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I'm no electronics engineer, but I don't think "drawing too much voltage" is a thing.

As current goes up, resistance has to go down, so there will be a voltage drop across the battery terminals.

That's my understanding, someone correct me if I'm wrong, or otherwise help clear this up?




Resistance is a simplification of characteristic curves, which themselves are simplifications. You can easily get a voltage drop across battery terminals, you just have to put the battery in series with another battery. This is dangerous. You could also hook up an active current sink. Again, dangerous. There is nothing preventing the internal battery resistance from causing the terminals to go negative except for the simple fact that most equipment we connect to batteries will stop working before that point is reached (unless you hook batteries in series—which is why batteries in series must be close to identical, so they run out at the same time).


I just want to point out that batteries are often connected in series in regular devices.

Many devices where there are two AA batteries (TV remotes, etc...) usually have the batteries wired in series. They need to do it to get the voltage high enough for some of the components.

Additionally, many batteries are actually made from multiple smaller batteries wired in series internally. For example, some 9V batteries are actually just 6 regular 1.5V batteries wired in series.

Of course, in these configurations, they generally discharge at a similar rate.


I believe the resistance is sort of constant (for most normal usage cases) for passive circuits and in solid state electronic it would be a function of how many parts of the chip are engaged.

This being said anything short-circuiting the loop, can create a new pathway with less resistance (think molten metal, sparks etc). But modern batteries can send a lot more current then, magnifying the effect.




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