reb78 Posted December 10, 2015 Share Posted December 10, 2015 I have a fused relay that feeds my spotlights. It's switched from the feed to the full beam headlights. It's a 15 amp fuse and lately it's been blowing occasionally. It doesn't go straight away and the fuse casing looks like it gets hot as it's melted a little. I've had a quick look at the wiring tonight. Nothing appears to be shorting. I'll check again in the light at the weekend and also clean up the earth connection. What else should I be looking at? Quote Link to comment Share on other sites More sharing options...
western Posted December 10, 2015 Share Posted December 10, 2015 examine the power wire inside the light as well, it maybe chaffed & touching the light case which shorts it out & pops the fuse. Quote Link to comment Share on other sites More sharing options...
reb78 Posted December 11, 2015 Author Share Posted December 11, 2015 Thanks Ralph. Will do. If it was a straight short though, do you think the fuse would blow straight away? This seems to work for quite a while get hot and then blow. Quote Link to comment Share on other sites More sharing options...
pete3000 Posted December 11, 2015 Share Posted December 11, 2015 how many watts are the spots?, just thinking if you have a bad earth and the voltage is lowered the current will try and increase to keep up. Hence the fuse getting warm, especially if you are near the limit. If the cable is up to it between the fuse and the spotlights you could use a 15A fuse to each. how thick are the wires feeding the lamps 55w/12v= 4.6A~ each lamp if they are on with lower voltage 55w/9v= 6.1A ~each lamp Quote Link to comment Share on other sites More sharing options...
Bowie69 Posted December 11, 2015 Share Posted December 11, 2015 Err, no it won't. V = IR, therefore I = V/R, with an increase in resistance or a drop in voltage, current will fall. As P = I2R we see that power is relative to the current and resistance, so your bulb will operate at a lower power. Also, with P = V2/R , you can see that as voltage falls, the power falls with the SQUARE of the fall in voltage, assuming fixed impedance of 2.6 Ohms we get: 14V = 75W 12V = 55W 10V = 38W Resistance changes as a bulb heats up, but for this demonstration it can be safely ignored. Reb, what bulbs are you running? If 100W you are right on the limit, in fact on 55W, and potentially 15V from the alternator you are still quite close. Have you tried using a different brand of fuse? Some are complete pap... Quote Link to comment Share on other sites More sharing options...
reb78 Posted December 11, 2015 Author Share Posted December 11, 2015 Thanks chaps. Its been a long time since I looked at the bulbs but i think they are 55W in each lamp. These lamps are generally really reliable and have never needed a lot doing with them. I didn't install this setup - mum had it put in when she owned the 110 in about 1995 - its worked perfectly until this year when I noticed the fuse had gone a couple of months ago (same as above - a bit melty). I just replaced it and its been ok for the last couple of months but last night i noticed it had gone again. I imagine the fuse came from Halfords, but i will try another, check the wiring (including inside the housing) carefully in the daylight and clean the earth. Quote Link to comment Share on other sites More sharing options...
pete3000 Posted December 11, 2015 Share Posted December 11, 2015 Just wondering where you got 2.6 Ohms from Bowie? agreed with the rest of it but I=P/V i.e watts/volts or I=V/R, but as r needs to be measured?........? or am i being thick Quote Link to comment Share on other sites More sharing options...
Bowie69 Posted December 11, 2015 Share Posted December 11, 2015 Well... P = IV and... I = P/V = 4.6A With V at 12V, and P at 55W we can use.... V = IR ...or... R = V/I = 2.6 Ohms. You can also use P = I2/R to get there directly, which goes to... R = P/I2 = 55/4.62 = 2.6 Ohms. If you can't remember the power formulae, then you can get there from just P = VI and Ohm's Law quite quickly. Think I may have just broken the internet with that last post. Quote Link to comment Share on other sites More sharing options...
landroversforever Posted December 11, 2015 Share Posted December 11, 2015 Think I may have just broken the internet with that last post. You will have done.... you're not allowed to post actual facts on the internet!! Quote Link to comment Share on other sites More sharing options...
muddy Posted December 11, 2015 Share Posted December 11, 2015 I would check the contacts for the fuse, I often find they get slack and cause resistance that ends up melting stuff. Quote Link to comment Share on other sites More sharing options...
Peaklander Posted December 12, 2015 Share Posted December 12, 2015 Just wondering where you got 2.6 Ohms from Bowie? agreed with the rest of it but I=P/V i.e watts/volts Bowie69 is correct. The power is a result of the voltage and the current - it doesn't create either of them. So as he says, if the voltage across the bulb is reduced because of a high resistance elsewhere and the available volts from the battery can't change then then the current has to reduce because I=V/R I would check the contacts for the fuse, I often find they get slack and cause resistance that ends up melting stuff. and now that same lower current passing through the bulb and another resistance, heats-up the additional resistance at a place that isn't designed to cope with heat (as it shouldn't have resistance) and it melts. Quote Link to comment Share on other sites More sharing options...
Hazza Posted December 12, 2015 Share Posted December 12, 2015 I had similar symptoms with the relay kit I used for my headlights. Cheap 'waterproof' fuse holder was letting in moisture and corroding the terminals leading to more resistance and a melted fuse. All replaced now and seems much better! Quote Link to comment Share on other sites More sharing options...
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