For a resistor to burn out you must pass too much CURRENT (not voltage) through it.
In a simple resistive circuit, current is a simple geometric function of voltage.
V = i x R So too much current can only happen as a result of too much voltage.
If the gauge has an internal problem that doesn't limit the current...
The only limits on current in the factory gauge circuit are:
- the battery's current output (usually on the order of 600-1200 CA, considering modern batteries)
- the alternator's peak output (usually 45-130A, considering common upgrades)
- the wiring's capacity (probably <25A for a second or 2) since there is no factory fuse on the gauge circuit
...I can't wrap my head around why that wouldn't also be the case with higher voltage...
It WOULD (for small increases in voltage, on the order of 101%-10,000% ). The differences are that water doesn't cause pipes to overheat; nor does it induce a magnetic field around the pipe as it flows; and water can spill over or leak out much more easily. But otherwise, the comparison is reasonably useful. Electrons DO move through wires similarly to fluid molecules moving through a pipe; and the effect of voltage to motivate current in electrons is comparable to the effect of pressure to create current in fluids. That's why several fluid terms are applied to electricity, like "current" and "eddy", and the mathematical formulas are nearly identical.
...(like too much pipe pressure bursting a pipe, seems like too much voltage would have the same effect).
That's a different phenomenon, and the electrical equivalent is dielectric breakdown of the insulation around the wire (which occurs on the order of 1,000,000% nominal voltage), at which point a spark jumps out of the wire like water bursting through a pipe. But the similarity ends there because water doesn't need anywhere specific to go - an electric spark MUST always jump to another conductor of lower voltage. If there's not one around, there will never be a spark. The distance is a direct function of the voltage, which is why lightning bolts have such high voltage, and why high-voltage wires are always high above the ground, which isn't allowed to have trees growing up on it.
...amperage also goes up. So clearly I'm missing something.
No, you seem to have that down.
...you'd think the sending unit coil is designed to handle 12v...
That's an oversimplification of the (VERY complicated) actual facts. What it's "designed to handle" is a function of MANY considerations, including what constitutes "failure" for the particular application, and how likely the circuit is to experience overvoltage for more than a fraction of a second (RFI). A much more-accurate simplification is to think of how much POWER a component is designed to handle (or physically, how much HEAT it can dissipate without changing the relevant properties of its materials). If more power is consumed (converted to heat) within a component than it can get rid of (via conduction/convection/radiation), then the component melts or burns (either oxidizes or carmelizes). When its chemistry changes, it's not likely to perform the way it was supposed to. So you can't simply talk about voltage in a design, OR current.
W (Watts of power or heat)
= V x i So both voltage AND current must be considered. And since either V or i can be expressed in terms of R,
W = R x i^2 = V^2 / R
At zero resistance, Amps become... Undefined? It's obviously asymptotically infinity but that can't be right. Mathematically the limit has to be infinite amps but obviously I'm missing something.
Yes - you're missing that current has no apparent mathematical limit, but it DOES have obvious physical limits (listed above) related to both the capabilities of the voltage source which is causing the current, and to the physical construction of the circuit (wire sizes).
The second thing I've never really figured out is why don't gas tanks catch on fire when a sending unit fails? If you put too much current thru a wire, it gets hot and melts. How is this not a risk with fuel sending units?
Several reasons...
The most-important (which is also the least-obvious to most people) is that gasoline DOESN'T BURN. It must first be vaporized (which is why carburetors have venturi and fuel injectors have nozzles), but even gasoline vapor doesn't burn. It ALSO has to be mixed with Oxygen. And even that's not enough... It has to be mixed in the correct proportions. So even if you put a heating element into a gas tank AND increased the pressure substantially (like firing a tracer bullet into it), it will neither explode NOR even ignite. (Just watch MythBusters.)
The next reason is that the circuit feeding the level sender can't carry enough power into the gas tank to vaporize the gas. Same as dumping gas onto a burning wick - if you dump it fast enough, you put the wick out. (DON'T TRY IT!!!) The wiring is too small; the heat necessary to melt the copper and open (kill) the circuit is MUCH lower than the heat necessary to boil the gas; and being a fluid, the warmed gas will convect away from the (failing/overheating) sender, carrying cooler gas onto it.
And if anything was going to blow up gas tanks, it
wouldn't be the level senders; it would be the fuel pump motors, which have gasoline flowing through & around their carbon brushes which continually spark against the rotating commutators. But for the same reasons I just described, those sparks don't ignite the gasoline. Ever. Not even with a few hundred million vehicles buzzing around the world with electric fuel pumps submerged in their tanks. The number of occurrences is ZERO, which is the same as the risk.
