2382 question

[BS3036]

I got this sample question wrong, and I'm not understanding the answer given. So can anyone explain how to get the right answer?

If a short circuit occurs in a low voltage installaton between a Line conductor and the Neutral conductor, the voltage between the other line conductors and the neutral can reach what value?

a) √3 U0
b) 1.45 x U0
c) 3 U0
d) √U0

 

[ColJack]

sqrt3 x Uo..
with a short on one phase to neutral, the neutral ends up at whatever the voltage is on that phase..
so it's effectively the 3 phase, phase to phase voltage.. which is sqrt3 x Uo..
of course I'm assuming that it's a theoretical circuit with no SC protection and it doesn't melt the wiring..

and of course I may be wrong.. it happens sometimes..

 

[BS3036]

Yes that's what I chose. They said that the correct answer was b) 1.45 x U0. I just needed some reassurance.

 

[plugwash]

It depends on your assumptions.

If you assume the phase and neutral conductors involved in the fault have equal impedance then 1.45 x U0 sounds about right (though I CBA to do the phasor calculations to check right now)

If you assume the neutral conductor may have a much higher resistance than the phase then it could be as high as √3 U0

 

[matt1e]

link here for ya clicky

 

[BS3036]

matt1e: Nice one! Thanks. A page I need to study then. (But I wish they explained why, instead of just saying it.)

 

[ColJack]

oh yeah.. 2382 is the regs isn't it..

that exact wording is in the regs.. should have looked it up.. was taking a stab in the dark..

 

[plugwash]

matt1e: Nice one! Thanks. A page I need to study then. (But I wish they explained why, instead of just saying it.)

It's actually a good question, right now i'm trying to figure out

If we assume that the live and neutral involved in the fault have the same impedance and the supply transformer has a much lower impedance than either than the voltage at the fault will be half the normal supply voltage.

The other live is 120 degrees out of phase so we need to do a little phasor calculation to find out the voltage between them, that is the magnitude of the difference between the two phasors

|0.5Uo-Uo(cos(120)+isin(120))|
=|0.5Uo-Uo(-0.5+0.866025404i)|
=|0.5+0.5-0.866025404i|Uo (since Uo is positive)
=(√(1²+0.866025404²))Uo
=(√1.75)Uo
=1.322875656Uo

So clearly that is not thier assumption, lets try assuming a half-sized neutral (not very common any more but still present in older wiring afaict). Now the voltage at the fault point will be higher, two thirds of the supply voltage.

|0.6667Uo-Uo(cos(120)+isin(120))|
=|0.6667Uo-Uo(-0.5+0.866025404i)|
=|0.6667+0.5-0.866025404i|Uo (since Uo is positive)
=(√(1.1667²+0.866025404²))Uo
=(√2.1111)Uo
=1.452962491

GOT IT!

 

[matt1e]

Plugwash wrote loadsa techy stuff

cheers plugwash! I was about to reply to BS that the need to get into phasors was needed to understand the equation which was way above my head at the mo (little old wine drinker me),well would be sober too with out a little swotting up! nice to see calcs like those again, cheers!
On another note I got this question right but got the answer via a different assumption.
my thinking was;
"145 is sticking out for some reason" (bad memory nearly 50)
a) √3 U0 - doubt its sqr3 because one phase is shorted,3 wont come into it
b) 1.45 x U0 - I'm sure its this
c) 3 U0 - same reasons as in (a)
d) √U0 - squareroot again of single figure! cant be' that presumes harmony which this situation can't be
so answer must be (b)
mmm just check.
What has 145 got to do with it again?
right' one shorted phase,
which leaves two that are ok
so sqr2?"
= 141
"mmm close enough because im sure you have to factor in other things due to the shorted coils impedance,
and
145 might be a misprint!

see sometime you just know the answer,you just somehow cant remember (or understand for some) why
Always good to revise though

matt

 

[BS3036]

I appreciate everyone's input on this. It's nice to understand it, and as a bonus, I now know which page to turn to, in the exam!