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Hi Guys,

could anyone explain to me the 1.2 volt drop rule of thumb? i understand the 3% for lighting and 5% for other circuits.
any help would be welcome studying for 2391.


Thanks in advance.
 
Hi Guys,

could anyone explain to me the 1.2 volt drop rule of thumb? i understand the 3% for lighting and 5% for other circuits.
any help would be welcome studying for 2391.


Thanks in advance.


I'd forget that rule of thumb job. Just do the calculation...length of cable run x design current x mV/m drp(from the appropriate big book table)

your %s are right.

Make up some imaginary circuits...100m run with 300Watt load, RFC in less than 100mm2(??)
then post your results.
 
You are right re the volt drop percentage.

The rule of thumb is for calculation ZS.

In the BGB it will give you ZS values for fuses mcb's ect you then multiply by 0.8 which will give you a lower figure.
This takes into account the the increase of resistance of the conductor with the increase of temperture due to load current and errs on the side of safety.
 
Hi mate volt is 3% lights and 5% power ie 6.9v lights and 11.5v power


asfor 1.2 do you not mean MF multiplying factor for increase in resistance due to operating temp for 70C cables when bunched or a core in a multicore Ie T/E
and 1.28 for xlpe 90C swa
 
the old 1.2 factor, I think he's getting confused with working out Zs sss, sod all to with voltage drop calcs

Depends how you're working it out.

If using volt drop tables you don't need it.

If you're working out by resistance then you do i.e ((mΩ/m x 2 x 1.2) x L x Ib) ÷ 1000
 
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The 1.2 is the multiplying factor used in volt drop.
If you measure resistance R1 + R2 at 20C or use tables measured at 20C you have to multiply the figure by 1.2 to give you the resistance at operating temperature 70C.
This R1+R2 *1.2 is then used in the equation V=IR.
.....I..... being the design current for the circuit or preferably the measured load current.

V is the volt drop along the cable.
 
The 1.2 is the multiplying factor used in volt drop.
If you measure resistance R1 + R2 at 20C or use tables measured at 20C you have to multiply the figure by 1.2 to give you the resistance at operating temperature 70C.
This R1+R2 *1.2 is then used in the equation V=IR.
.....I..... being the design current for the circuit or preferably the measured load current.

V is the volt drop along the cable.

Isn't that what I just said in post #6 :smiley2:

Depends how you're working it out.

If using volt drop tables you don't need it.

If you're working out by resistance then you do i.e ((mΩ/m x 2 x 1.2) x L x Ib) ÷ 1000
 
Guys thanks for all the input a big help.. bit clearer, im i right in saying the multiplying by 1.2 converts from ambient temp 20c to operating temperture of 70c?
 
The 1.2 is the multiplying factor used in volt drop.
If you measure resistance R1 + R2 at 20C or use tables measured at 20C you have to multiply the figure by 1.2 to give you the resistance at operating temperature 70C.
This R1+R2 *1.2 is then used in the equation V=IR.
.....I..... being the design current for the circuit or preferably the measured load current.

V is the volt drop along the cable.

The verification of voltage drop will require a measurement of the 'circuits impedance'. Therefore it will be R1&Rn, and not R1&R2.
 
ive left my electricians design guide in the van, and i know that there is a formula for working out resistance of a conductor from mV/A/m figures. Does anyone have the calc to hand please?
 
ive left my electricians design guide in the van, and i know that there is a formula for working out resistance of a conductor from mV/A/m figures. Does anyone have the calc to hand please?

The volt drop figures are the values of resistance of the line and neutral conductors per meter at 70°C.

The formula to work out the volt drop figures is (R"1 + R"n) x Ct

R"1 = resistance per meter of line conductor at 20°C.
R"n = resistance per meter of neutral conductor at 20°C.
Ct = correction factor for operating temperature of cable.

Example

Resistance per meter of 2.5mm² is 7.41mΩ/m so the formula would be (7.41 + 7.41) x 1.2 = 17.78 (rounded up to 18).

This is the value of the mV/A/m for 2.5mm² from the volt drop tables.
 
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Does this wrk with cables larger than 16mm2? will this be afected by reactance?ive transposed tha above to get miliohm/m = mV/A/m X 0.833
2

is this correct? ive onl;y got my BGB to test this with, and i dont think that it lists the resistance of cables in it.

Can somebody post me a few values for resistance of single core cables?

Thanks,

John
 
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