How do I work out what size fuses to put on my savage switches and gauges? The gauges only say 12V DC 10-180 Ohms and the switches, well nowt, though I suppose I could measure the resistance across these?

They should have the current rating written on them somewhere surely or in the packing info etc, if not it would be worth trying to get in contact with Savage or a supplier.

Remember the switch might not be the weakest/lowest amperage rated part of the circuit, so don't just base your fuse rating on them.

The switches have nothing to do with the fuse rating. What you have to do is work out (or measure ) the current taken by a particular circuit.

If your headlight main beams were rated at 48W (say, to keep the figures simple) each bulb takes 4A, a total of 8A. A 10A fuse would probably not have enough margin so you would probably use a 15A

Then, you think about the switch which in this case is carrying 8A so it must be rated higher than that and you might want to go to a relay capable of carrying 15A. Remember the surge current when you switch on a cold lamp will be considerably higher than the steady state current.

So simplest is to measure he actual current in the circuit and then fuse rating is that nearest to 1.5 times the measured current.

^Yeah I see what you mean there and I reckon I can stumble through that one but for the gauges? I bought them off here and there was a sticker on the sides which has peeled off when I mounted them. Anyone help. They're Smiths white faced gauges. Just confused as they will have a varied resistance when the needle moves eh?

No I do't think so. As far as I know gauges work by heating effect of the current passing through them, which is why they move slowly to the position when you switch on ignition. So you have 12V connected to one side of the gauge and the other side is connected to earth through the sensor - fuel, temp or whatever.

I can only guess that the fiigures 10-180 Ohms mean that when the sensor is presenting a resistance of 10 Ohms, the gauge gives Full or Max reading and when it presents 180 Ohms, you get empty or min reading. The resistance of the gauges will stay fairly constant.

So if this guess is correct each gauge will pass 1.2A when reading Full or max. If you have 4 gauges then they take 4.8A if all reading max. I would fuse at 10A.

Of course. I could be talking utter nutsack and somebody will be along to say so

No, again I agree and it all sounds good to me
edited to say I=V/R which is 12/180 which is erm? 0.054 amps or roundabout is it not?

[Edited on 21/10/07 by mistergrumpy]