G

Gringoking88

Hi all,

I am currently doing my 2391-52 and have been given a mock paper, one of the questions is around vold drop as per below, can someone please help me understand how to get the answer as I just cant seem to work it out!

Thank you advance!

Voltage drop of a single-phase distribution circuit supplying a power distribution board in a remote building is to be verified as part of the periodic inspection and testing within a workshop complex. The installation forms part of a public 400/230 V TN-S system.
The circuit has a measured R1+Rn value of 0.15 Ω and an Ib of 60 A. The circuit protective device has an In of 80 A, see Figure 5.
What is the maximum acceptable voltage drop for this distribution circuit if the highest circuit voltage drop on DB-3B is 5.0 V?
IGZvA3KxNPILPwjwr8XDN6mu-02bn3XflSgHbYpnFZi6bk5z1YYIdwA4J_53ocLYeLgwWyhtD-eV5K0sjWKMOKxRdhTVTf9WlmLKhnV35PdzPiOGmVnEBkMK8S2DwOLQudvhiNiVb77KGBz5sQ

a) 11.5 V
b) 6.9 V
c) 6.5 V
d) 1.5 V
 
How far have you got? Have you found the formula for voltage drop?
 
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Total Maximum permitted VD from Main DB to end of final circuit from DB3 = (say5%)
Therefore max permitted VD on distribution circuit = 5% - 5.0volts (from question) =
 
How far have you got? Have you found the formula for voltage drop?
Hi,

Thanks for the reply.

I get the formula e.g. ((mV/A/m) x current x cable length) / 1000, but I dont get a result of 1 of the 4 possible answers given;

a) 11.5 V
b) 6.9 V
c) 6.5 V
d) 1.5 V

Thanks again!
 
Total Maximum permitted VD from Main DB to end of final circuit from DB3 = (say5%)
Therefore max permitted VD on distribution circuit = 5% - 5.0volts (from question) =
Im sorry this is probably the bit that is confusing me, could you explain please?
 
Kind of a trick question really. You're given much more information than you need. The solution is to weed out all the unnecessary info. What is the question actually asking?
Is it asking you to calculate the volt drop?
 
Im sorry this is probably the bit that is confusing me, could you explain please?
The maximum permissible VD at the farthest point of any circuit (except a lighting circuit) is 5% or 11.5 volts
The volt drop on one of the circuits from DB3 is already calculated to be 5volts (in question)
Therefore 11.5 - 5 = 6.5vots permitted to be 'dropped' on distribution circuit.

As #6, too much information given beforehand.
 
So it's asking for maximum permissable v.d and its single phase so 5% of 230v = y (This is the total v.d you are allowed). You are told that a circuit from DB3 has a v.d of 5v so y - 5v = max v.d allowed for the distribution circuit.
 
This is one of the reasons stated in the inspectors report on 2391 failures, not reading the question properly.
 
simple. you are allowed 5%, or. 11.5V in total . as the DB3 has a VD of 5V already, your distribution circuit max. VD id 11.5 - 5.0 =6.5V. answer (c),
 
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Any advice for equations and calculations i know the basics but always trip up on them failed 4 times on multiple choice
 
People have already responded to your other thread.
 
Last edited:
Many years ago I set the questions for the Fellowship entrance to one of my Institutes, it was always the older, supposedly more intelligent candidates that failed and always for the same reason, not reading the question correctly, ergo if you fail you are more intelligent. ?
 
Many years ago I set the questions for the Fellowship entrance to one of my Institutes, it was always the older, supposedly more intelligent candidates that failed and always for the same reason, not reading the question correctly, ergo if you are old you are more intelligent. ?
corrected for you.
 
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The problem is that I have so much information stored in my Brain that some of it has to leave to allow new information to come in, that's my excuse for being forgetful and I am sticking to it.
 
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I was taught to RTFQ
 
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our teacher was illiterate illigitimate.
 
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