Transformer sizing
A transformer is sized on apparent power, and its impedance decides both the voltage sag and the fault current everything downstream has to survive. This converts the load to kVA, rounds to a standard rating, and reports the currents, regulation, losses and fault level for that size.
| Quantity | Value | Note |
|---|---|---|
| Connected load | 500 kW | at pf 0.90 |
| Load apparent power | 555.6 kVA | kW ÷ power factor |
| Demand factor | 1.00 | Diversity across the connected load |
| Spare capacity | 25% | Room for growth |
| Required rating | 694.4 kVA | Before rounding to a standard size |
| Selected rating | 750.0 kVA | Next standard dry-type size |
| Loading at demand | 74% | Design aim is 70–80% |
| Primary current | 31.4 A | at 1.380e+4 V |
| Secondary current | 902 A | at 480 V |
| Full-load current check | 902 A | I = kVA × 1000 / (√3 V) |
| Percent resistance | 0.95% | from %Z and X/R = 6.0 |
| Percent reactance | 5.67% | √(%Z² − %R²) |
| Voltage regulation | 3.43% | ASA approximation |
| Voltage drop at full load | 16.5 V | Regulation × secondary voltage |
| Symmetrical fault current | 1.569e+4 A | Secondary FLA ÷ (Z%/100) |
| Magnetising inrush | 1.083e+4 A | About 12× full-load current, first cycle |
| No-load loss | 1200 W | Constant whenever energised |
| Load loss at demand | 4390 W | Scales with the square of loading |
| Total losses | 5590 W | Core plus winding |
| Efficiency | 98.89% | Output ÷ (output + losses) |
The fault current is what the secondary switchgear has to interrupt, and it comes straight from the impedance. A lower-impedance transformer is more efficient but demands more expensive breakers downstream.
S = P / cos φ · The reason kVA, not kW, sizes the transformer.
I = S / (√3 V) · Three-phase; drop the √3 for single-phase.
I_sc = I_FL / (Z%/100) · Infinite-bus assumption, symmetrical RMS.
%R cos φ + %X sin φ + (%X cos φ − %R sin φ)²/200 · ASA approximation, good to a few tenths.
A clear path from inputs to results.
A transformer is sized on apparent power, not on the load's nameplate kilowatts, and its impedance decides both the voltage sag and the fault current everything downstream has to survive. This converts the load to kVA, rounds to a standard rating, and reports the currents, regulation, losses and fault level for that size.
How to use
- Enter the connected load in kW and the load power factor.
- Set the demand factor and the spare capacity you want to leave.
- Enter the primary and secondary voltages and whether the service is single or three-phase.
- Add the percent impedance and X/R ratio from the nameplate.
- Read the selected rating, the currents, the regulation and the fault current.
The formula
Reference: ANSI C57 and IEEE C57.12 for ratings and impedance; the ASA regulation approximation from IEEE Std 141 (Red Book).
Worked example
500 kW at 0.9 power factor, 13.8 kV to 480 V
- Connected load 500 kW at pf 0.90, demand factor 1.0.
- 25% spare capacity.
- 13,800 V primary, 480 V secondary, three-phase.
- 5.75% impedance, X/R of 6, 1,200 W no-load and 8,000 W load loss.
- 1
Apparent power is 500 / 0.90 = 555.6 kVA.
- 2
With full demand and 25% spare, the required rating is 694.4 kVA.
- 3
The next standard dry-type size is 750 kVA.
- 4
Secondary full-load current is 750,000 / (√3 × 480) = 902 A; the primary is 31.4 A.
- 5
Fault current is 902 / 0.0575 = 15,689 A, or 15.7 kA symmetrical.
- 6
Regulation at 0.9 pf is 3.43%, so the secondary sags 16.5 V at full load.
- 7
Losses are 1,200 W of core plus 4,390 W of copper at 74% loading, giving 98.89% efficiency.
750 kVA · 902 A secondary · 15.7 kA fault · 3.43% regulation · 98.89% efficiency.
Interpreting the result
The selected size is 74% loaded at demand, which is where a distribution transformer should sit: enough headroom for a motor start and for growth, without paying for copper that never gets used. The fault current is the number the downstream switchgear is bought against, and it comes straight from the impedance — a more efficient, lower-impedance transformer is a more expensive breaker.
Assumptions & limitations
- The load is modelled at a single power factor; a mixed load should be summed in kW and kVAR and converted once.
- The standard sizes are the common dry-type distribution ratings and may differ from a specific manufacturer's catalogue.
- Fault current uses the infinite-bus assumption, which ignores source impedance and is therefore conservative.
- Regulation uses the ASA approximation, accurate to a few tenths of a percent for distribution transformers.
- Efficiency uses the stated no-load and load losses, scaled with the square of loading.
- Harmonics, inrush duration, temperature rise and altitude derating are not modelled.
Frequently asked questions
Why is a transformer rated in kVA and not kW?
Because the winding losses depend on current, and current depends on apparent power. The insulation and conductor sizing are set by the voltage and current, not by how much of that current does useful work, so the rating has to be in kVA.
Why does lower impedance mean higher fault current?
Because the fault current is the full-load current divided by the per-unit impedance. A 4% transformer passes 25 times its rated current into a bolted fault; a 6% one passes 16.7 times. Lower impedance also means better regulation, so it is a direct trade against switchgear cost.
How accurate is the regulation formula?
The ASA approximation is good to a few tenths of a percent for distribution transformers at typical power factors. It becomes less accurate at very low power factors, where the reactive term dominates and the exact phasor solution should be used.
What loading should I design for?
70–80% at demand is the usual target. Below 50% you have paid for capacity you are not using and the transformer runs at poor efficiency; above 90% there is no headroom for a motor start, and the winding runs hotter than its insulation class wants.
What is the inrush current?
The magnetising current on energisation, typically 8–12 times full-load current for the first cycle or two, decaying within a few cycles. It is why the primary protective device has to be sized for inrush rather than for full load.
Does altitude affect the rating?
Yes. Above about 1,000 m the thinner air reduces cooling, so a transformer has to be derated — typically 0.3–0.5% per 100 m above the threshold, depending on the insulation class. The model does not apply that derating.
Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.