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Home / Technical Articles / Quick estimation of the short circuit current at the end of a feeder
Quick estimation of the short-circuit current at the end of a feeder
Quick estimation of the short-circuit current at the end of a feeder (photo credit: elpro-kriznic.si)

Rule of thumb for quick estimation

If we consider a branch circuit going away from the point of supply and follow it radially along all the connections and branches, we shall find that further away we are from the transformer, lower is the value of the maximum possible short-circuit current.

Each length of conductor or each device in the circuit provides an impedance which helps to reduce the short-circuit current.

Example of Isc at the end of feeder

A prospective short-circuit current of 50 kA at the secondary terminals of a transformer at 400 V, will be limited to about 10 kA at the end of a connecting lead with a length of 10 m and cross-section of 10 mm2. If the same feeder has a cross-section of 25 mm2, the length of the wire is to be 25 m for reducing the current down to 10 kA.

Rule of thumb for a quick estimation of the short-circuit current at the end of a feeder
Figure 1 – Rule of thumb for a quick estimation of the short-circuit current at the end of a feeder

Specimen example //

Arrangement of a section of an installation
Figure 2 – Arrangement of a section of an installation

The expected short-circuit current at the end of the feeder is expressed by the relation :

Short-circuit current at the end of the feeder

Where:

  • IK” – Short-circuit current [kA]
  • UNTrafo – Rated voltage of the supply transformer on the low-tension side [V]
  • ZTrafo – Impedance of the transformer
    Impedance of the transformer
  • ZConducting – Impedance of the conducting lead

However, short-circuit at the point 5 is according to the above mentioned formula:

Short circuit calculation formulas

The value calculated above is rather conservative and includes a factor of safety as other sources of current reduction like the arc voltage, the contact resistance and the internal resistances of the different devices in the path of the short-circuit were not considered.

Reference: Basics of circuit breakers – Rockwell

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Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.

6 Comments


  1. Andy
    Apr 10, 2018

    Why is the I_shortcircuit for transformer different when calculated using kVA and different while using this relation you mention?
    For example, Xfmr_FLA=kVA/(1.73 x kVsec) = 630/(1.73×0.4) = 909A then Xfmr_I_shortcirc=Xfmr_FLA/0.06=15.15kA
    While I_sc_xfmr = kV_sec/(1.73 x Zxfmr) = 0.4/(1.73 x (3.1^2+13.5^2)^0.5) = 16.694kA
    Why are these two giving mismatch? what am I missing?


  2. Mustafa
    Nov 26, 2014

    Good Example but I dont understand some values. How are they calculated Uk , all R/mohm , X/mohm and all cable length and cross section best regards.

    could you help me pls

    best regards


  3. T.N.Ramesh
    Nov 25, 2014

    Yes I’m an Electrical Engineer


  4. Girish
    Nov 22, 2014

    Nice article. The procedure is similar to IEC 60909 method, only the correction factor in case of the transformer and voltage is not considered. Nevertheless, it is a good way of estimating the short circuit current.


  5. Vicente Martinez
    Nov 21, 2014

    Those are very instructive articles, they help me a lot Thank you


  6. LALO EL HIPPIE
    Nov 21, 2014

    La simpleza hace hermoso los desarrollos matemáticos. Calculo con Melshort de Mitsubishi dio el mismo resultado

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