Upgrading shower circuit from 6mm to 10mm

Also, if its a wylex cartridge fuse board, and not rewireable, then the plug in MCB's will not fit.
 
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Thanks for the replies and my apologies for the incorrect terminology.
:oops:

I presume that a fuse 'blows' and a circuit breaker 'trips'. I just need to flick the switch back up to restore power.

I am somewhat alarmed to read the shower may be dangerous. To clarify, I don't have an RCD but the main 'on/off' switch incorporates an "earth leakage circuit breaker".

I believe this has a similar function to an RCD but is not sufficient for new installations. (It has a 'trip' button to test the system).

The circuit breakers themselves are not the type you can plug in - they are part of the unit itself.
 
To clarify, I don't have an RCD but the main 'on/off' switch incorporates an "earth leakage circuit breaker".

I believe this has a similar function to an RCD but is not sufficient for new installations. (It has a 'trip' button to test the system).
Can you post a picture?
 
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That's good (that you have an ELCB = RCD). Just test it now and then and ensure it kills the shower? And do the upgrade on both the ELCB, shower MCB and wiring to ensure the contacts inside can cope with the full load.

8500W/230V = Almost 37Amps

Happy days ..........
 
It doesn't make any difference to the thread but the 8500 will likely be at 240V so a bit lower amperage.

Even less at 230V.
 
Nope. The resistance of the element is the only constant.

8500W @ 240V = 6.8Ω - 35.4A
7800W @ 230V = 6.8Ω - 33.9A
 
Nope. The resistance of the element is the only constant.
8500W @ 240V = 6.8Ω - 35.4A
7800W @ 230V = 6.8Ω - 33.9A
[well, it is Saturday night :)] .... even the resistance of the element won't be completely constant. The element will be presumably be cooler (hence lower resistance) with less current flowing through it - so, if it's resistance is 6.8Ω with 240V applied, it will presumably be less than 6.8Ω with 230V applied, hence the current will be somewhat more than 33.9A (but less that 35.4A).

Kind Regards, John
 
Nope. The resistance of the element is the only constant.
8500W @ 240V = 6.8Ω - 35.4A
7800W @ 230V = 6.8Ω - 33.9A
[well, it is Saturday night :)] .... even the resistance of the element won't be completely constant. The element will be presumably be cooler (hence lower resistance) with less current flowing through it - so, if it's resistance is 6.8Ω with 240V applied, it will presumably be less than 6.8Ω with 230V applied, hence the current will be somewhat more than 33.9A (but less that 35.4A).

Kind Regards, John

OMG!!! talk about perdantic! are you related to my wife?!, If resistance is proportional to 'kelvin, then the difference would be ~2'k = 1/150th, why am I even replying to this!!!
 
OMG!!! talk about perdantic! are you related to my wife?!, If resistance is proportional to 'kelvin, then the difference would be ~2'k = 1/150th, why am I even replying to this!!!
I don't know - you tell me!! As I said, it's Saturday night! Do I take it that you are a teetotaller?

Kind Regards, John
 
Lads, lads, lads ..............

Are we going to argue over 10 volts?

If someone down the road make a brew (electric kettle) whats the line voltage going to be?

Chill out, after all its not money

No tolerance, it has to be dead on ........ :)
 

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