A transformer is a passive electrical device that transfers energy between two or more coils through a shared magnetic field, changing voltage and current in proportion to the turns ratio while keeping the frequency unchanged.
Learning objectives
- State the transformer voltage and current ratios
- Calculate secondary voltage and current from the turns ratio
- Explain why apparent power is conserved in an ideal transformer
- Describe the losses and limits of a real transformer
- Explain impedance reflection
Progressive explanation
Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.
Beginner The core idea
If the secondary has fewer turns than the primary, the voltage is lower: it is a step-down transformer. More turns give a step-up transformer.
Intermediate How it behaves in a circuit
A real transformer has winding resistance, leakage flux, core loss and magnetising current, so efficiency is below 100 % and the output voltage falls slightly with load. The core must not saturate, and the supply frequency and voltage must match the design.
Advanced Limits, assumptions, engineering detail
Formulas and variables
Units of measurement
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage | V | volt (V) |
| Current | I | ampere (A) |
| Apparent power | S | volt-ampere (VA) |
| Turns ratio | a | dimensionless |
Worked examples
Step-down transformer
Problem. A transformer has 460 turns on the primary and 46 turns on the secondary. The primary is connected to 230 V. Find the secondary voltage.
Solution. V2 = V1 × N2/N1 = 230 × 46/460.
Current and power in the same transformer
Problem. The 23 V secondary supplies 2 A. What primary current flows, assuming an ideal transformer?
Solution. S = 23 × 2 = 46 VA. I1 = S/V1 = 46/230.
Step-up transformer
Problem. A transformer with a 1:2 turns ratio is supplied at 120 V and delivers 2 A to a load. Find the secondary voltage, the primary current and the apparent power.
Solution. V2 = 120 × 2 = 240 V. S = 240 × 2 = 480 VA. I1 = 480/120 = 4 A.
Practical applications
- Power transmission and distribution
- Isolation and safety separation
- Battery chargers and power supplies
- Audio and signal transformers
- Instrument and current transformers
- Electric-vehicle on-board chargers
Automotive applications
The ignition coil is a pulse transformer that steps up 12 V to a high voltage, and DC-DC converters in electric vehicles use transformers for isolation and voltage conversion. On-board chargers and auxiliary supplies also rely on transformer action.
Industrial applications
Distribution transformers step medium voltage down to usable levels, isolation transformers protect sensitive equipment, and instrument transformers scale high voltage and current down to measurable values for meters and protection relays.
Measurement procedure
- Use a turns-ratio tester to verify the winding ratio without energising at full voltage.
- Measure winding resistance and insulation resistance to detect faults.
- Perform no-load and short-circuit tests to separate core loss from copper loss.
- Confirm the supply frequency and voltage match the transformer rating before energising.
Common mistakes
- Connecting a transformer to DC; there is no changing flux to induce a secondary voltage and the primary may overheat.
- Ignoring efficiency and assuming the output power equals the input power exactly.
- Reversing primary and secondary connections beyond the design voltage.
- Ignoring the frequency rating; a 50 Hz transformer is not necessarily suitable at 60 Hz or the reverse.
Safety considerations
- Transformer terminals can be at hazardous voltage; isolate and verify dead before working.
- Never backfeed a transformer unless the installation and protection are designed for it.
- Isolation transformers reduce shock risk only when used correctly; they do not make a circuit safe to touch.