Cooker on its own ring

[ed-ectrician]

Hello there,

You lot have always been helpful in the past and generally not too condescending so I'm going to ask what I feel could be seen as a dumb question (the dumbest being not asking it!)

I went to do an inspection on a kitchen the other day and the spark that had done the work had a kitchen ring on a 32 amp breaker. All good, then he put the cooker on a 32a ring circuit of its own, now my guess on this is that it is not OK becasue the idea of a ring is that the load is distributed on the first and second halves of the ring. With one appliance on the ring this isn't the case so doing a cooker on a 32a ring circuit of it's own with 2.5mm twin and earth cable is not right because the cooker will draw all its current from the "half" of the ring that is the least resistance right? So it's like basically putting a cooker on a 2.5mm 32a radial which is not ok.

The bloke I spoke to said the electrician who did it didnt have 4 or 6mm cable so did it this way. Often the case!!

Am I right here? Missing something?!

Ed

 

[Darkwood]

If a current has 2 paths it will go down the path of least resistance and this is always the case... but as the current flows down this path it will generate heat which gives resistance and at a certain point it will match the resistance of the other leg so current will then find 2 paths of roughly equal resistance to share the remaining current flow, the resistance fluctuates at a quantum level at near light speed trying to maintain balance as the voltage cycles from 0v to approx 325v peak at 50hz.

In a nutshell the current takes the path of least resistance until enough current flows to generate a heat rise in turn adding resistance, when it matches the resistance of leg 2 then the remaining current is roughly shared down both legs.

 

[Deleted member 26818]

Think about what your saying.
If that was the case, then one circuit would not operate untill another circuit had reached sufficient temperature for the resistance to be equal.
Nothing on my 32A RFC would work untill my 6A lighting circuit had heated up enough.
Be just my luck, that it would never reach the correct temperature.

 

[Knobhead]

If you're woried about it get the clamp meter on each leg. (The cable NOT you're legs, before the daft buggers start!)

 

[Darkwood]

Think about what your saying.
If that was the case, then one circuit would not operate untill another circuit had reached sufficient temperature for the resistance to be equal.
Nothing on my 32A RFC would work untill my 6A lighting circuit had heated up enough.
Be just my luck, that it would never reach the correct temperature.

Im refering to parallel paths on a RFC and how the current is divided between both paths regardless of the position of the load just for the OP's insight, im a little lost as to what you mean regarding the lighting circuit as thats nothing to do with my post and nor is it part of the parallel paths of the RFC (well until it adds together at the dist' board).

 

[Deleted member 26818]

You say that current takes the path of least resistance and won't take the path of higher resistance untill the path with least resistance heats up and becomes as resistant as the other path.
If that's the case, how is it that I can have ten circuits off of a CU, and they all work all the time, even though they have different resistances?

 

[Knobhead]

My head hurts!

 

[Darkwood]

You say that current takes the path of least resistance and won't take the path of higher resistance untill the path with least resistance heats up and becomes as resistant as the other path.
If that's the case, how is it that I can have ten circuits off of a CU, and they all work all the time, even though they have different resistances?

Your misreading my post im talking about a ring with a randomly positioned load and how the current is divided between the left and right leg, a seperate circuit will have nothing to do with the ring i refer to so don't understand your response.... your seperate lighting circuit will be a radial (normally) and will see it path of least resistance as its own supply cable from the board it dosn't have any alternative routes like a load does in a RFC.

 

[Deleted member 26818]

It doesn't make any odds whether its two legs of an RFC or two separate circuits.
The principles are the same.
I = V / R.
Current will flow down any available path.
More down a path of less resistance than down a path of high reistance, but it will still flow.
The amount flowing is dependant on the resistance, and can be ascertained using the formula above.
Ther's no waiting the the two resistances to eqalise before current will flow.

 

[Darkwood]

Yer canno change the law of physics captian!:smart:

 

[Deleted member 26818]

Exactly.

 

[Des 56]

It doesn't make any odds whether its two legs of an RFC or two separate circuits.
The principles are the same.
I = V / R.
Current will flow down any available path.
More down a path of less resistance than down a path of high reistance, but it will still flow.
The amount flowing is dependant on the resistance, and can be ascertained using the formula above.
Ther's no waiting the the two resistances to eqalise before current will flow.

If I heard this discussion in a site cabin I would certainly agree with the above

The only problem being Darkwood seems to have a very in depth understanding of quantum mechanics and no doubt,in the world of the activity of the electron,he will provide an explanation that very few will understand including myself :blush2:

 

[Engineer54]

On a parallel cabled circuit following the accepted criteria, the differences in resistances, and therefore current paths would be minuscule. And this is basically what Spin is trying to explain, that current will ALWAYS pass down each leg of a ring circuit regardless of minor resistance differences....

 

[Darkwood]

It doesn't make any odds whether its two legs of an RFC or two separate circuits.
The principles are the same.
I = V / R.
Current will flow down any available path.
More down a path of less resistance than down a path of high reistance, but it will still flow.
The amount flowing is dependant on the resistance, and can be ascertained using the formula above.
Ther's no waiting the the two resistances to eqalise before current will flow.

I agree with you and i only stated why it is so , the resistance you measure fluctuates when current flows and so does the cable temp, your talking mathematic formula that are fine but can't explain why it shares the current how it does i gave a deeper insight as current has no IQ so has to follow a rule and its a fundamental well tested and proven law of physics what you are confusing here is tabled values that are fixed which work and are accurate.... what i expressed was why they work and when i say it waits until the resistance is the same before the 2nd leg shares the remaining current i did express that this occurs at close the speed of light.
Yes it will flow down both paths but this only occurs after the first path becomes equal and this will meet this condition in a fraction of a second too fast to measure and to quick for it to effect the equation I = V/R

Im expressing how current division works and was never arguing with any of your posts, although i didnt read until now that you did say it isnt the path of least resistance which i will strongly disagree with.
Ill avoid taking this thread down the Quantum theme but im well knowledged in the how electricity actually works at quantum level and I = V / R works well in the macro world but is too basic a sum to be used to prove me wrong.

Take a super conductor with no resistance and a normal conductor then create a parallel path and attach a load... no current will flow down the normal conduct it will all flow down the super conductor as their is no electrical resistance and no heat will be generated to give it such.
These experiments were performed many decades ago and still havent been contradicted.

 

[1shortcircuit]

I daren't ask how this would affect a radial circuit with branches off in all directions lol

 

[Darkwood]

This is somewhat of a false debate as i agree with what you say spin and and think i might be going a little deeper although im also correct i feel it may take a while to explain my approach and why im correct, i think ill back off on this one though only for my own sanity if not other readers.....

I find im just on of those who dosn't just learn equations but find out why the equations work and that all i did i expressed what happens but this is all so fast its not measurable or noticeble to you or i with our humble MFT or clampmeter.