Work out turns ratio, primary and secondary full load current, efficiency at any load fraction, voltage regulation for lagging or leading loads, the load point of maximum efficiency and the through-fault current — all from the nameplate. Handles both three-phase and single-phase units.
| Symbol | Meaning | Unit |
|---|---|---|
| a | Turns ratio, primary over secondary | — |
| S | Rated apparent power | VA |
| I₁, I₂ | Full load current, primary and secondary | A |
| %Z | Percentage impedance from the nameplate | % |
| %R, %X | Percentage resistance and reactance | % |
| x | Load fraction of rated kVA | — |
| PFe | Iron or no-load loss, constant | W |
| PCu | Copper loss at full load | W |
| η | Efficiency at the stated load | % |
| Isc | Through-fault current | A |
A 1000 kVA distribution transformer steps 11 kV down to 415 V. The nameplate gives 5 % impedance, 1 % resistance, 1500 W iron loss and 9000 W copper loss at full load. It is loaded to 100 % at 0.8 power factor lagging.
| Load | Output at 0.8 PF | Copper loss | Total loss | Efficiency |
|---|---|---|---|---|
| 10 % | 80 kW | 90 W | 1590 W | 98.051 % |
| 25 % | 200 kW | 562.5 W | 2062.5 W | 98.979 % |
| 40.82 % | 326.6 kW | 1499.6 W | 2999.6 W | 99.090 % |
| 50 % | 400 kW | 2250 W | 3750 W | 99.071 % |
| 75 % | 600 kW | 5062.5 W | 6562.5 W | 98.918 % |
| 100 % | 800 kW | 9000 W | 10500 W | 98.705 % |
| 125 % | 1000 kW | 14062.5 W | 15562.5 W | 98.468 % |
| Input | Unit | Accepted range | Default |
|---|---|---|---|
| Phase system | — | Single or three phase | Three phase |
| Rating | kVA | > 0 | 1000 |
| Primary voltage | V | > 0 | 11000 |
| Secondary voltage | V | > 0 | 415 |
| Impedance %Z | % | > 0, typically 4–6.25 | 5 |
| Resistance %R | % | > 0 and ≤ %Z | 1 |
| Load level | % | > 0, up to 200 | 100 |
| Iron loss | W | > 0 | 1500 |
| Copper loss | W | > 0 | 9000 |
| Power factor | — | > 0 and ≤ 1 | 0.8 |
| Rating | %Z | Iron loss | Copper loss | Peak η load | I₂ at 415 V |
|---|---|---|---|---|---|
| 100 kVA | 4.0 % | 260 W | 1760 W | 38.4 % | 139 A |
| 250 kVA | 4.0 % | 520 W | 3300 W | 39.7 % | 348 A |
| 500 kVA | 4.5 % | 900 W | 5500 W | 40.5 % | 696 A |
| 630 kVA | 4.5 % | 1100 W | 6500 W | 41.1 % | 877 A |
| 1000 kVA | 5.0 % | 1500 W | 9000 W | 40.8 % | 1391 A |
| 1600 kVA | 6.0 % | 2200 W | 13000 W | 41.1 % | 2226 A |
| 2000 kVA | 6.0 % | 2700 W | 16000 W | 41.1 % | 2782 A |
| 2500 kVA | 6.25 % | 3200 W | 19500 W | 40.5 % | 3478 A |
For a three-phase transformer, divide the rating in volt-amperes by √3 times the line voltage on that side. A 1000 kVA unit at 11 kV draws 52.49 A on the primary and delivers 1391.21 A at 415 V on the secondary. For a single-phase transformer the √3 is omitted, which makes the current higher by a factor of 1.732 for the same rating and voltage. The kVA rating is the same on both windings, since apart from losses the power in equals the power out.
Percentage impedance is the percentage of rated voltage that must be applied to one winding, with the other short circuited, to drive full load current. A 5 % impedance means 5 % of rated voltage produces full load current, so a bolted fault at the terminals draws about 20 times full load current. Low impedance gives better voltage regulation and easier motor starting but higher fault current, which is why distribution transformers cluster between 4 and 6.25 %.
Maximum efficiency occurs where the variable copper loss equals the fixed iron loss, which happens at a load fraction equal to √(PFe/PCu). For a unit with 1500 W of iron loss and 9000 W of copper loss the peak sits at 40.8 % of rated kVA. This is deliberate: distribution transformers are designed to peak well below full load because they spend most of their life lightly loaded, so the average annual efficiency is what matters rather than the full load figure.
Voltage regulation is the drop in secondary voltage from no load to full load, expressed as a percentage of the no-load value. It is approximately %R × cos φ + %X × sin φ. At a lagging power factor both terms add and the voltage falls. At a leading power factor the reactive term subtracts and regulation can become negative, meaning the secondary voltage rises under load, which is a real effect on lightly loaded cable networks and capacitor-corrected installations.
Divide the full load current by the per-unit impedance, so IFL ÷ (%Z/100). A 1000 kVA transformer at 5 % impedance with a 52.49 A primary gives about 1049.73 A of primary fault current, equivalent to about 27824 A on the 415 V secondary. This assumes an infinite source behind the transformer, so it is the maximum the transformer alone can pass. A full study adds the finite grid impedance, the cable to the fault and any motor back-feed, all of which change the answer.
Results are for estimation and preliminary design. Through-fault current assumes an infinite source and ignores grid impedance. Verify against the transformer test certificate and a qualified engineer before construction.
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