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Standard electrical-machines formulas (Kapp's approximation)10 min read

Worked Example: Transformer Efficiency and Voltage Regulation from OC/SC Test Data

A 1000 kVA transformer's own open-circuit and short-circuit test results, used to find its full-load efficiency, the loading point where efficiency actually peaks, and its voltage regulation at rated load.

Scenario

Rated capacity1000 kVA
No-load loss (open-circuit test)1800 W
Load loss at rated current (short-circuit test)11,000 W
Impedance (nameplate, from SC test)6.0%
Loading condition100% of rated load, 0.85 power factor lagging

Step-by-step calculation

Step 1: Compute total loss and output at full load

totalLoss = Pfe + x² x Pcu output = x x S x PF
totalLoss = 1800 + 1² x 11,000 = 12,800 W output = 1 x 1,000,000 x 0.85 = 850,000 W
totalLoss = 12.8 kW, output = 850 kW

Step 2: Compute efficiency at full load

eff% = output / (output + totalLoss) x 100
850,000 / (850,000 + 12,800) x 100
eff = 98.52%

Step 3: Find the loading point where efficiency actually peaks

Maximum efficiency occurs where variable (copper) loss equals fixed (iron) loss — not necessarily at full load.

x(maxEff) = √(Pfe / Pcu)
√(1800 / 11,000)
x(maxEff) = 0.405 (about 40.5% loading) -> peak efficiency = 98.96%, slightly higher than the full-load figure

Step 4: Compute resistive and reactive voltage-drop components

vr% = Pcu / (S x 1000) x 100 vx% = √(z%² - vr%²)
vr = 11,000 / 1,000,000 x 100 = 1.1% vx = √(6² - 1.1²) = √34.79
vr = 1.1%, vx = 5.90%

Step 5: Apply Kapp's approximate voltage regulation formula

reg% = x(vr·cos(phi) + vx·sin(phi)) + (x²/200)(vx·cos(phi) - vr·sin(phi))²
reg = 4.14% at full load, 0.85 PF lagging

Result summary

CheckRequirementActualStatus
Full-load efficiencyn/a (informational)98.52%✓ PASS
Peak efficiency (occurs at 40.5% loading)n/a (informational)98.96%✓ PASS
Voltage regulation at full loadn/a (informational)4.14%✓ PASS
This 1000 kVA transformer runs at 98.52% efficiency at full load, but is actually most efficient (98.96%) when loaded to only about 40% of rating — and its voltage sags 4.14% between no-load and full-load secondary voltage at 0.85 PF lagging.

Key insight: A transformer's efficiency curve peaks where fixed loss equals variable loss, not at 100% loading — which is why a transformer that's oversized relative to its typical load (running well below full rating most of the time) isn't necessarily wasting efficiency, and can sometimes be operating closer to its actual efficiency sweet spot than a tightly-sized unit running near full load constantly.

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Frequently asked questions

Why does annual energy cost care about average loading separately from peak efficiency?

No-load loss (1800 W) is constant 24/7 regardless of load, while load loss scales with the square of loading fraction — so a transformer's annual loss cost depends on its typical average loading over the year (this example uses 60% as a representative average), not just its instantaneous efficiency at any one snapshot like full load or the maximum-efficiency point.

What does the impedance percentage (6%) actually limit?

Nameplate impedance directly determines available short-circuit current on the secondary side during a fault (roughly rated current / z%, before any source impedance is added) and also sets the reactive voltage-drop component (vx) used in the regulation calculation — a lower-impedance transformer gives tighter voltage regulation under load but also allows higher fault current, which is exactly the kind of trade-off protection studies have to account for.

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