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CT effective accuracy-limit factor with actual burden

Rn=SnIsn2ALF′=ALF·Rct+RnRct+RbIlim=ALF′·Ipn
where
Rn
rated burden resistance (Ω)
Sn
rated burden (VA)
Isn
CT rated secondary current (A)
ALF′
effective accuracy-limit factor
ALF
rated accuracy-limit factor
Rct
CT secondary winding resistance (Ω)
Rb
actual connected burden resistance (Ω)
Ilim
primary current up to which accuracy is held (A)
Ipn
CT rated primary current (A)
Explain this equation

Recalculates a class P CT’s accuracy-limit factor for the burden actually connected. Use it to show that a 5P20 CT with a light numerical relay stays accurate well beyond 20 times rated current.

Rn is the rated burden expressed as a resistance. Rct is the CT winding resistance and Rb the real external burden of leads and relay. ALF is the nameplate figure and ALF′ the figure that applies in service.

A CT reaches its accuracy limit at a fixed internal EMF, ALF × Isn × (Rct + Rn). With a different burden, the same EMF drives a current in inverse proportion to the total loop resistance, which gives the ratio shown.

Sense check: a 400/1 5P20 15 VA CT with 4 Ω winding resistance and 2 Ω actual burden reaches ALF′ ≈ 63, so it holds accuracy to about 25 kA primary.

Watch out: leaving out Rct exaggerates the gain from a light burden.

Recalculates the accuracy-limit factor of a class P CT for the burden actually connected, since a lighter burden lets the CT stay accurate to a higher fault current.

ratio
VA
A

Enter your values to see the result.

Good to know

The CT’s internal EMF at the accuracy limit is fixed by its design: ALF × Isn × (Rct + Rn). Dividing by the real loop resistance gives the new limit. Modern numerical relays draw well under 1 VA, so leads dominate the burden. Compare the primary current limit with the maximum fault current (and the relay’s required multiple).

Common mistake

Ignoring Rct. Without it, the formula overstates the gain from a light burden, sometimes by a large factor on low-ratio CTs.

Limits

Resistive burden assumed; inductive burden at a different power factor changes the result slightly. Applies to IEC 61869-2 class P/PR; class PX CTs are specified by knee-point voltage instead.

Sources

For competent high-voltage engineers. Results are calculations from the published methods shown and depend entirely on the data you enter. They are not a substitute for a system study, manufacturer data, the applicable standards and Engineering Recommendations, or your organisation’s safety rules. Never use a result to justify working on or near live HV equipment.

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