Voltage drop/2391

[Pretty Mouth]

So this would be as if both cpc and line conductors are the same Size but sometimes its not the case

CPC, line conductor, neutral conductor, it doesn't matter what they're called in the table. They're all copper, and they all have the same resistance per meter for a given cross sectional area. 1 meter of 2.5mm² twin and earth is 2 meters of live conductor: 1 meter of 2.5mm² line and 1 meter of 2.5mm² neutral.

Read back over my post carefully, it should explain things

 

[Zerax]

To the OP: Focus on understanding the principle.. NOT a formula. The OSG & BS7671 & all the IET books that I've looked at... try to make things easier by giving you the same thing in a different way. But IMHO this just makes things far more confusing...

I think all the standard publications could be made half the size they are if they removed all the variations. Oh, and added page numbers to the index !!

 

[Pretty Mouth]

To the OP: Focus on understanding the principle.. NOT a formula. The OSG & BS7671 & all the IET books that I've looked at... try to make things easier by giving you the same thing in a different way. But IMHO this just makes things far more confusing...

Completely agree with this. It's messy, and a nightmare when trying to learn. Eg the table on p196 should have just 3 columns: conductor CSA, resistance/m copper, resistance/m Aluminium. No need to mention line conductor, protective conductor, R1 or R2.

 

[Lister1987]

VD itself is as we know mV/A/m * Ib * L

Max VD is either 6.9v (3%) or 11.5v (5%) for lighting & power respectively

We can rearrange the formula (using Max & Actual VD) to find Max circuit length for the calculated VD



Let's say we have a 6 mm² T&E with pvc insulated and sheathed cable feeding a a lighting circuit which has a mV/A/m value of 7.3. The actual circuit current is 12.5 A, and the length of run is 14 m.
Volt drop = 7.3 x 12.5 x 14 / 1000
= 1.28 V


If a 14 m run gives a volt drop of 1.28 V,
the length of run for a 11.5 V drop will be:
11.5 x 14m = 105m

Completely agree with this. It's messy, and a nightmare when trying to learn. Eg the table on p196 should have just 3 columns: conductor CSA, resistance/m copper, resistance/m Aluminium. No need to mention line conductor, protective conductor, R1 or R2.

I'd drop the individual rows for line and CPC, as you say you just want a value for 2.5 and 1.5, doesn't matter if it's line neutral or CPC.

I disagree on dropping R1+R2, I find that useful for determining the length of a circuit (or an implausibly low/high reading), dividing my R1+R2 value by the listed value

 

[Pretty Mouth]

I'd drop the individual rows for line and CPC, as you say you just want a value for 2.5 and 1.5, doesn't matter if it's line neutral or CPC.

I disagree on dropping R1+R2, I find that useful for determining the length of a circuit (or an implausibly low/high reading), dividing my R1+R2 value by the listed value

Problem is, you need both line and CPC columns present in order to have an R1+R2 column. Can't drop one without dropping the other!

Personally I think it clutters the mind when learning. And it's not difficult to calculate (R1+R2)/m if you have the resistance/m of each of the conductors: just add them together.

I would have the resistances in ohms/m as well. It would remove the annoying step of dividing by 1000 to convert from milli ohms to ohms.

 

[AJshep]

The other basic method of working out the voltdrop they tend to use on the 2391 is using ohms law, I believe they use this to test the candidates basic knowledge

You would need to use your R1+RN mutipled by the design current and then mutilpying by 1.2 to allow for the circuit to be running at 70°c

A basic question would be
"Whats the volt drop on a 13 amp immersion circuit with an R1+RN of 0.30 ?"

A=4.68v
B=3.59v
C=4.45v
D=5.68v

Answer
13x0.3x(1.2) = 4.68v