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GeniusSimple. 5 + 5 is 10 so your appliance 20 metres away would not reach.
GeniusSimple. 5 + 5 is 10 so your appliance 20 metres away would not reach.
Is not the most obvious way to measure voltage drop to use a voltage meter (e.g. a multimeter)?The way to test volt drop is to use a loop impedance meter, but that is not something the normal house holder will have ....
To be fair, they might be assuming that the fixed wiring has been designed, installed and tested such that the VD at a socket is acceptable (probably within the BS7671 guidance of 5% max) but, since it might be 'only just' acceptable, adding any (unknown) extension might take it 'over the line'. Hence, if VD is as crucial as eric suggests, then that wouldn't be an unreasonable approach.So the makers are sure that voltage drop on 20m of 1.5mm² T&E will be OK, but the drop along 5m of 2.5mm² T&E plus 5m of 1.5mm² flex will not be. How do they know this?
Dont be stupid, the appliance would be 10 metres awaySimple. 5 + 5 is 10 so your appliance 20 metres away would not reach.
I think you have maybe missed the point of my last two posts in this thread ...As to measuring volt drop, one would need to measure incoming supply and socket at the same time to work it out, and that would be hard ...
I'm not quite sure how you get that figure [by my reckoning, 26 x (0.94 - 0.35) is 15.34V, not 11.5V] but, in any event, such a calculation requires knowledge of the external L-N loop impedance. With TN-C-S, it should not be over 0.35Ω (but could be lower), and if it's not TN-C-S, then I think one would probably have to measure it.... however with a loop impedance meter if incoming is say 0.35Ω and line - neutral at socket in centre of a ring final is 0.94Ω we can calculate the volt drop with a 26A load (20 at centre and 12 even spread) that the volt drop will be 11.5 volt ...
Indeed, but unless one has fairly sophisticated measuring equipment, one probably has to rely on calculation from knowledge (or an estimate) of the start-up current and the R1+R2, since the start-up current is probably too brief in duration to measure directly with just a meter.One to measure the volt drop with a refrigeration device would need you to record the volts as one is only interested in the volts at start up, once running there is no real problem with volt drop, it is only volt drop at start which is important.
'Correction' for what - temperature?Two there is a correction factor of 0.917 when working out the volt drop for the final ring, so worked out at 16.5197 mV/A/m not 18 hence the slightly different figure ....
In a way suppose it is temperature, the it is all to do with cable not being loaded to the max, the calculation was in the appendix of the 17th Edition I remember it used sq root etc, so a bit complex. To do without using excel or java script all set up to do it automatic would be hard, and as with any complex calculation it is easy to make an error, so having it pre-set in java script allows me to test knowing what the results should be, before entering figures where the result is unknown. Sorry not got book to hand and too complex a formula to remember.'Correction' for what - temperature?
Kind Regards, John
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