Module 3.15 Transformers
719 words · checked 2026-10-03 · Pro
Knowledge levels
| Sub-module | A | B1 / B2 / B2L | B3 |
|---|---|---|---|
| 3.15 Transformers | — | 2 | 1 |
Content: construction and operation; losses and how they are reduced; action on load and no load; power transfer, efficiency, polarity marking; line and phase voltages and currents; three-phase power; primary/secondary current, voltage, turns ratio, power and efficiency; auto-transformers.
How a transformer works
- Two (or more) windings on a common laminated iron core. AC in the primary produces an alternating flux in the core, which induces an EMF in the secondary by mutual induction.
- A transformer works only with changing current; DC produces no continuous output (and would overheat the primary, whose only opposition to DC is its low resistance).
- There is no electrical connection between primary and secondary (except in an auto-transformer) - so it also gives isolation.
The transformer equations (ideal transformer)
Vₚ / Vₛ = Nₚ / Nₛ = Iₛ / Iₚ
- Step-up: more secondary turns, higher secondary voltage, lower secondary current.
- Step-down: fewer secondary turns, lower voltage, higher current.
- Ideal transformer: power in = power out (Vₚ Iₚ = Vₛ Iₛ). The frequency is unchanged.
Worked example - step-down
A transformer has 1 150 primary turns and 260 secondary turns; primary 115 V, 400 Hz. The secondary supplies 5 A.
- Vₛ = 115 × 260 / 1 150 = 26 V
- Iₚ = 5 × 260 / 1 150 = 1.13 A (ignoring losses)
- Output frequency: 400 Hz
Trap answers: inverting the turns ratio (508 V) or multiplying current by the voltage ratio the wrong way (22 A).
Worked example - efficiency
Input 1 000 VA at unity PF, output 960 W.
- η = 960 / 1 000 = 96 %; losses = 40 W.
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