High Volt Drop On Lighting Circuit 2A Mcb?

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darren23

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Hi I'm in the process of installing a small bd in a out house it has a exsiting 2.5 3 core swa supply of a 16A mcb which I would like to utilise I have calculated the Vd for the swa and it's 6.08v I want to run 2 sockets and 3 florescent lights by using a small DB with a 16A mcb for sockets and a 6A fir lighting both protected by a rcd main switch the problem is after I calculated the volt drop for the lights it goes overthe 3% as you calculate it against the size of mcb not the demand.

If I was to use a 2a mcb volt drop goes just under 3% so all is good in the world again.

Can any one see a problem with this.

Also I should say that the out building has a large steel lintel which will need to be bonded so installation will have to be a tt

 
The VD for supply is 6.08v.

If I used a 6A mcb and calculated VD for the whole 6 amps it takes the total over the 3% drop and as there are only 3 florescent lights a 2a mcb keeps the drop just under 3%.

Thanks in advance

 
Yes after small DB would like to put new supply cable in but after digging up a 10m stretch of tarmac I then would have to go around the out side of house this would too toss and as it's there and it's insulation resistance in over limit and good r1 and r2 seams a shame to not use it.

 
Yes but don't you add the distribution Vd to the light circuit Vd to give a total Vd from source to end of circui.

Only really need to know if using a 2a mcb is ok for lighting circuit (do florescent lights have much start up current as the three together only add to 0.75 A

 
As NozSpark said, VD is calculated using the load, it says "The VD is determined from the demand of the current-using equipment, applying diversity factors where applicable, or from the value of the design current of the circuit". It also says in Note 1 that greater VD is acceptable for high inrush currents. So it should be sufficient to use 0.75ohms in your VD calc.

So I would use 6a mcb for the lighting circuit, at least there are references to max Zs etc for a 6a mcb in bs7671 and no references to a 2a mcb in there.

 
Last edited by a moderator:
As NozSpark said, VD is calculated using the load, it says "The VD is determined from the demand of the current-using equipment, applying diversity factors where applicable, or from the value of the design current of the circuit". It also says in Note 1 that greater VD is acceptable for high inrush currents. So it should be sufficient to use 0.75ohms in your VD calc.

So I would use 6a mcb for the lighting circuit, at least there are references to max Zs etc for a 6a mcb in bs7671 and no references to a 2a mcb in there.
So, if a 2a MCB is used, is that a deviation from 7671,?

I'd say it is,

As said, stick with a 6a MCB and use actual load for your calcs. 

If I done my shed on MCB sizes it would be 70amp, :eek: for a 12x8 shed :slap  

 
I did go down this route first then read 7.2.3 in the on site guide

lighting circuit

a maximum volt drop of 3% of the 230v nominal supply voltage has been allowed

the circuit is assumed to have a load equal to the rated current (In) of the circuit protective device, evenly distributed along it circuit. Where this is not the case the lengths will need to be reduced where voltage drop is the limiting factor . .....?

 
Why this obsession with a 2A  MCB  ?  ;)

Calculate the VD  on the design load in the outhouse ,  Lighting is sod all  ( well 1A )   what will be plugged in to the sockets.  ? One 2KW fan heater ?

And why not just bond the lintel if it worries you .  Is the main supply PME ?    Do you think it may become live ?

I assume the supply cable is miles long as the VD is so important ?

 
I did go down this route first then read 7.2.3 in the on site guide

lighting circuit

a maximum volt drop of 3% of the 230v nominal supply voltage has been allowed

the circuit is assumed to have a load equal to the rated current (In) of the circuit protective device, evenly distributed along it circuit. Where this is not the case the lengths will need to be reduced where voltage drop is the limiting factor . .....?


If you are taking the time to read the On-Site-Guide...

PLEASE also read...

page 140  (green)  page 146 (new yellow)

Volt drop..

To calculate the voltdrop use ((mV/A/m) x Ib x L) / 1000

mV/A/m is found in table F5(ii) page 149 Green / page 154 Yellow.

Ib is DESIGN CURRENT....   (NOT In  Which is the over current device rating.)

L is cable length,

You don't calculate Volt drop using over current device rating (MCB rating)..

It is from the design current of your load!!!!

HTH.

Guinness

 
Neither are deviations.

The MCB values listed in BS7671 are merely those common on installations.

As long as the MCB meets the disconnection requirements of 60898 or 60947 then it complies with 7671.

Basically the same applies to MCCB's they meet 60947.

60947 is a recognised standard in 7671 thus equipment to 60947 meets the requirements of 7671 without a deviation.

Otherwise 7671 would be limited to installations meeting the OSG guidance as 60898 breakers only really "do" domestic up to 100A.

There is nothing wrong with using manufacturers data for Zs for example.

 
Couldn't we just shorten BS7671 a bit then,

One line should do it

"manufacturers instructions must be followed"

Most all electrical stuff comes with instructions, which normally say,

"to be fitted by a competent electrician" 

 
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