Lighting Current

[rantoftheday]

Ok this is a pretty basic question but can someone please refresh my memory

If I have a circuit with lets say 3 points

From DB to first point to second to third

As an example lets say that each point draws 5 a of current each

Will the cable from DB to point 1 be carrying 15A,
The cable from 1 to 2 carrying 10A
The cable from 2 to 3 carrying 5A

under full load

Is this correct?

 

[JUD]

Correct

 

[rantoftheday]

Thanks, So in a circuit of downlights the cable from the supply to the first downlight is handling most current?

 

[Octopus]

depends on the size of the cable and the MCB in the CU

 

[rantoftheday]

why would that matter?

 

[JUD]

Yep.

Also voltage drop decreases as you get closer to the end of the circuit.

 

[Octopus]

why would that matter?

If the consumer unit has a 6A MCB and you've got 1.5 T&E you've got a problem on your hands. I suspect your loads won't be as high as you indicate in your original post.

 

[rantoftheday]

the original post - values used hypothetically

 

[JUD]

depends on the size of the cable and the MCB in the CU

I think the 15A was meant as an example of 3 points pulling 5A each, not necessarily a lighting circuit.

The first length of cable would be carrying 15A as that would be feeding all 3 points.

The second length of cable would carry 10A as that would be feeding the last 2 points only.

The last length of cable would only be feeding the last point, so would only be carrying 5A.

 

[Octopus]

I think the 15A was meant as an example of 3 points pulling 5A each, not necessarily a lighting circuit.

The first length of cable would be carrying 15A as that would be feeding all 3 points.

The second length of cable would carry 10A as that would be feeding the last 2 points only.

The last length of cable would only be feeding the last point, so would only be carrying 5A.

Post #3 states downlighter.

Theres another post by the same individual asking another similar question.

 

[JUD]

OK, let's say 3 230V downlights -- 50W each.

50 ÷ 230 = 0.22A

so...

Cable 1 = 0.66A

Cable 2 = 0.44A

Cable 3 = 0.22A

 

[rantoftheday]

yeah i was using 5a as an easy number rather then a 0. something as in the case of downlights which would complicate things