Transformer rated and short-circuit current
From the rating, voltages and impedance voltage uk, calculate the rated currents and the short-circuit current at the LV terminals.
Transformer
Enter the nameplate data. The result is for a fault at the LV terminals.
Standard ratings
| Rating [kVA] | In HV [A] | In LV [A] | Ik LV [kA] |
|---|---|---|---|
| 100 | 3.8 | 144 | 2.4 |
| 160 | 6.2 | 231 | 3.8 |
| 250 | 9.6 | 361 | 6.0 |
| 400 | 15.4 | 577 | 9.6 |
| 630 | 24.2 | 909 | 15.2 |
| 800 | 30.8 | 1,155 | 19.2 |
| 1000 | 38.5 | 1,443 | 24.1 |
| 1250 | 48.1 | 1,804 | 30.1 |
| 1600 | 61.6 | 2,309 | 38.5 |
| 2000 | 77.0 | 2,887 | 48.1 |
| 2500 | 96.2 | 3,608 | 60.1 |
Symmetrical three-phase fault at the LV terminals with an infinite upstream fault level. The impedance of the HV network, cables and busbars lowers the real current; the voltage factor c from IEC 60909-0 raises it by 5–10%. Do a full short-circuit study before selecting switchgear. The result is for preliminary calculations and must be checked in the design and by measurement. It does not confirm that the installation complies with standards.
Method and assumptions
In = S / (√3 · U)
Ik = In · 100 / uk%
Check examples: 630 kVA, 400 V, uk 4% gives 909 A and 22.7 kA; 1000 kVA, 400 V, uk 6% gives 1443 A and 24.1 kA.