Transformer Inrush Current Calculator
Energising a transformer draws a magnetizing inrush that is far above full-load current — commonly 8 to 12 times, decaying over several cycles. It is not a fault, but an upstream protective device cannot tell the difference unless it has been selected to ride through it. Getting this wrong produces a transformer that trips its own feeder breaker every time it is energised.
Run the numbers
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The formula
Peak inrush ≈ M × FLA
| kVA | Transformer nameplate rating |
| V | Line-to-line voltage on the side being energised |
| M | Inrush multiplier, typically 8 to 12 for modern units |
Worked example
A 1000 kVA transformer energised from a 480 V bus:
FLA = 1000 × 1000 / (1.732 × 480) = 1,203 A
At a multiplier of 10, peak inrush is roughly 12,000 A for the first cycle.
The upstream device must not see that as a fault, which usually means an instantaneous setting above it or a deliberately delayed curve.
Which standard governs this
IEEE C57.12.00 covers transformer requirements. Overcurrent protection limits for transformers are set by NEC Article 450, which permits higher primary protection specifically to allow for inrush.
What this calculation does not account for
The multiplier is an estimate, not a measurement. Actual inrush depends on residual core flux and the exact point on the voltage wave at which the switch closes, and worst-case energisation can exceed the typical figure. Magnitude also decays over several cycles, so a single peak number does not describe the whole event.
Common mistakes
Setting instantaneous protection from full-load current alone. Assuming inrush is the same on every energisation when it depends on residual flux from the last de-energisation. Applying a low multiplier to a modern low-loss core, which tends to have higher inrush than older designs.