2.5mm swa from 4mm radial?

Fair enough - but do I not also often see that very same thing happening here in relation to (indoor) light switches? :)
That's not really a fair comparison.

You can tell someone to wire a light switch correctly with them having no knowledge although quite correctly Bas does not agree this is advisable.

Outdoor installations have so many options and cannot be done without considerable understanding.


Would you like to list the choices and processes from supply to outlets for (at best) an understanding novice?
 
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Fair enough - but do I not also often see that very same thing happening here in relation to (indoor) light switches? :)
That's not really a fair comparison. ... You can tell someone to wire a light switch correctly with them having no knowledge although quite correctly Bas does not agree this is advisable.
I wasn't really 'making a comparison' - I was merely responding to your comment:
...the person has to be told everything step by step.
... which seems to be much the same in both cases.
Outdoor installations have so many options and cannot be done without considerable understanding.
I think all you're really saying is that these are much more complex issues. I certainly agree that the more complex the situation, the smaller will be the proportion of 'DIYers' for whom tackling the task will be reasonable or appropriate, but I don't think it matters whether it relates to indoor or outdoor work.

When responding to questions (here or elsewhere, in whatever field), I do try to give explanations, hence hopefully impart some knowledge/ understanding (which is one reason why my posts are rarely brief!), rather than giving 'wiring by numbers' answers (since I largely share BAS's view) - that's one reason why you'll very rarely find me involved in, for example, 'light switch' questions!

Kind Regards, John
 
When I am talking about cable ccc, I am referring to its capacity mentioned on tlc's website without derating factors such as lenth and insulation taken into account. (i.e tlc specifies 24A for 2.5mm twin/earth)

And what I mean by it is hard to work with is it is time consuming.

Oh and for price difference, not that more expensive in reality.

3 core 4mm swa it is and is safe and sufficient from overload because the upstream mcb is 32A type B.
 
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A longer cable will cause a voltage drop, thus causing a increase in current, thus if current from the load and the current caused by voltage drop exceeds the cables capacity, a cable with thicker cores is needed.

(i.e. a cables current rating wont change, but the amount of current passing through it will will with length.)

And for ccc, maybe TLC specify 24A for 2.5mm over 27A as in the regulations for safety reasons such as unknown ambient temperature and other factors.
 
A longer cable will cause a voltage drop, thus causing a increase in current,
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thus if current from the load and the current caused by voltage drop exceeds the cables capacity, a cable with thicker cores is needed.
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(i.e. a cables current rating wont change, but the amount of current passing through it will will with length.)
Indeed it will.

It will decrease.

CLICK


And for ccc, maybe TLC specify 24A for 2.5mm over 27A as in the regulations for safety reasons such as unknown ambient temperature and other factors.
Ambient temperature and other factors should all be known by the designer, and the right derating factors applied by him. That's what designing involves.
 
I am a total idiot and a tad embarrassed ;/ Just used ohms law and you are right the current and power decreases as the impedance goes up.

I always assumed that as Power (W) = V x I , I assumed that as the voltage dropped the current increased to maintain the the specified Power (W), where as in reality as the voltage and current drooped, so would the power (W).

Think i need to drop my secondary school theories thinking that the intended power would be maintained regardless of voltage drop? (was a time where I hardly knew what current was let along resistance/impedance.)
 
I am a total idiot and a tad embarrassed ;/ Just used ohms law and you are right the current and power decreases as the impedance goes up. ... I always assumed that as Power (W) = V x I , I assumed that as the voltage dropped the current increased to maintain the the specified Power (W), where as in reality as the voltage and current drooped, so would the power (W). ... Think i need to drop my secondary school theories thinking that the intended power would be maintained regardless of voltage drop?
The passage of time has actually helped you a bit. A few decades ago, you would, indeed, have been totally wrong. In those days, nearly all domestic electric loads were 'passive' (usually resistive) ('dumb' if you like), and simply obeyed Ohm's law (the equations derived from Ohm's Law). Hence, if voltage dropped (e.g. due to a drop in the cable, due to its impedence), current would reduce and hence the power would reduce.

However, we now have many 'intelligent' loads - in particular the 'Switched Mode Power Supplies' that come with, or are part of, virtually all electronic devices we now have (from computers through Extra Low Voltage Lighting to TVs to whatever). They DO behave in the fashion you were thinking, automatically adjusting the current drawn according to the voltage in order to maintain the required power. Hence, with those sorts of loads, reducing the voltage (e.g. due to a drop in the cable) will result in an increase in current drawn, to maintain the same power.

... so you don't need to feel totally idiotic :)

Kind Regards, John
 
I was going to mention SMPSUs, but then they are typically such a small % of the overall load, most of which remains "passive" (lighting, heating, cooking, motors....).
 
I was going to mention SMPSUs, but then they are typically such a small % of the overall load, most of which remains "passive" (lighting, heating, cooking, motors....).
That's true (although I've heard it suggested that some induction hobs, and maybe even microwaves, may now have SMPSUs).

Of course, if one moves one's interest from current & power to energy, even many of the large loads are, in one sense, 'intelligent'. Although a reduction in voltage will reduce current (and power) in these 'passive' loads, if they are thermostatically (or whatever) controlled, they will merely increase the amount of time during which that (reduced) power is being expended, so as to maintain energy usage - hence the total folly in those 'money saving' devices which reduce voltage!!

Kind Regards, John
 
Some heaters are "very intelligent" yet are nothing more that temperature sensitive resistive material Just apply a voltage and they draw the current amount of current to keep themselves and that which they are heating at a predetermined temperature. They are not too fussy about the voltage either.

http://www.graybar.co.uk/track-heating-systems/ being just one example.
 
Isn't this a bit academic?

Surely, you would require the same amperage of SMPS supplied equipment as an applied load that caused voltage drop - before the current increased (because of SMPSs) to what it would be there were no voltage drop.

In other words, say, 2 amps of SMPS equipment affected by voltage drop from a 10A kettle would cause far less current increase than if the kettle caused no voltage drop.

I know what I meant.
 
Surely, you would require the same amperage of SMPS supplied equipment as an applied load that caused voltage drop - before the current increased (because of SMPSs) to what it would be there were no voltage drop.
I don't totally follow that ....
In other words, say, 2 amps of SMPS equipment affected by voltage drop from a 10A kettle would cause far less current increase than if the kettle caused no voltage drop.
You seem to be saying that the increase in current draw by SMPS equipment (due to voltage drop) would be less if there were less voltage drop - clearly true and 'obvious', but I'm not sure that I understand what point you're trying to make!
I know what I meant.
That I don't doubt, but ..... :)

Kind Regards, John
 
You seem to be saying that the increase in current draw by SMPS equipment (due to voltage drop) would be less if there were less voltage drop - clearly true and 'obvious', but I'm not sure that I understand what point you're trying to make!
I thought he was trying to point out that if the voltage drops because of current drawn by a large resistive load, e.g. a 2.5kW kettle, the increased draw from a few hundred watts of SMPSU would not cancel out the decreased draw from the kettle.
 

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