Transformer Nameplate Data
Calculated Values
Additional Values
Transformer X/R sets the peak current your switchgear has to make. Checking make and break duty across the whole network is a short-circuit study. Short-circuit and power system studies →
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Related: Turn this X/R into a peak make current with the fault waveform tool
Calculation Method
Per-Unit Resistance:
R = PCu / Srated
Where PCu is copper losses in MW and Srated is MVA rating
Per-Unit Reactance:
X = √(Z² - R²)
Base Impedance:
Zbase = V² / S
Where V is in kV and S is in MVA
Actual Impedance:
Zactual = Zp.u. × Zbase
This is the two-winding transformer method of IEC 60909-0:2016, clause 6.3.1, formulas (7) to (9), written per unit. For a maximum fault calculation the standard also applies the impedance correction factor KT of clause 6.3.3, which this tool does not.
Worked Example
An 11 kV, 2 MVA transformer with 6 % impedance and 20 kW load loss.
R = 0.020 MW / 2 MVA = 0.0100 p.u.
Z = 6 / 100 = 0.0600 p.u.
X = √(0.0600² − 0.0100²) = 0.0592 p.u.
X/R = 0.0592 / 0.0100 = 5.92
Zbase = 11² / 2 = 60.5 Ω, so R = 0.605 Ω and X = 3.579 Ω on the 11 kV side
Enter 11, 2, 6 and 20 above to reproduce it. That X/R gives a peak factor of κ = 1.02 + 0.98 e−3/5.92 = 1.61 (IEC 60909-0), so the first peak reaches 1.61 × √2 = 2.28 times the r.m.s. fault current. The fault waveform tool plots that current and the d.c. component at break.
A transformer X/R on its own does not give the X/R at a fault point: the source, cables and other transformers all add their own R and X. Building that network model is part of our short-circuit studies.
Minimum Impedance by Rating
IEC 60076-5:2006, Table 1 gives the recognised minimum short-circuit impedance for two-winding transformers. Use the nameplate value where you have one; these are floors, not typical values.
| Rated power (kVA) | Minimum impedance (%) |
|---|---|
| 25 to 630 | 4.0 |
| 631 to 1,250 | 5.0 |
| 1,251 to 2,500 | 6.0 |
| 2,501 to 6,300 | 7.0 |
| 6,301 to 25,000 | 8.0 |
| 25,001 to 40,000 | 10.0 |
| 40,001 to 63,000 | 11.0 |
| 63,001 to 100,000 | 12.5 |
| Above 100,000 | Above 12.5, by agreement |
Resistance does not scale with the impedance. IEC 60909-0 clause 6.3.1 notes that R/X generally falls as transformers get larger, so X/R rises with rating. Take the load loss from the test report.
Frequently Asked Questions
How do you calculate the X/R ratio of a transformer?
Take the per-unit resistance as the load loss at rated current divided by the rating, and the per-unit impedance as the nameplate impedance divided by 100. The reactance is the square root of Z squared minus R squared, and X/R is X divided by R. This is the method of IEC 60909-0:2016 clause 6.3.1. A 2 MVA, 6 % transformer with 20 kW load loss gives R = 0.010 p.u., X = 0.0592 p.u. and X/R = 5.92.
Why does transformer X/R matter?
X/R sets how slowly the d.c. component of a fault current decays, and so the peak make current. IEC 60909-0 takes the peak factor as kappa = 1.02 + 0.98 e^(-3R/X). At X/R = 5.92 kappa is 1.61; a higher X/R gives a higher peak for the same r.m.s. fault current.
Does X/R increase with transformer size?
Yes. IEC 60909-0 clause 6.3.1 states that R/X generally decreases with transformer size, and that for large transformers the resistance may be neglected when calculating the r.m.s. short-circuit current. It still has to be included for the peak current and the d.c. component.
Which losses should I enter?
The load loss, also called copper loss: the total loss in the windings at rated current, P_krT in IEC 60909-0. Take it from the factory test report. The no-load (iron) loss does not enter the short-circuit impedance.
What impedance should I use if I do not know it?
The nameplate or test report value. Without one, IEC 60076-5:2006 Table 1 gives the recognised minimum impedance by rating, for example 6.0 % for 1,251 to 2,500 kVA. A minimum is a floor, not a typical value, and a lower impedance gives a higher fault level.