Short circuit

[Wilson12]

Sorry just learning but with a short circuit.
V=i x r so dosnt that mean that the maximum short circuit in a circuit with even no resistance would be 230 Amps.
How comes I hear of higher currents in faults and breaking capacity for the breakers are 6ka.

 

[Darkwood]

Your confusing yourself.

If you measure your Zs at 0.1 and your voltage is 230 then your short circuit would be 2300amps (230/01).. when your install is next to the sub station and say you have a S/C then you could be looking at 0.03 then you are looking at 7k +... noting that a short between phases would be more than double this value and then some (installation dependent).

 

[davesparks]

The resistance would have to be 1 ohm for a current of 230A
The earth fault loop impedance at the supply terminals for a new PME supply will typically be below 0.35 ohm.
At 0.35ohm you get a fault current of 657A
At a possible (and pretty realistic) 0.1 ohm the fault current would be 2300A
If the installation is very near to the substation then the fault current can be even higher.

The DNO will usually state a maximum prospective fault current of 16kA for a single phase supply as this is the upper limit they usually work to when designing supplies.

But you also need to consider that 230V is a myth which only exists in the textbooks, in actual fact the substation will be fixed at an output of 250V so the fault currents will be a little higher than that.

Also consider that the fault current could be a short circuit rather than an earth fault, so the potential could be theoretically 400V (433V in reality) if it is a three phase supply



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[Marvo]

Sorry just learning but with a short circuit.
V=i x r so dosnt that mean that the maximum short circuit in a circuit with even no resistance would be 230 Amps.
How comes I hear of higher currents in faults and breaking capacity for the breakers are 6ka.

I think your maths might be flawed.

Using the formula, if the resistance is zero then the fault current would be the voltage divide by zero. (I=V/R >> I=V/0)

The actual voltage would in this case be a moot value because any voltage divided by zero would give you an infinite fault current.....

In the real world of fault currents the limiting factor is actually the impedance rather than purely the resistance.