Mm2

[resybaby]

Good spotting CP, i missed that difference between the two's calculators options.

Its quite the minefield with so many variables involved, my electrical knowledge is (obviously from my posts) basic, so i tend to look at the 'load' applied and work back from there using a basic rule of thumb for what cables can supply based on what i see in my house/experience. For instance my two existing (unused) immersion wires are 2.5mmTE each off separate mcbs. I know 2.5mm is good enough for 3kw immersions as my house hasnt burnt down yet, and its used for radial circuits etc generally, so that cable would be good for that load - but i wouldnt have put them in in that manner at home myself as they are 20m long and smothered in insulation - id have gone up a size.

Similarly, the sparky who put my Sunsynk in used 4mm cable back to the consumer unit for the AC feed. I noticed it getting rather warm to the touch in my loftspace, so i had a conversation with him. He explained that it was adequate for carrying the load and could infact carry a fair bit more (i forget the figures) and he kindly demonstrated all the calcs backing up his installation. The installation manual however suggests 6mm, so as My loft gets very very hot in the summer, i decided to have it replaced in 6mm. Better safe than sorry.

Well thats how this daft plumber mostly puzzles it out, embarrasing hey, but am always prepared to ask questions rather than just take a flyer and screw it up.

[Joeboy]

Joeboy,

If I want to check the reality of a result, I compare doubling or halving one or both.

To halve the resistance, one would need to double the cross sectional area, which would double the current carrying ability (and power) at constant voltage.

The analogy would be of having two conductors in parallel, each carrying the same current and delivering the same power. However this may not always quite hold, for calculators/tables, simply because conductors come in standard incremental increases, and the conductor size has to always be rounded up! (ie minus zero, plus to the next higher size).

Also some calculators may not include every possible scenario. Perfect insulation of the conductor would mean the only heat loss could be at the ends, so the temperature at the centre of the run would continue to rise - likely past the maximum permissible (where the resistance will rise and the insulation sheaths will degrade).

So yes, the resistance would fall by 40%. A graph of resistance v cross sectional area is a straight line, so a proportional relationship (double one, halve the other in this case). We have not doubled the cross sectional area so the resistance has been reduced, but not by half.

Imagine three 0.5mm^2 conductors (equivalent to one 1.5mm^2 conductor), each carrying 333W. 5 of those 0.5mm^2 conductors will be able to carry 5*1/3kW = 1 2/3kW = 1.67kW

The relationship between parallel resistors and actual overall resistance is the sum of the reciprocals (whereas for the total of resistors in series is the sum of those resistors).

Without setting out all the Ohm’s Law related maths, increasing the cross sectional area from 1.5mm^2 to 2.5mm^2 conductors is a 67% increase of the former to the latter. The reciprocal of the 1.667 is 0.60 which is 60% of the original - a 40% decrease in resistance.

Someone else could likely explain it better than I can….

That's brilliant Oliver, thank you. I'll find the excellent ohm/volts/Amps cartoon and post too.

[Oliver90owner]

That’s a good one. I used to compare volts to water pressure, resistance to a valve or different diameter pipes and flow through a hydro-generator relating to power.