question about ring calculation...
- Thread starter dingodeano
- Start date
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And why is that (r1+r2)/4?
my hundredth post!!! was my explanation of r1 + r2 > R1 + R2 ok??
No - fault current will flow in the entire ring...
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im thinking the shortest distance back to suppliers transformer back up through live to opperate o.c.device
im going to bed to read guidance note 3
Current won't only flow along the shortest path.
And forget Ze for now - why is R1+R2 = (r1+r2)/4?
The ratio is to divide by 4 when resistors wired in parallel / series
I got the answer from the AMICUS book
I got the answer from the AMICUS book
Is it always 4?
Dingodeano,
It's just ohm's law. Resistors in series and parallel. Draw out an equivalent circuit and work it out.
Then to see if you understand, add a spur part way round the ring and see if you can work out the R1+R2 at the spur.
It's just ohm's law. Resistors in series and parallel. Draw out an equivalent circuit and work it out.
Then to see if you understand, add a spur part way round the ring and see if you can work out the R1+R2 at the spur.
He's been an electrician for 15 years - is he playing games with us?
Depends if there's any spurs, but in a ring final circuit with identical conductors yes, it will be 4.
As for the reason, I know but I'll keep
We got an Ohms sign now too
ΩThe R1+R2 value is only exactly 1/4 of the r1+r2 value if the conductors are of equal CSA. If they are not then it is approx. 1/4.
Well raise my rent!We got an Ohms sign now too
What else we got?
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No - it's always ¼.
What you measure at any point when cross-connected for testing is indeed only approximately ¼, but when not connected like that, i.e. in normal use, R1+R1 will vary dramatically, just like it would with a radial, depending on how close you are to the origin.
But the value you need to know is the worst case, which will be at the mid point of the ring and there R1+R2 is exactly (r1+r2)/4 no matter how different r1 and r2 are.
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