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CT Burden Calculation: Why the Secondary Leads, Not the Relay, Decide the Verdict

A 2.50 VA relay on a 400/5 CT looks like a light load — until you add 20 m of 2.5 mm² copper and the leads alone draw 6.90 VA. How the lead-burden square law works, where the verdict flips, and why the same circuit on a 1 A secondary carries exactly 25 times less.

Published 2026-08-02 · EI Portal

The most common CT burden calculation mistake is adding up only the relay. A current transformer drives the whole secondary loop — the connected device plus the wiring back to the panel — and on a 5 A secondary the wiring is usually the bigger load. Count only the device VA and you can sign off a CT that is quietly running over its rated burden. A CT secondary burden check exists to make the leads explicit, and to tell you how much cable run you have left before the core is full.

What rated burden means

A CT's rated burden is the VA it is designed to drive at rated secondary current. The burden check is a budget: does everything hanging on the secondary fit inside that VA? The verdict is a three-state comparison of utilisation against the rating — >100% is "Over burden", ≥85% is "Near limit", anything else is "Within rating".

The lead-burden method

Every number below comes from one short chain, referred to the secondary:

lead loop           = 2 × one-way length
      lead resistance     = ρ × loop / area        (ρ Cu 0.017241 Ω·mm²/m at 20 °C)
      lead burden VA      = Isec² × lead resistance
      total burden        = connected + lead
      utilisation %       = total / rated × 100
      max one-way length  = (available VA ÷ Isec²) × area ÷ (2ρ)

The term that carries the surprise is Isec² × lead resistance. Lead burden scales with the square of the secondary current — which is why the same length of copper is a non-event on a 1 A circuit and the dominant load on a 5 A one.

Worked: the 400/5 feeder circuit

The tool's default case is a feeder protection circuit: a 400/5 CT with a 15.00 VA rated burden, a numerical relay drawing 2.50 VA, and 20 m one-way of 2.5 mm² copper to the panel. The chain runs:

Verdict: "Within rating", with 5.60 VA of margin, an equivalent burden of 0.376 Ω, a secondary loop voltage of 1.88 V, and a maximum one-way lead length of 36.3 m. Do the last division yourself: 6.90 ÷ 9.40 puts the leads at 73.4% of the total burden. The copper, not the relay, is the load on this core — count only the device and you would have logged about a quarter of the truth.

Where the verdict flips

Hold everything else and stretch only the lead run, and the three states walk past in order:

The hero circuit's own readout told you where the edge was: the maximum one-way length was 36.3 m, and 40 m is past it.

The 1 A rescue — the 25× law

Keep the identical circuit — same 20 m, same 2.5 mm² copper, same relay — and swap the secondary to 1 A. The lead resistance does not change; the current in it does:

That 25 is not a coincidence: (5/1)² = 25, straight from the Isec² law. Same cable, same device — a twenty-fifth of the lead burden, and twenty-five times the reach. This is why large substations and long secondary runs specify 1 A CTs.

When the leads alone bust the rating

The square law cuts the other way on the tool's long-run preset: an 800/5 CT, 15 VA rated, a 3.00 VA relay, and 45 m of 2.5 mm² copper. The leads alone draw 15.52 VA — more than the entire rating before the relay is even connected. Total 18.52 VA = 123.4% → "Over burden", margin −3.52 VA. The maximum run on that core was 34.8 m; 45 m busts it. Nothing about the relay caused this — the cable did. The full walk-through — the feeder circuit both ways and this failure — is in the worked case, the copper was the load.

Different circuits, same method

Big device or long leads, 5 A or 1 A — the budget is the same; only the term that dominates it changes.

Check it live

The CT Burden Quick Checker puts the whole chain on one surface: the lead resistance, the lead burden, the total against the rated VA, the utilisation, the equivalent burden and loop voltage, the maximum one-way lead length you have left — and the verdict, "Within rating", "Near limit" or "Over burden". Change the lead length, the conductor size or the secondary current and watch the decision move.

It is a rated-burden quick check, and it says so. It does not assess CT accuracy class, ALF, knee-point voltage, excitation current, remanence, saturation, transient offset or relay operating thresholds — that depth is the EI Portal CT Selection tool. Lead burden is estimated from a two-wire loop at 20 °C; resistance rises with temperature, so use project cable data where you have it, and confirm the core against the manufacturer's / IEC 61869-2 ratings. It is not a CT certification, and there is no PDF or report export.

Frequently asked questions

How do I calculate CT secondary burden?

Total everything the CT drives: the connected device's VA plus the lead burden. Lead burden = Isec² × lead resistance, where the resistance is taken over the two-wire loop (2 × one-way length) at ρ Cu 0.017241 Ω·mm²/m at 20 °C. Compare the total against the rated VA: a 400/5 CT with a 2.50 VA relay and 20 m of 2.5 mm² copper totals 9.40 VA on a 15.00 VA core — 62.6% utilisation, "Within rating".

Why do the leads dominate CT burden on a 5 A secondary?

Because lead burden scales with the secondary current squared. At 5 A, every ohm of loop resistance costs 25 VA — so 0.276 Ω of copper draws 6.90 VA while the relay it feeds draws only 2.50 VA. Work the division and the leads are 73.4% of the total burden on the worked feeder circuit. On a 5 A secondary the wiring, not the device, usually fills the core.

What is the difference between 1 A and 5 A CT burden?

For identical leads, the 5 A circuit carries exactly (5/1)² = 25 times the lead burden of the 1 A circuit. The worked 20 m run draws 6.90 VA at 5 A and 0.28 VA at 1 A, and the maximum one-way lead length jumps from 36.3 m to 906.3 m. That single square law is why long runs and large substations specify 1 A secondaries.

How long can CT secondary leads be?

Rearrange the burden budget for length: max one-way length = (available VA ÷ Isec²) × area ÷ (2ρ), where available VA is the rated burden minus the connected device. The worked 400/5 circuit has 36.3 m of reach on 2.5 mm² copper; the same circuit at 1 A has 906.3 m. On the 800/5 long-run preset the limit was 34.8 m — which is why its 45 m run reads "Over burden".

Is a rated-burden check enough to select a CT?

No — it is the VA budget only. It does not assess accuracy class, ALF, knee-point voltage, excitation current, remanence, saturation, transient offset or relay operating thresholds; that depth is the CT Selection tool. Use the quick check to catch an over-burdened secondary early — especially a long lead run — then confirm the core against manufacturer / IEC 61869-2 ratings and the protection study.

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