Transformer Sizing & Fault Level

Full load current

A transformer is rated in kVA, not kW, because it does not care about your power factor — it cares about the current in its windings. On a three-phase transformer:

I = kVA × 1000 / (√3 × V)

Use the winding voltage on the side you are asking about. A 630 kVA 11 kV / 433 V transformer carries about 33 A on the HV side and 840 A on the LV side — same power, twenty-five times the current, which is the whole reason distribution is done at 11 kV.

Note the 433 V. Indian distribution transformers are wound for 433 V at no load so that the voltage has somewhere to fall to as load comes on, landing near 415 V at the board. Size the incomer on 433 V and you are on the safe side.

What per cent impedance actually means

The nameplate figure — 4.5 %, 5 %, 6.25 % — is the fraction of rated voltage you have to apply to the primary, with the secondary shorted, to drive full load current. Turn that around and it tells you the fault current:

I_fault = I_full load / (Z% / 100)
Fault MVA = kVA / (Z% / 100)

At 5 % impedance a transformer will push twenty times its full load current into a bolted fault on its terminals. For the 630 kVA set above, that is roughly 17 kA. Every device on that busbar must be able to make and break it.

This is why impedance is not a number to minimise. Low impedance means better voltage regulation and a worse fault level; high impedance means the opposite. IS 1180 fixes it by rating precisely so that switchgear selection stays predictable.

Peak versus rms

The kA figure above is symmetrical rms. In the first half cycle the current is offset by the DC component and peaks far higher. IEC 60909 uses a factor of about 2.5 for LV systems, and that peak is what bends busbars — it sets the mechanical bracing, while the rms value sets the breaking capacity. Two different numbers for two different failure modes.

Assumptions, and where they break

• Infinite source. The calculation ignores the impedance of the 11 kV network behind the transformer. Real networks lower the fault current, so this figure is conservative — which is the direction you want to be wrong in when rating switchgear.
• Terminal fault only. Add a few metres of cable and the fault current drops fast. Do not use the terminal figure to check discrimination at a downstream board.
• Motor contribution ignored. Running motors feed into a fault for the first few cycles. On a motor-heavy plant add roughly 4 times the connected motor full load current to the peak.

How much load should a transformer carry?

Remember that the loading figure here is kVA, not kW. Poor power factor loads the transformer without doing any more work — fix the power factor and you may find you do not need the bigger transformer at all.

Vector group, and why Dyn11 is the default

The nameplate carries a code like Dyn11: delta primary, star secondary, neutral brought out, secondary lagging the primary by 330 degrees. Each letter earns its place. The delta on the HV side gives third-harmonic circulating currents somewhere to go instead of distorting the supply, and it means an LV earth fault does not reflect straight back into the 11 kV network as an unbalanced load.

The star on the LV side is what gives you a neutral at all, and with it the 240 V single-phase supplies every small load on the site expects. Earth that star point and you have fixed the system voltage reference — which is what makes an earth fault a large, detectable current rather than a slow rise in touch voltage on every enclosure.

The number matters when two transformers have to run in parallel. Group numbers are clock positions, and only transformers with the same displacement can be paralleled — a Dyn11 and a Dyn1 differ by 60 degrees, and connecting them puts that difference across the winding impedance as a permanent circulating current. Matching impedance to within about 10 per cent matters too, or the lower-impedance unit takes more than its share of the load and reaches its temperature limit first.

Inrush, and the protection setting it forces

Energise an unloaded transformer at the wrong point on the voltage wave and the core saturates. The first peak of magnetising inrush reaches 8 to 12 times full load current, decaying over a few hundred milliseconds to nothing. No fault is present, and the transformer is fine.

The protection does not know that. An HV fuse or overcurrent relay set purely on full load current will clear on every energisation, which is why the HV protection on a distribution transformer is deliberately slow and set high — and why the fault level you calculate here does not translate directly into a relay setting. Inrush is rich in second-harmonic content, so differential relays on larger units restrain on it rather than time-grading around it.

It also explains a fault that looks electrical and is not: a transformer that trips only on a Monday morning is usually being energised into a cold, fully connected load rather than suffering anything wrong with its windings.