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]

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

It wouldn't is the short answer the current needs more than one path back to apply to my post, a branched radial only offers one path from the load to the DB unless its been wired dodgy

 

[1shortcircuit]

It wouldn't is the short answer the current needs more than one path back to apply to my post, a branched radial only offers one path from the load to the DB unless its been wired dodgy

I'm only playing darkwood

 

[Darkwood]

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....

And ive agreed with this but he has trouble realising what my post is about and why i am also correct hes applying the depth of my post and trying to compare to his 'correct!' views and he is thinking they conflict but actually its the very reason his post is correct with what i have written.
Hes quoted the formula and i in a previous post explained why the formula is so.

 

[Rampage]

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.

Could you simulate this by using parallel conductors to a load using say a long reel of PVC single as one conductor and a 30cm PVC single for the other? (assuming both conductors) are rated way above the load)

 

[Engineer54]

And ive agreed with this but he has trouble realising what my post is about and why i am also correct hes applying the depth of my post and trying to compare to his 'correct!' views and he is thinking they conflict but actually its the very reason his post is correct with what i have written.
Hes quoted the formula and i in a previous post explained why the formula is so.

Fair comment, i can see where paths are getting mistakenly crossed here

 

[Darkwood]

Could you simulate this by using parallel conductors to a load using say a long reel of PVC single as one conductor and a 30cm PVC single for the other? (assuming both conductors) are rated way above the load)

You can do simple table experiments to predict load distribution using ohms law and instead of using loads of cable just set up 2 parallel paths joined at the supply and load and put variable resistors inline on each although i would suggest using a safe voltage.... the problem comes if you want to actually witness the current rise in the least resistive leg before it balances and shares as it happens so quick that you may need to ring CERN and ask to use their lab

 

[Darkwood]

Fair comment, i can see where paths are getting mistakenly crossed here

I think what you mean is me and spin went down different parallel paths here to arrive at the same answer

 

[Engineer54]

I think what you mean is me and spin went down different parallel paths here to arrive at the same answer

Yes.... lol!!

 

[44ALLAN]

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.

That's exactly what I thought happened :biggrin5: