Why would "a bit of bondage" be locked?

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Paul,

The resistance of all of the other copper cables in the circuit is decreasing whilst the copper cpc is increasing???

I seem to be answering the same questions almost simultaneously on two different forums, but oh well.......

No, I'm not saying that at all. What I'm saying is that the huge increase in current which flows during a fault is because the overall resistance of the circuit as a whole is very much lower.
 
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OK,

During a fault of negligable impedance to earth, the cpc will get hotter than the line circuit conductor.
This process of heating the cpc takes time, this time will affect the automatic disconnection time of the over-current protective device.
Decrease the resistance of the cpc (Table 41C) and the cpc temperature will not need to increase as much, therefore logic suggests that the automatic disconnection time will not be affected quite so much.


Conclusion:

The R2 values (for fuses) given in Table 41C where measured not calculated.
 
Hi Paul,

The fourteenth Edition was approximately when MCB's where invented and when the Regulations were first expanded to their current format, therefore I assume that this was when Table 41C was extended.
This was also the Edition where the word radial was first misused!
 
The fourteenth Edition was approximately when MCB's where invented

They were around a good while before the 14th edition.

when the Regulations were first expanded to their current format

Not quite sure what you mean there - The format of the 14th edition is quite different from today.

This was also the Edition where the word radial was first misused!

In what way?
 
Please stop this madness!!!!

Cockburn, you have a fundamental, poor understanding of physics and this is where your problem firmly rests. There are many things which appear "intuitive" yet when one "crunches the numbers" one can get a much different answer than "intuitively" expected.

When I was 13 I thought that if you connected a generator to a motor then the generator would power the motor and what was all this fuss about perpetual motion being impossible!! When I was 15 I did the maths and realised why it did not work and could not work.

If you want to save the last remnants of your dignity then I encourage you to stop your debate right now. Go and study the physics formally at a college then come back in a year and try again.

If you do not do that then you will continue down this no hope path.

Please GO AWAY thank you.
 
OK,

During a fault of negligable impedance to earth, the cpc will get hotter than the line circuit conductor.
This process of heating the cpc takes time, this time will affect the automatic disconnection time of the over-current protective device.
Decrease the resistance of the cpc (Table 41C) and the cpc temperature will not need to increase as much, therefore logic suggests that the automatic disconnection time will not be affected quite so much.


Conclusion:

The R2 values (for fuses) given in Table 41C where measured not calculated.

Conclusion:
:?: :?: :?:
Is there any point to any of this, at least answer that
 
Paul_C said:
You are being asked to believe that during a fault of negligable impedance to earth the temperature of a cpc increases, but during this time the resistance of the cpc is increasing as the current flow is also increasing.
Which contradicts ohms law!

No it doesn't. The current which flows during the fault is determined by the overall resistance of the circuit, not by the resistance of just the earth conductor. It's the reduction of that overall resistance of the circuit due to the fault which results in the increase in current.

Let's try a practical example of this very basic electrical theory.

Here's Part 1 of the problem:

2egd7qc.gif


R1 and R2 are the resistances of the two supply conductors, RL is the load resistance of the connected appliance. You may assume that the internal resistance of the supply source and all other interconnections shown by solid lines are so low as to be of no consequence and may be ignored.

David, would you care to calculate, from the information given, the current which flows in the circuit, the total power dissipated, the power dissipated by the load, RL, and the power dissipated in each of the two supply conductors represented by R1 & R2?

Part 2 will follow upon correct answers to part 1 being given.....
 
John,

An increase in heat must be caused by an increase in power, therefore the resistance must be less.

Otherwise you are telling me that to make something hotter takes less energy, which obviously makes no sense.

What a load of twaddle! Think about how an electric kettle works, it doesn't take more power when hot, it takes time to transfer the heat to the water.
Forget ohms law and go and play about with the equation Q=mc/\T :LOL:
 
Conclusion:
:?: :?: :?:
Is there any point to any of this, at least answer that
I believe there is a point to this conversation.

Mr Cockburn is obviously dedicated to promoting safety in electrical installations. He appears to be promoting the need for care and attention to detail when designing and installing and using electrical equipment.

But his understanding of how safety circuits and equipment function is not yet complete or accurate.

This conversation could be a means to fill the gaps and correct the errors in Mr Cockburn's knowledge. Then his campaign using re-written literature will be more effective as the content will be respected by learned bodies.

It just requires an open mind and some polite but firm implanting of the correct information.
 

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