David, if Mr Sheds and Mr OMS treat you unfairly then just remember that one day you might be something.
Why would "a bit of bondage" be locked?
- Thread starter david911cockburn
- Start date
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Every Electrician that I have met agrees with every word I say.
I assume that when you say "met" you must mean personally and not via the forums (although I find even that explanation difficult to believe)
"Originally posted by: david911cockburn
OMS,
Do you mean P = (I x I) R:
If so why does this suggest to you that an increase in 'P' calls for an increase in 'R'?"
This was a "Classic" False Logic statement
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ebee,
Do I need to do the maths for you?
V = IR therefore:
240v = 60amps x 4ohms or;
240v = 120amps x 2ohms.
Therefore P = (I x I)R
(60 x 60) x 4 = 14400watts or;
(120 x 120) x 2 = 28800watts.
The resistance is lower therefore the current increases along with the power!
Do I need to do the maths for you?
V = IR therefore:
240v = 60amps x 4ohms or;
240v = 120amps x 2ohms.
Therefore P = (I x I)R
(60 x 60) x 4 = 14400watts or;
(120 x 120) x 2 = 28800watts.
The resistance is lower therefore the current increases along with the power!
H
holmslaw
..
holmslaw,
Do you understand 'ohms law' as it is being used above?
Do you understand 'ohms law' as it is being used above?
echoes,
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!
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!
Although it's basically against my instincts to engage in this silly discussion...
Kind Regards, John.
Yes, that would contradict Ohm's Law - but who is asking us to believe that the current would be increasing as the cable/CPC heats up (since it won't be)?
Kind Regards, John.
@cockburn Thankyou.
So, assuming that the cpc impedance has a positive temperature coefficient (like a tungsten filament in a light bulb). The relationship between voltage and current is no longer linear, because the resistance is not constant.
That does not contradict ohms law though.
@john - yes it is a silly (and rather basic, O-level one at that) discussion!
So, assuming that the cpc impedance has a positive temperature coefficient (like a tungsten filament in a light bulb). The relationship between voltage and current is no longer linear, because the resistance is not constant.
That does not contradict ohms law though.
@john - yes it is a silly (and rather basic, O-level one at that) discussion!
The current has to increase to a point where the speed of automatic disconnection of supply is met.
John,
The equation above proves that the more power you use, the more current you must use; and you need to use power to heat the cpc!
The equation above proves that the more power you use, the more current you must use; and you need to use power to heat the cpc!
echoes,
Ohms law is the basis of prospective fault current and loop impedance.
Ohms law is the basis of prospective fault current and loop impedance.
Do (briefly) go on... why would that not be achieved with correctly designed cbales and devices?
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