Calculating the burden: how to size the VA rating of a current transformer
The burden is the load a current transformer has to drive in the secondary circuit – stated as apparent power in VA at rated secondary current. It consists of the consumption of the connected meters, instruments or relays, the resistance of the measuring lead (outgoing and return conductor) and the contact resistances at terminals.
The rated output of the transformer must cover the burden. If the burden is higher, the core saturates earlier and the transformer reads low. If it is much lower, the accuracy class is not guaranteed either: for classes 0.1 to 1 it applies between 25 % and 100 % of the rated output.
Worked example: secondary current 5 A, one-way cable length 10 m, copper 2.5 mm². The lead resistance is R = 2 · 10 m · 0.0178 Ω·mm²/m / 2.5 mm² = 0.1424 Ω, the cable loss P = 5² · 0.1424 = 3.56 VA. With a 0.5 VA meter this gives 4.06 VA – a transformer with 5 VA rated output fits. With 1 A secondary the cable loss would be only 0.14 VA.
Cable burden per 10 m one-way length (copper, 20 °C)
| Cross-section | Resistance out + back | Burden at 5 A | Burden at 1 A |
|---|---|---|---|
| 1.5 mm² | 0.237 Ω | 5.93 VA | 0.24 VA |
| 2.5 mm² | 0.142 Ω | 3.56 VA | 0.14 VA |
| 4 mm² | 0.089 Ω | 2.23 VA | 0.09 VA |
| 6 mm² | 0.059 Ω | 1.48 VA | 0.06 VA |
| 10 mm² | 0.036 Ω | 0.89 VA | 0.04 VA |
Calculated with ρ = 0.0178 Ω·mm²/m and 20 m conductor length (10 m out, 10 m back). The burden grows linearly with length: at 20 m one-way the values double. Instrument and terminal consumption come on top.
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FAQ
What is the burden of a current transformer?
The burden is the total load in the secondary circuit: consumption of the connected devices, resistance of the measuring lead and contact resistances at terminals. It is stated as apparent power in VA at rated secondary current or as impedance in ohms.
How do I calculate the burden?
Cable loss P = I² · 2 · L · ρ / A (I secondary current, L one-way length in m, ρ = 0.0178 Ω·mm²/m for copper, A cross-section in mm²) plus the device consumption from the datasheet. Example: 5 A, 10 m, 2.5 mm² gives 25 · 2 · 10 · 0.0178 / 2.5 = 3.56 VA.
What happens if the burden exceeds the rated output?
The core saturates earlier, the secondary current becomes too small and the accuracy class is no longer met. Remedies: a larger cross-section, a shorter lead, a transformer with higher rated output or 1 A instead of 5 A secondary.
May the burden be lower than the rated output?
Yes, but not arbitrarily: for classes 0.1 to 1 accuracy is guaranteed between 25 % and 100 % of the rated output. In addition, the instrument security factor FS is less effective at low burden – the transformer only saturates at higher currents. So choose the rated output to match rather than generously.
1 A or 5 A – which is better for long leads?
1 A. The cable loss rises with the square of the current; at 1 A it is one twenty-fifth of the value at 5 A. In the example with 10 m and 2.5 mm² it is 0.14 VA instead of 3.56 VA.
How do I convert the burden from VA to ohms?
Z = S / I². A burden of 5 VA equals 0.2 Ω at 5 A secondary and 5 Ω at 1 A secondary. Conversely, a lead resistance R at 1 A gives exactly R in VA.
Where do I find the consumption of my instrument?
In the datasheet under “burden” or “power consumption of the current circuit”, usually in VA per phase. Only the current circuit counts for the burden, not the power supply of the device.