Presets
Playback
Load and transformer data
The series resistance and reactance come from the copper loss and the short-circuit impedance. A lagging load makes the secondary voltage fall; a leading (capacitive) load can make it rise. Efficiency peaks where the load-dependent copper loss equals the constant iron loss.
Model checks
The science
What is modelled
A single-phase-equivalent, per-unit model of a 1400 kVA 11 kV / 400 V transformer: a series resistance R and leakage reactance X (from the copper loss and short-circuit impedance) between the primary voltage and the load, plus a constant iron loss. The load is a current of fixed magnitude at a given power factor, and the secondary voltage follows from the phasor loop V1 = V2 + (R + jX) I.
Equations
R = Pcu/Sn,X = √(z² - R²).V2 = √(V1² - [a (X cosφ - R sinφ)]²) - a (R cosφ + X sinφ).- Regulation ≈
a (R cosφ + X sinφ) + ½[a (X cosφ - R sinφ)]². - Efficiency
η = P2 / (P2 + a² Pcu + Pfe), maximal ata = √(Pfe/Pcu).
Checks run in the page
- Numerical efficiency peak against the equal-loss rule.
- Exact regulation against the textbook approximation.
- Short-circuit impedance and power balance.
- Phasor loop with complex arithmetic.
Honest simplifications
One phase-equivalent, a single series impedance with the magnetising branch folded into a constant iron loss (scaled with the square of the primary voltage), no saturation, harmonics, temperature or cooling model, and a load held at constant current magnitude. The 1400 kVA rating and loss figures are illustrative, not a manufacturer specification, and nothing here guides work on live equipment.
What to try
- Press Light load and see the efficiency fall because the iron loss dominates.
- Press Capacitive load: the secondary voltage rises above the primary.
- Move the tap changer to bring the secondary voltage back to nominal.