Sorry, hopefully not too nerdish, but we are all here to learn and consequently, I cannot resist tackling some of the technical aspects for accuracy, through these threads.
Let’s step through the electrical theory, it a stage at a time:
We have Ohms Law illustrated by volts divided by amps equals resistance. Also if we multiply volts by amps we get power in watts.
So all these quantities are inter related and proportional to each other BUT the point that I wish to make is that Resistance is the ONLY constant, (only affected marginally by temperature, but lets keep it simple.)
Referring to the earlier thread No. 791827, when we look at an electrical load in Watts, that is the power consumed at the NOMINAL RATED voltage of the equipment (12 volts in automobile cases)
So if 12 v is applied to this load of 30 watts as quoted, it will consume 30 / 12 = 2.5 A
By virtue of Ohms Law, this means that the actual Resistance of the load, which is fixed, is 12 / 2.5 = 4.8 ohms.
Now, If under normal running, the system voltage is actually 14v (due to alternator etc) , the current that will flow, again by virtue of Ohms Law will be (Volts / Resistance)14 / 4.8 = 2.91 amps. ; and similarly we are now actually getting almost 40 watts in total from the same bulbs. (e.g. with headlights switched on, rev the engine from tick over, and you see the lights brighten)
As we can see this is almost 0.8 amps above the calculation in the earlier posts.
I stated in an earlier thread that the selection of fuses for auto use should be subject to a load of about 70% or so maximum for reliability. So here is just one reason why that margin is necessary, as the current flowing will increase with alternator charging.
Also, if a fuse is run constantly at its full rating, although in theory, it should be able to do that forever, in practice, due to variations in ambient temperature, current spikes, etc. and the general heating of the fuse element working at full tilt, one could find that the fuse “gets tired” and melts at some time in the future yet can be replaced with a fresh one that will in turn, carry on for a long time before the process repeats.
Coming back to Monte’s case, if the 30 w total load is correct, then we can see that his 5A fuse is loaded to about just under 60% so happily should be ok in this case.
Just as a final note, bear in mind that as I stated earlier, fuses are not that great at protecting for overload conditions. Their prime function is to protect against short circuits and safely disconnect the circuit before the whole thing burns out. A car battery under a dead short can easily produce several THOUSAND amps. So the wiring causing the short circuit would virtually vaporise in seconds, catch fire and burn our beloved car to a cinder. So although it is important to select a fuse with a rating in excess of the total load, it is essential to never choose a fuse rating that is above the rating of the wire that it is protecting.
Here endeth!
