Cable losses and cable length in the secondary circuit

How many VA does the measuring cable between current transformer and instrument consume? Calculates cable burden, total burden against the rated output and the maximum cable length – with a recommendation on 1 A or a larger cross-section.

Input values

Distance from transformer to instrument – the calculator adds the return conductor itself.

Current circuit from the datasheet; leave empty = cable only.

Result

Cable resistance (out and back)
0.142 Ω
Cable burden
3.56 VA
Total burden
3.56 VA
Share of rated output
71 %
Maximum cable length with this cross-section
14 m
Same cable with 1 A secondary
0.14 VA

Fits: the total burden is between 25 % and 100 % of the rated output – the transformer keeps its accuracy class.

Recommendation

  • With a transformer for 1 A secondary the total burden drops to 0.14 VA – the cable losses are one twenty-fifth.

How it is calculated

P = 2 · L · I² · ρ / A

Example: 5 A, 10 m, 2.5 mm² → 2 · 10 · 25 · 0.0178 / 2.5 = 3.56 VA

ρ = 0.0178 Ω·mm²/m (copper at 20 °C). Terminals and contacts are not included; the VA calculator adds a flat 0.1 VA for them.

Chart of cable burden in VA versus one-way cable length for 2.5 mm² copper: with 5 A secondary 3.56 VA at 10 m, with 1 A only 0.14 VA; a 5 VA transformer is sufficient at 5 A up to 14 m
The cable burden grows linearly with length and with the square of the secondary current: P = 2 · L · I² · ρ / A. With 1 A secondary it is one twenty-fifth of the value at 5 A.

Related pages

FAQ

How do I calculate the cable losses in the secondary circuit of a current transformer?

P = 2 · L · I² · ρ / A. L is the one-way cable length in m (the factor 2 accounts for outgoing and return conductor), I the secondary current, ρ = 0.0178 Ω·mm²/m for copper at 20 °C and A the cross-section in mm². Example: 5 A, 10 m, 2.5 mm² gives 2 · 10 · 25 · 0.0178 / 2.5 = 3.56 VA.

How long may the measuring cable be?

Long enough that cable burden and device consumption together do not exceed the rated output of the transformer: L_max = (rated output − devices) · A / (2 · ρ · I²). At 5 A, 2.5 mm² and a 5 VA transformer without devices this is 14 m, with 1 A secondary 351 m.

Why is 1 A better than 5 A for long cables?

Losses rise with the square of the current. At 1 A the cable burden is one twenty-fifth of the value at 5 A – in the example 0.14 VA instead of 3.56 VA.

What do I do if the total burden is too high?

Choose a larger cross-section, shorten the cable, use a transformer with 1 A secondary or a higher rated output. The calculator shows which standard cross-section is sufficient and which rated output fits.

Why should the burden not be much lower than the rated output?

Under IEC 61869-2 the accuracy class applies at a burden of 25 % to 100 % of the rated output (classes 0.1 to 1). Below that the class is no longer guaranteed and the instrument security factor is less effective.

Are terminals and contacts included?

No, the calculator considers cable and devices. The VA calculator adds a flat 0.1 VA for terminals and contacts.

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