Recent content by mbrooke

I know, but the 5% is applied to the middle, end or evenly spaced apart as an algebraic sum?

But do I assume a full 20amps of load lumped at one end?

I did find this: http://cablesizer.com/iec/

This is a 20amp radial in 2.5mm2 T&E. My apologies for not stating it earlier.

The regs say no more then 5% voltage drop, however how do I apply (calculate) this for socket circuits?

OK, doing more research I came across this, not sure if it helps though, the tables used for countries that use 2.08mm2 wire. But it has me confused. At 75*C, 2.08 is listed at 20amps. I take it the conductor insulation is rated higher...

That makes sense. Now my understanding is the factor is less restrictive for 1.5mm2 because smaller conductors can dissipate heat better? Or is it the other way around? Correct to think that heat dissipation of 2.08mm2 is in between 1.5mm2 and 2.5mm2? Edit: just to confirm, by 1.35 you mean...

Thanks! That is what I was looking for. :) I think it does hold some value in my opinion. However this equation only works for clipped direct I take it? What would it be reduced to when using other methods like method A? And the power curve is based on a 70*C conductor operating temperature?

Multi core. Live and neutral plus PE in conduit (3 conductors total).

I dont think so. I did some digging and found this. 12 gauge is 3.31 mm2. This wire here is 2.08mm2.

I dont think those are accurate, or would comply with the regs. For example 2.08 seems to be listed at 15 amps, but that is obviously based on method 101 as 1.5 is already 15 amps for installation method A. Dont the regs base the values off a certain equation? Either that use the current value...

How would one determine the current carrying capacity of 2.08mm2 conductor?

Thanks! :)

How do earth fault loop impedance requirements and disconnect times change when the live to earth voltage is 133 instead of 230? Thanks!

Do you have a wiring diagram for the unit? That usually has detailed info.