The short-circuit (SC) test determines the series equivalent impedance of a transformer (as referred to one side, usually the primary). In the test the transformer’s secondary terminals are shorted and a reduced voltage is applied to the primary so that rated (or required) current flows. Measured quantities are the applied voltage VSC, the input current ISC and the input power PSC (real input power).
Because applied voltage is small (a few percent of rated), core losses are negligible and the magnetizing branch can be ignored; therefore the circuit reduces to the series leakage impedance of the transformer.
- R’e = total series resistance referred to primary (R₁ + R’₂)
- X’e = total series leakage reactance referred to primary (X₁ + X’₂)
So the primary sees an equivalent complex impedance Z’e = R’e + j X’e.
Measurements and basic formula for Short Circuit Test of Transformer
During the test, measure:
- VSC — applied primary voltage (r.m.s.) required to drive the chosen current (often rated current)
- ISC — primary current (equal to short-circuit current referred to primary)
- PSC — input real power (losses in windings at that current)
The magnitude of the equivalent impedance referred to primary is
The equivalent series resistance is obtained from measured power:
Hence the equivalent leakage reactance (magnitude) follows from
Derivation for Short Circuit Test of Transformer
Start with the measured complex input: VSC applied across series impedance Z’e carrying current ISC. By Ohm’s law:
The average (real) power absorbed by a series R’ at current ISC is:
Solve for R’e: R’e = PSC / ISC2. Then compute X’e from the magnitude relation Z’e2 = R’e2 + X’e2.
Note: vector sign of X’ is positive (inductive). The derivation assumes magnetizing current is negligible (true because applied V is small) and measured PSC is mostly copper (I²R) loss.
Percent (per-unit) impedance
It is common to express the measured VSC required to produce rated current as a percentage of the rated primary voltage V1,rated:
This percent impedance is important for voltage regulation and short-circuit calculations: a lower %Z gives higher short-circuit currents and worse regulation.
Step-by-step test procedure
- Short the secondary terminals (through a thick link or instrument). Keep protection in place.
- Connect a variable AC supply (autotransformer) to the primary. Insert an ammeter in series, a voltmeter across the primary, and a wattmeter to measure input power.
- Increase applied voltage slowly until the primary current equals the rated current (ISC=Irated) or desired test current.
- Record VSC, ISC and PSC.
- Compute Z’, R’, X’ using the formulas above.
Uses & limitations
Uses: obtaining leakage impedance for short-circuit studies, fault current estimation, voltage regulation calculations, and matching impedance in power systems.
Limitations: The test ignores core losses (valid because V is low) and measures combined series resistance and leakage reactance referred to the tested side. If you need separate R₁ and R’₂, additional data (e.g., DC resistance measurement or open-circuit test) is required.
What the test measures
The short-circuit test determines the transformer’s equivalent series impedance (referred to the side being energized, usually primary). It gives:
- Magnitude of series impedance Z’ = VSC/ISC
- Equivalent series resistance R’ = PSC / ISC2 (copper losses)
- Leakage reactance magnitude X’ = √(Z’2 − R’2)
Because only a small fraction of rated voltage is applied (to get rated current through the shorted secondary), core (iron) losses and magnetizing branch can be neglected.
Key formulas
R’ = PSC / ISC2
X’ = √( Z’2 − R’2 )
%Z = ( VSC / Vrated(primary) ) × 100%
How to perform the test (brief)
- Short the transformer’s secondary terminals securely.
- Connect an adjustable AC source to the primary with ammeter, voltmeter and wattmeter in appropriate places.
- Increase primary voltage slowly until desired primary current (often rated current) flows.
- Record VSC, ISC, PSC. Compute Z’, R’, X’.
Worked Example 1 — small single-phase
Measured: VSC = 12.00 V, ISC = 100.00 A, PSC = 800.00 W. Primary rated voltage Vrated = 230 V.
0.1200² = 0.014400 , 0.0800² = 0.006400
difference = 0.014400 − 0.006400 = 0.008000
X’ = √0.008000 = 0.0894427191 ≈ 0.08944 Ω
Result: Z’ = 0.1200 ∠θ Ω where θ = arctan(X’/R’) = arctan(0.08944 / 0.0800) ≈ 48.37°. Equivalent series: R’ = 0.0800 Ω, X’ = 0.08944 Ω.
Worked Example 2 — different rating
Measured: VSC = 8.00 V, ISC = 200.00 A, PSC = 800.00 W. Primary rated voltage Vrated = 400 V.
0.04000² = 0.001600 , 0.02000² = 0.000400
difference = 0.001600 − 0.000400 = 0.001200
X’ = √0.001200 = 0.03464101615 ≈ 0.03464 Ω
Result: Equivalent series impedance referred to primary: R’ = 0.02000 Ω, X’ = 0.03464 Ω, |Z’| = 0.04000 Ω, %Z = 2%.
Interpretation & practical notes
• The short-circuit test isolates winding (copper + leakage) effects. The measured input power PSC is primarily copper (I²R) loss — hence R’ comes directly from power.
• A low %Z means larger short-circuit currents on the secondary for a given system voltage (important for protection and fault calculations).
• If you need per-phase values for three-phase transformers, use the per-phase measured quantities (test each phase or use delta/wye relations depending on connection).