 100 W transformer feeding a 10 W load spends most of its life at the low-efficiency end of the curve. (Theory note: maximum efficiency is reached when copper loss equals core loss — for this transformer that would be at about 244% of rated load, beyond its rating, so in practice its efficiency simply keeps rising up to full load.)

Key points to remember

• Turns ratio: V₂ / V₁ = N₂ / N₁. For 230 V to 12 V the ratio is 19.17 — 460 primary turns need 24 secondary turns.
• Ideal transformer: power in = power out, so currents scale inversely — 12 V at 2 A needs only 104 mA at 230 V.
• Worked example: 24 W out at 92% efficiency needs 26.09 W in; 2.09 W is lost — 0.30 W copper loss (I²R) and 1.79 W core loss; real primary current 113 mA.
• Efficiency rises with load because core loss is constant while output grows: 76.8% at quarter load, 92.0% at full load.
• Don't oversize a transformer for a small load — it will spend its life at the low-efficiency end of the curve.