The following is from an offline discussion between me and slant-6 Dan. He suggested I post here, so here we go:


The fusible link should NOT cause a noticeable voltage drop under normal conditions, and here's why:
OK, the resistivity per foot of the wire you mentioned would be 0.2/1000. The total resistance of your fuse link, R, would be the resistivty per foot times the number of feet. So if you have a 6" link,

R=6/12 * 0.2/1000, or 0.0001 ohms (100 micro-ohms... this should be the first warning that what we're going to calculate may be kinda worthless information!).

For a nominal system voltage of 14 volts (round number), a 2% drop would be 0.28 volts across the link.

Rearranging Ohm's law to solve for current, I=V/R, so I(allowable)=0.28/.0001 or 2800 amps! So in order to produce a noticeable voltage drop the current would have to be astronomical. And by the way, that would mean that the fusible link under those conditions would be absorbing almost 800 watts (Vsquared/R).

This also points out why fuse links are so imprecise- you're multiplying tiny numbers by huge numbers, so ANY variation in the resistance, such as at the connector or if the wire has a broken strand at some point, will make a huge difference in the amount of current that it would take to blow the link.

The other thing neglected so far is that "R" is not constant as the link heats up. You know that it really doesn't take 2800 amps to blow the wire you are talking about, and in fact 100 amps will blow it, right? I mean, you can't run your starter through that wire without melting it, and that draws order-of 100 amps.

So, let's solve for how much power that link absorbs at 100amps:

P=Isquared *R, or 100*100*.0001, =1 watt.


Now waittaminute! 1 watt won't blow a 6" wire either, right? So what's wrong? Why does it, in fact, BLOW when this calculation says it should only absorb 1 watt?!? Well, that 1 watt doesn't blow the wire, but it HEATS it, and R starts increasing dramatically, so in a fraction of a second you have a runaway condition where the resistance is skyrocketing, and the link absorbs enough (hundreds of watts) power to melt copper and thus blow.

That's why its just about impossible to do a simple paper-and-pencil calculation to answer his question, and everyone uses rules-of-thumb (or just fuses!). Yeah, you could put together the simultaneous equations that link reisistivity to temperature, temperature to power absorbed, and power absorbed to current through the link and solve for the room-temperature resistance you should start out with (and that's what SOMEONE did and discovered it always comes out about 4 gauges smaller than the wire to protect)... but SHEESH! If your loads add up to 20 amps, buy a 30 amp breaker and call it a day. And if that one "nuisance trips," buy a 50 amp instead, it'll still be fast enough to protect the feed wiring from going up in flames. That's the real beauty of true fuses and circuit breakers versus fuse links- breakers and fuses respond fast enough and consistently enough that sizing them is not super critical.


Also- ALWAYS size the main wiring on the CALCULATED LOAD! Don't rely on the fuse or breaker to protect the wiring from YOU being an idiot and turning on too many loads at once. The breaker or fuse should only protect from a failure (short circuit) or accidental gross overload (a bad device plugged into a power outlet, for example) not from any load that you could conceivably turn on under normal conditions.

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