Voltage drop and short circuit

[lyledunn]

If a number of single phase installations are connected to a distribution transformer such that their respective loads cause successive voltage drops in the distribution cable, will a short circuit in the last installation on the distribution line be affected by those voltage drops? In other words, will the fault experience the open circuit voltage of the transformer or some proportion of it?

Regards,

Lyledunn

 

[jghrist]

The fault current will be lower than at the transformer because of the impedance of the distribution cable. The voltage will be lower than at the transformer because of the impedance of the distribution cable. There is a common cause, but the voltage drop does not effect the fault current directly.

 

[davrom]

I agree to jghrist. The voltage drops do not affect directly the shc current value.

Basically and very much simplifying, the short-circuit reduces to a circuit formed by an equivalent impedance connected to the source (it is a "short" circuit). The shc current is basically I=U/z. The value of z depends of distance between the source and point of shc; the distance is not to be understood as "length" but as number and arrangement (series, parallel) of impedances that give/compose the equivalent impedance.

For your case, the impedance is not only a series impedance given by the cable's impedance (long length, z=L*z0) but also by the impedances of the succesive loads along the distribution cable which are to be considered in parallel connection (if the loads are connected to the distribution cable/network in the moment of shc). However, the impedance at the last point is bigger than the impedance at transformer side, even considering the parallel connected impedances (zpar=sum(zi)/n, i=1...n), so the shc current will be lower.