Fluorescent tube light help

[plugsandsparks]

The correct method for calculating the protective device is as follows.

30 x 58 x 1.8 = 3132 W

3132 Divide be 230V = 13.61A

The protective device should be 16A

The normal type this type or lighting should be type C.

Cable calculations should be based on a load of 13.61A

Doesn't always work, mate. You can easily run out of ZS using cables rated for 13a e.g. 1.5. Its nearly always ZS i am chasing not load or volt drop when using C and D type breakers.
Worst case scenario is when the Db is at one end and the switch is at the other end

 

[steveberry11]

Doesn't always work, mate. You can easily run out of ZS using cables rated for 13a e.g. 1.5. Its nearly always ZS i am chasing not load or volt drop when using C and D type breakers.
Worst case scenario is when the Db is at one end and the switch is at the other end

If you follow this method and the correct cable calculations the Zs will be correct. It may mean you need a 2.5 or greater.
But it I'll be correct.