Quote:
Originally Posted by Tmacmi
Its a Yanmar 2qm15. The manual says that based on a voltage drop of .2volts/amp with a cable with a cross section of 20 mm squared (approx 4 gauge a circuit length of 2.5 meters is acceptable.
The description of the starter says the no-load current is 60 amps or less.
But then there is this that suggests the in-rush amps are 300!
There is a significant disparity between the sizing recommended in the manual and what would be calculated at a 300 amp inrush
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Being very conservative (i.e. assuming that each light bulb is an old fashioned 25 volt incandescent bulb) I figure that if everything was turned on at the same time it would draw approximately 75 amps.
I then estimate the average length of the lines to be 30 feet. I know estimating isn't optimal, but I am not able to measure each line.[/QUOTE]
I think you need to review your units. "a drop of .2v/amp would be a 12 volt drop at 60 amps.
Likewise your bulbs are 12 volt, not 25 volt.
I think you meant 25 watts not volts.
A 25 watt bulb draws 25watts/12volts = 2.08amps.
the equation is
Power (watts) / voltage =
current (amps)
also useful is
Voltage = resistance * current
A 25 watt bulb is really bright and really hot. I'd be surprised if any of your bulbs are more than 10 watt bulbs. 10W/12V= .9 A , so just use one amp per bulb in your calculations.
The calculations don't have to be exact you already know that everything works without blowing
fuses right?
Battery cables are expensive, but the difference between just ok and really good isn't that much. I strongly suggest making sure your
battery cables are big enough. That one day when it's cold and your
batteries are low you don't want to waste energy
heating up battery cables that are a size too small.