Lighting circuit

[Yoda]

But what do I know !

 

Thought we'had done this one?

[ChrisPy]

It is if even one boater creates a fire risk through utilising bad advice on cable sizes.

A bit melodramatic, surely?

 

6sq.mm. cable is OK for 42A, that's about 0.5KW, way above anything considered here. The maximum load considered in the first post equates to about 240watts. Where is the fire risk in the examples quoted above?

 

I thought we had already established that the cable was fine for the load capacity, but that volt drop may be excessive.

[Sir Nibble]

Julian

 

In the evening when you are most likely to need your lights, you will doubtless have an amount of other equipment running such as TV, laptop, inverter etc etc as well as lights..

 

These will pull the battery voltage down to maybe below 12v. That doesn't mean the battery is greatly discharged, it's simply the total current² x the internal resistance of the battery. So if you are drawing say 20A, a fully charged (ie: nominal 12.6v) battery will actually be showing 12.2v or even less than 12v depending on its actual state of charge.

 

Remember too, that 12v domestic halogens are normally run off a 13v AC transformer which is hitting the lamp with over 18v pulses!

 

Chris

Snibble

 

Power is indeed IV or (the same thing) I²R. But since Power P = IV and I = V/R (from Ohms law) then substituting for I gives P = V²/R

 

ie: Power is proportional to Voltage². Thus, if the voltage drops by say 20% (ie: the voltage is now 80% of what it was) the power drops to 0.8² which equals 0.64 (ie: 64%) of what is was on full voltage. That's why a small voltage drop produces a larger brightness drop since brightness (of a lamp) is proportional to power.

 

Chris

Thanks Chris, but the confusion was that your post APPEARED to suggest that P=Vsqd, read it again and you will see what I mean. Edited December 7, 2007 by snibble

[chris w]

Thanks Chris, but the confusion was that your post APPEARED to suggest that P=Vsqd, read it again and you will see what I mean.

 

I can see that it could be ambiguous so I have edited the original post.

 

Chris