What the Transformer Current Calculator does
This calculator converts a transformer's kVA rating into primary and secondary full-load current, for single-phase or three-phase units. It gives the voltage ratio, the current at any loading percentage, an estimate of the secondary fault current from the nameplate impedance, and the magnetising inrush band the upstream protection has to tolerate.
Full-load current is the number almost everything else hangs on: the cable on each side, the protective device ratings, the current transformer selection and the busbar. It is a short calculation, and the value of doing it here is having the primary, the secondary, the partial-load figures and the fault estimate on one page rather than four.
How to use it
- Choose single or three phase, then enter the rating and both voltages. Use the nominal voltages from the nameplate, not the measured ones.
- Enter the nameplate impedance if you have it. It is what produces the secondary fault estimate; without it no fault figure is claimed.
- Set the loading percentage to see the currents at the load the transformer actually carries rather than at its rating.
- Read the inrush band and treat it as a protection-setting input, not a calculated value - the multiples are adjustable because different units and different switching instants give very different peaks.
- Export the loading table to size cables, current transformers and metering on both sides.
Reading the results
Full-load current is the rated current, not the current flowing. A transformer at 40% load draws 40% of it.
The voltage ratio equals the winding turns ratio for star-star and delta-delta transformers, but not for delta-star ones: there the winding ratio differs from the voltage ratio by root-three, which matters when working with the windings themselves rather than the terminals.
The secondary fault estimate is the infinite-bus figure: full-load current divided by per-unit impedance. It ignores everything upstream, so the real fault current is lower.
The inrush band is a planning convention for protection settings. It is a switching transient that depends on residual core flux and the point on the voltage wave at which the breaker closes.
Worked example: a 500 kVA 11 kV / 400 V three-phase transformer at 4% impedance
Secondary full-load current is 500,000 / (root-three x 400) = 500,000 / 692.82 = 721.7 A. Primary full-load current is 500,000 / (root-three x 11,000) = 26.24 A.
The voltage ratio is 11,000 / 400 = 27.5 : 1, and the current ratio is its inverse - 721.7 / 26.24 = 27.5, as it must be, because the same kVA passes through both windings.
At 75% loading the secondary carries 541.3 A and the primary 19.68 A. That is the figure to use for metering and for a load study; 721.7 A is the figure to use for the busbar and the main device.
The infinite-bus secondary fault estimate is 721.7 / 0.04 = 18,042 A, or 18.0 kA. The switchgear on that board therefore needs at least a 25 kA breaking capacity from the standard series. Adding the real source impedance upstream would bring the figure down somewhat, which is the safe direction to be wrong in.
Magnetising inrush on the primary is estimated at 8 to 12 times 26.24 A, so 210 to 315 A, for a few cycles. A primary device set to trip at 200 A would trip every time the transformer is energised - which is why a transformer feeder gets a type D curve or a deliberately time-delayed setting.
Formulas and scoring rules
- Three-phase full-load current
I = kVA x 1000 / (root-three x V_line)- Single-phase full-load current
I = kVA x 1000 / V- Voltage ratio
n = V_primary / V_secondaryEquals the winding turns ratio for star-star and delta-delta; differs by root-three for delta-star.- Current ratio
I_primary / I_secondary = 1 / nThe same apparent power passes through both windings.- Current at partial load
I = I_full-load x loading% / 100Linear; the magnetising current is a small addition that does not scale with load.- Secondary fault estimate
I_sc = I_full-load / (%Z / 100)The infinite-bus case: it ignores the source impedance and so overestimates.- Inrush
8 to 12 times primary full-load current for a few cyclesA convention for protection settings, not a calculated quantity.
Why the current ratio is the inverse of the voltage ratio
A transformer transfers apparent power with only a small loss, so the volt-amperes entering the primary very nearly equal those leaving the secondary. If the voltage is stepped down by 27.5, the current must step up by the same factor to keep the product the same. That is the whole idea, and it is why an 11 kV cable to a 500 kVA transformer is a modest thing while the 400 V cable leaving it is enormous.
It also explains why faults look so different on the two sides. A 20 kA fault on the secondary appears as roughly 727 A on the primary - well within the range of an ordinary medium-voltage device, and quite invisible to anyone who only looks at the primary current.
Magnetising inrush and why protection has to tolerate it
When a transformer is energised, the core can be driven briefly into saturation depending on how much residual flux it retained and where on the voltage wave the breaker closed. While it is saturated the winding presents almost no impedance beyond its own resistance, so the current is enormous for a few cycles before decaying over several hundred milliseconds.
It is a genuine transient, not a fault, and protection must ride through it: a type D curve on a low-voltage feeder, a time-delayed or restrained setting on a medium-voltage relay, and second-harmonic restraint on differential protection, because inrush is rich in second harmonic and real faults are not. The 8 to 12 times band is a planning figure; large modern cores with low-loss steel can exceed it.
Limitations: what the result does not prove
- Full-load current is the rated current, not a measurement. The actual current depends on the load.
- The fault estimate treats the supply as an infinite bus, so it is higher than the real fault current. A proper study under IEC 60909 includes the source impedance and the motor contribution.
- The inrush figures are a convention for setting protection, not a calculation. The real peak depends on residual flux and switching instant, and it varies from one energisation to the next.
- Nothing here covers the vector group, the tap range, the neutral earthing arrangement or the losses, all of which are part of specifying a transformer.
- Nothing here is a compliance certificate. A qualified engineer must verify the design against the applicable standard and the local regulations.
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Standards and sources
- IEC 60076-1 - Power transformers: general - checked 19 Sep 2026
- IEC 60909-0 - Short-circuit currents in three-phase AC systems
- NFPA 70 - National Electrical Code, Article 450
Frequently asked questions
How do I calculate transformer full-load current?
Divide the rating in VA by the line voltage, and for a three-phase transformer also by root-three. A 500 kVA 400 V three-phase transformer gives 500,000 / (1.732 x 400) = 721.7 A on the secondary. The same formula with the primary voltage gives the primary current.
Why is the primary current so much smaller than the secondary?
Because the same apparent power passes through both windings, so the current is inversely proportional to the voltage. Step the voltage down by 27.5 and the current steps up by 27.5. It is why the high-voltage cable into a substation is small and the low-voltage cable out of it is not.
What is the fault current on the secondary of a transformer?
As a first estimate, the full-load current divided by the per-unit impedance: 721.7 A at 4% impedance gives about 18 kA. That treats the supply as infinitely strong, so the real figure is somewhat lower once the network impedance upstream is included - which is the conservative direction for choosing switchgear.
Is the turns ratio the same as the voltage ratio?
For star-star and delta-delta transformers, yes. For a delta-star transformer they differ by root-three, because the primary winding sees the line voltage while the secondary winding sees only the phase voltage. That matters if you are working with the windings themselves, and not at all if you are working with terminal voltages.
How much inrush current does a transformer draw?
Typically 8 to 12 times the primary full-load current for the first few cycles, decaying over several hundred milliseconds. It is a switching transient that depends on the residual flux in the core and the point on the voltage wave at which the supply is connected, so two energisations of the same transformer can look quite different.
What current does an unloaded transformer draw?
Its magnetising current, typically well under 2% of full-load current for a modern distribution unit, plus the no-load loss. It flows continuously whenever the transformer is energised, which is why switching off a lightly used transformer in a pair can be worth real money over a year.
Can I use full-load current to size the secondary cable?
Yes, and you generally should - the cable and the main device on the secondary are sized for the transformer's rating rather than today's load, because the load will grow to fill it. What it does not do is size the cable for its installation conditions, which needs the derating factors from the cable size calculator.
What size CTs do I need?
Metering CTs are normally chosen at or just above full-load current, so 800/5 A for a 721.7 A secondary. Protection CTs are chosen for their accuracy limit factor and burden at fault current instead, which is a different specification entirely - the same primary rating with a very different core.
Last reviewed by the A2Z.Tools team against the sources listed above.
