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Determine the maximum permitted value of voltage drop for each of the following circuits.

a) A 230V lighting circuit supplied directly from a public low-voltage distribution system. (1 mark)
b) A 400V three-phase motor circuit supplied from a factory transformer. (1 mark)
c) A 230V immersion heater circuit in a domestic installation.

a) 3% 6.9v
b)
c) 5% 11.5v

i dont know the answer to b, i am assuming its 5%, but why the difference between a and c? i would have thought all 230v circuits would be 3%?

there is not a lot of info on vd in gn3, i am going to have a look through amberleafs sticky

Have a look at Appendix 12 in BS 7671 it covers VD only.
 
Hi

for single phase 3% for lighting 6.9v
5% for power circuits and distribution 11.5v
and for the answer to b) its 20V which is 5% of 400v

no sorry i am wrong its not 20V as its there own supply not sure

I'm guessing it's 5% per phase so it's 11.5V at each phase where the equipment is installed, not measured between phases.
But I'm guessing.
Sorry it's 8% from a private supply source (This is a new one for me) so I am guessing it is 8% per phase = 18.4 V
But it's just a guess so hopefully I'll be put right by someone.
 
Last edited:
Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.
 
Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.

Spot on, the percentage is actually of 400 Volts ie. 32 Volts
 
the loop length for circuit 1 is 50 m long the ze is is 0.11 ohms

determine showing all calculations the r1 + r2 and the expected zs

7.41 x 50/1000 = 0.37 12.10 x 50/1000 = 0.60

zs = ze + (r1 = r2)

0.11 + (0.37 + 0.60) = 1.08

have i done this correctly?
 
Last edited by a moderator:
the loop length for circuit 1 is 50 m long the ze is is 0.11 ohms

determine showing all calculations the r1 + r2 and the expected zs

7.41 x 50/1000 = 0.37 12.10 x 50/1000 = 0.60

zs = ze = (r1 = r2)

0.11 + (0.37 + 0.60) = 1.08

have i done this correctly?

I can't confirm the mOhm values but the maths is correct.

You MUST use (R1+R2) when doing these calculations and remember that r1, rn and r2 are reserved for ring continuity figures, this exam is very big on correct terminology.
 
Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.

just ordered a set of regs from fleabay so they should be with me in the next couple of days - the only problem i found with the regs in the past was navigating them, i always favoured the on site guide for obvious reasons
 

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