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Calculation of earth fault loop impedance.

Nothing on the box, so I thought new starters would like an idea how the above is obtained.

                                                                R ohms          X ohms



1 x 500 kVA transformer                            0.005            0.017

100 m of 2 x185 mm2 CNE                       0.008            0.004

   aluminium (Ø)              

100 m x 2 x 185 mm2 CNE

   aluminium (neutral/earth)                        0.008            0.001

Total                                                           0.021            0.022


Earth fault loop impedance

Zs(A) = V(R2 + X2)

Zs(A) = V(0.021sq + 0.022sq)

     = 0.030 ohms (at distribution board  A)

A 630A fuse(supply intake) requires an impedance of less than 0,054 ohms.


50 m x 185 mm2 swa 

aluminium (Ø)                                             0.008            0.004

aluminium (armouring)                                0.032

Total                                                            0.061             0.026


Earth fault loop impedance 

Zs(B) =V(R2 + X2)

Zs(B) =V(0.061sq + 0.026sq)

        = 0.066 (at B)

A 250A fuse( at distribution board B) requires an impedance of less than 0.16 ohms.


5 m 6mmsq pvc single cores

in 25mm conduit                                         0.023

5 m of 25mm conduit (light gauge)             0.007              0.008

Total                                                            0.091              0.034


Earth fault loop impedance

  Zs(C) = V(R2 + X2)

  Zs(C) = V(0.091sq + 0.034sq)

       = 0.097 ohms (at C)

A 32A fuse requires an impedance of less than 1.8 ohms.


Regards UKPN



  • UKPN


    I think you forgot to put a square route sign in to your equations when adding the R squares to the X squares.
  • But the idea of posting a worked example from time to time is a very good one - not every one in this business knows where the magic numbers come from, nor indeed what are the common DNO cable sizes (I do not  for example) and Amtech and the like do not encourage the use of imagination - I recall it took me a while to realise that in a typical housing estate the 10 metres or so of wiring up the drive of the individual house is far larger effect than the 100m or whatever back to the transformer, and indeed what are sensible no.s for a transformer.

    I agree Z= sqrt (R2 + X2 ) but the actual numbers look OK..

    Mike.

  • Using √ in Source mode will get you a proper √ too.

      -  Andy.
  • Thanks John Peckham, I am aware of the sq root missing sign, but I wanted to post the calculation so I made the best of it!



    Thanks UKPN.
  • Can I ask what is the volt drop like on this size of cable? I ask because where I am our voltage is around 242 max and 233 minimum well within limits assuming the TX is set on its 250 tapping im just curious our TX is 500 KVa with 400 amp fuses protecting 5 outgoing ways a fairly standard set up I expect.