Please correct me if I'm wrong. During short circuit fault, the fault current that flows in a cable become twice as much as the load current.
Now using Method 2 Installation if I'm estimating cable size for plugs with estimated power of 3000W, I get 13.04A which I multiply by 2 to get 26.08A Ic current. The table 6.2(a) on SANS 10142 leads me to 4mm.
Does this mean that if I use a 2.5mm cable, it will get burnt immediately in the event of short circuit with an overcurrent/overload protection (circuit breaker) failure?
Why do we need to multiply load current by 2 during Load Estimation?