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Current transformer burden

Add up the lead resistance, the relay and the other devices of the secondary circuit, compare the total with the CT rated burden and see the longest cable run it allows.

IEC 60228conductor resistance, class 2IEC 61869-2rated burden in VAIEEE C37.110burden by connection and fault

Example cases
Current transformer burden
E1Current transformer
Ω

Optional, from the test report. It is not part of the burden; it only gives the loop for the saturation check.

E2Leads
m
°C
E3Connected devices
Ω

Relay: VA at rated current, as the manufacturer gives it. Other: terminal blocks, test switches, meters or transducers in the same circuit.

Use a decimal point. The calculation runs in your browser and updates as you type.

Against the rated burden

— %

Maximum lead length

— m

Connected burden ZB

— Ω

Lead, one way

— Ω

Leads share of the burden

— %

Loop with RCT

— Ω

Burden against lead length
  • Selected size
  • Next size up
  • Rated burden and present length
The burden rises linearly with length. Where the magenta line crosses the rated burden is the maximum length; the grey line shows what the next conductor size gains.
Burden breakdown
  • Leads
  • Relay
  • Other
  • Rated
In ohms at rated current. The yellow mark is the CT rated burden.

Step-by-step calculation

What it means for the CT

    What it is for

    A CT only guarantees its accuracy class when the connected burden does not exceed the rated burden. For protection, a high burden also brings the CT closer to saturation. Use this tool to check an existing secondary circuit, choose the lead size or the relay location in a design, and compare a 1 A CT with a 5 A one.

    How it is calculated

    IEC 60228table 2 and annex B

    Lead resistance

    Rθ = R20 · (1 + 0.00393 · (θ − 20)) · RL = Rθ · L

    For metric sizes, R20 is the maximum DC resistance of IEC 60228 class 2 stranded copper conductors (1.5 mm² = 12.1 Ω/km; 2.5 = 7.41; 4 = 4.61; 6 = 3.08; 10 = 1.83), the usual construction of a control cable. For 14, 12, 10 and 8 AWG the tool uses the uncoated stranded copper values of NEC Chapter 9, Table 8 (3.14, 1.98, 1.24 and 0.778 Ω per 1000 ft at 75 °C), referred to 20 °C. 16 AWG is calculated from its area with ρ20 = 1/58 Ω·mm²/m, which corresponds to a solid conductor and gives slightly less than real stranded wire.

    IEEE C37.110burden by connection and fault

    Secondary burden

    ZB = kL · RL + kR · Zrelay + Zother
    • Single CT, or phase-to-ground fault on wye-connected CTs: the current goes out and back, kL = 2 and kR = 1.
    • Wye with neutral and balanced three-phase fault: the return currents cancel, kL = 1.
    • Delta-connected CTs and three-phase fault: each CT sees three times the branch burden, kL = kR = 3.
    • Delta-connected CTs and phase-to-phase fault: kL = kR = 2.

    IEC 61869-2rated burdenIEEE C57.13standard B burdens

    Conversions and criterion

    Z = VA / In² · S = In² · ZB · Lmax = (Zn − kR · Zrelay − Zother) / (kL · Rθ)

    The IEC 61869-2 rated burden is given in VA at cos φ = 0.8 (1.0 below 5 VA); the IEEE C57.13 burden in ohms, for example B-0.5. Here resistances and impedances are added in magnitude, which gives a slightly higher value than the phasor sum and keeps the result on the safe side. The indicator stays green up to 100 % of the rated burden and turns red above it.

    Limits of the calculation

    • Use the expected operating temperature of the conductor, not the ambient: at 75 °C copper has 22 % more resistance than at 20 °C.
    • The length is the one-way distance; the connection factors already count the return path.
    • Metering CTs guarantee their accuracy between 25 % and 100 % of the rated burden (IEC 61869-2), so a very low burden can also be a problem.
    • This does not replace the saturation check of a protection CT, which depends on the fault current, the X/R ratio and the excitation curve. For that, use the CT saturation tool with the loop calculated here.

    Frequently asked questions

    How is the secondary burden of a CT calculated?

    Add the lead resistance, the relay and the other series devices: Z = kL · RL + kR · Zrelay + Zother, with RL the one-way lead resistance at operating temperature. The factors kL and kR depend on the connection and the fault type: with a single CT, or a phase-to-ground fault on wye-connected CTs, the lead counts out and back (kL = 2). A burden in VA converts to ohms with Z = VA / In².

    Why does a 1 A CT allow longer leads than a 5 A CT?

    Because the power the lead consumes grows with the square of the current: the same cable draws 25 times more VA at 5 A than at 1 A. That is why substations with long cable runs prefer 1 A secondaries, and with 5 A the leads usually dominate the circuit burden.

    Does the CT winding resistance count in the burden?

    Not in the rated burden: the nameplate VA (IEC 61869-2) or the IEEE C57.13 B burden cover only what is connected to the terminals. It does count for the voltage the CT has to develop during a fault, so it is added to the loop when saturation is checked.

    We measure the actual burden on site

    When we test instrument transformers we check ratio, polarity, excitation curve, winding resistance and the actual burden of the secondary circuit, with a test report per unit. Field work in Colombia, or elsewhere on request.

    Request a quoteSee remote relay configuration

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