Sparkies: Electrical wiring conundrum

[braailus]

And now....... On to brewing

[vacant]

P is not a constant.

3kW is the maximum power the element is rated to safely take. The element doesn't insist on always drawing 3kW. It would be good if it did as you wouldn't have to connect it to the leccy to heat your water

[themadhippy]

P = VI

I = P/V

yep,but dont forget R=V/I.To keep things simple well treat it as a DC circuit as a heating element is almost purely resistive, the resistance of load is the same regardless of the voltage, yea ok, put too many volts up it and it might get upset and go open circuit so you have infinity ohms, but lets pretend were in an ideal world.If you change the voltage you also change the current flowing in the circuit,so the power of the load (or the amount of work done) changes.
A 2500 watt element rated at 250 volts,the the current is 2500w/250v = 10A and the resistance ,which is fixed is 250v/10A = 25 ohms,now change the voltage to 220v across the same element,the current changes to 220v/25ohms or 8.8A ,and the power is now 1936 watts (8.8 x 220) .Going the other way increase the voltage to 500v and we get a current of 500/25=20A,and 20Ax500v is gets you 10000watts,or to put it another way the more energy(volts) you put in to a job the more work (watts) done, less energy the less work.

but this is what happens with lighting

what sort of lighting ?if its rock n roll have a look at how moles or ACLs are wired,8 x 28v lamps in series on a 240v supply,the over voltage makes em brighter and more white,the downside is reduced lamp life

[Jim]

........................ I may be wrong to approach it from total load point of view but this is what happens with lighting. The wattage remains constant and is made up by a factor of the voltage and ampere. If the voltage drops the current increases and visa versa. This is why you always leave headroom in the design of a system to cater for fluctuations is supply voltage.

That's completely wrong unfortunately.

What stays constant (well for all practical purposes) is the impedance of the load - if you put 110 volts into a 240V kettle, it will NOT draw extra current to make up for the shortfall in voltage!

The current that flows depends on the voltage applied and the impedance (or resistance) of the device it's applied to. The power dissipated in the load (wattage) is a result of that current flow, not a property of the load.

EDIT: Oops, should have read to the end of the thread! That's already been said!

[Andy]

Can anyone recommend a supplier for good quality crimps?

[Pinto]

City electrical/Denmans/Eyre & Ellistons/Rexel Senates/Edmundsons.....

Use the ones from an electricians wholesaler. You'll know quality ones when you feel them - solid and heavy.

[AnthonyUK]

Amphenol are pretty decent