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Ok I have another VD question. I'm installing 9 LED floods around my friends garden which totals at just over 100m in length. Obviously for each branch the VD varies so for argument sake I worked It out as if all the cable would be taking all the load (rather than dropping relevant current after each light, worse case scenario) the load is minimal being LEDS but to be safe I'm putting 2.5mm 3c hightuf in to cope with the distance and to help for the zs reading. Now I am also using a wise box control box (wireless switching) now am I right in thinking I can supply this box from the CU with a length of 1mm t+e as this is a very short distance and is my thinking correct that as long as no voltage has been dropped through the 1mm to the wise box then I have a full 230v (give or take) for the garden circuit and as its run in 2.5mm it will stop any VD issues. Basically what I'm asking is, whatever voltage you have at a certain point regardless of cable size, then that's what you have from that point onwards. I.e if I lost 3volts through initial 1mm cable then I'm down to 227v at wise box (for example)
 
your thinking appears logical, but for what it costs, i'd run from CU to wise box in 2.5mm T/E.
 
Yeah I thought that would be the general consensus. I was just trying to get my head round it. You could technically run it in 1mm until the voltage started to drop then up the cable size to reduce resistance and VD. Ohms law. It's just as a design point it would be best to keep it all the same.
 
If you want to reduce volt drop then increase the CSA from the origin of the circuit, for example I added a heavily loaded 10amp lighting circuit to a very big house the other week, DB was at opposite end of house, and ran the feed in 2.5mm to an adaptable box then came off that in 1.5mm to the individual parts of the circuit. The load on the 2.5mm was terminal, so the whole length was subjected to the full effect of the voltage drop, whereas the individual branches were not.
 
Yeah like I was saying your voltage at any given branch is subject to what has been lost from the origin. So providing there is no major loss from origin you have 230v(ish) for all branches and then subject to what current each branch is taking x length of cable, each branch will have its own VD subject to those factors. So you could run supply to garden in 1mm with no real VD. Then use 2.5 for long lengths to combat any VD. it's a backwards way of doing it and not great from a design point of view. But it would work the same. Ohms law.
 
Ok I have another VD question. I'm installing 9 LED floods around my friends garden which totals at just over 100m in length. Obviously for each branch the VD varies so for argument sake I worked It out as if all the cable would be taking all the load (rather than dropping relevant current after each light, worse case scenario) the load is minimal being LEDS but to be safe I'm putting 2.5mm 3c hightuf in to cope with the distance and to help for the zs reading. Now I am also using a wise box control box (wireless switching) now am I right in thinking I can supply this box from the CU with a length of 1mm t+e as this is a very short distance and is my thinking correct that as long as no voltage has been dropped through the 1mm to the wise box then I have a full 230v (give or take) for the garden circuit and as its run in 2.5mm it will stop any VD issues. Basically what I'm asking is, whatever voltage you have at a certain point regardless of cable size, then that's what you have from that point onwards. I.e if I lost 3volts through initial 1mm cable then I'm down to 227v at wise box (for example)

You need to work out the volt drop across the entire circuit, so you need to know how much power the LED floods will use combined and then any power for the control box, then you can calculate the volt drop correctly and accurately. Guesstimates are for bodgers.
 
Most sparks just prefer to use the same CSA cable throughout a circuit. As do I, but there are times when I've upped a cable size for a specific reason just for part of a circuit.....voltage drop is an example, but also when there is a reduction in CCC of the cable due to external factors for part of the run, e.g. cable passing through insulation of through a higher temperature area.
 
Outspoken I understand that completely and I've gone fully worse case scenario so I know everything is fine and no VD problems with this particular situation. I was just seeing if my thinking was correct and providing no VD from origin cable then you have full 230v for branch cables of circuit.
 
you will get VD on the 1.0mm once you load the circuit
 
I know that too but If you worked it under full load conditions and as run to wise box was minimal (very low resistance) so at the wise box we were still getting 230v(ish) then you would have that 230v(ish) for all your branch circuits. Then they would VD individually, respectively of their load/length factors.
 
Haha you guys can be so particular. Are we allowed to be hyperthetical and just slightly brush over precise figures to just get a grasp on the bigger picture. We all know voltage varies from one house to another. Just fundamentals.
 
we're pulling your chain, monkey, and you sure is dancing.

and just to be pedantic it's 240 ish.
 
Haha! I'm all up for the banter mate. I love it. Ha. And I know my thought processes are very weird. I just like to know that my thinking is right. I hate building knowledge on rubbish foundations. I'm just a beginner with hunger to know more.
 
On top of all this can someone please explain this to me. So you have a lighting circuit which over a long run gets down to 220v from supply to junction box. You have a cable which goes to a light and another cable which carrys on the feed. Now say the next run to the next point drops a further 2 volts. You take the 2 from the already dropped 220v correct? What I'm trying to get at is there will be a further VD from first junction box to the light I mentioned but as we T'd off at junction box both the light and the next feed take a refrence point of 220v to work from correct?? We don't include VD to light in circuit calc as it would be wrong. Because where we T'd off it is at 220v not 220 - the VD to light.... So the feed out will also start at 220 not including VD to first light. I find this all very confusing. Does anyone follow what I'm trying to get at.
 
the cable to the first joint/light will have a VD depending on the total load of the complete circuit. at the 2nd, the VD is dependent on the load of the 2nd light , 3rd light and so on. at the 3rd, it's lights 3,4, etc. so the further down the circuit you go, the VD decreases.
 
I understand that. Haha I'm so confusing. What I'm finding hard to work out is. The voltage at the joint is 220. Then to the light it drops a little more, now thats the total VD for that light. But to the next joint we don't include the first lights VD we take it from supply, to first joint, to second joint. Missing first light. Any clearer?
 
sketch it out. select a hypothetical load and cable size, then calculate the VD at each light. you will see a progressive fall in voltage at each load.

you work out the first one using the total load and the cable length from the CU. then the next point is the remaining loads and cable length from CU and so on
 
You wouldn't calculate vd that way unless you add extra load to an existing cct, vd is calculated for the full load at it's furthest point. The vd is calculated from the origin of supply in all cases where you can expect to have no more than 3 or 5% vd, a good example is where you have sub mains.
 

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