According to an answer in this thread on Skeptics: If you take one of the little

Question

According to an answer in this thread on Skeptics:

If you take one of the little 12V garage door opener batteries and short out (directly connect) the two terminals with a piece of wire or something else. You'll get a light current flow through the wire or metal. It may get a little warm. This battery is only capable of supplying a small amount of current.

If you take a 12V car battery and short out the two terminals (don't do it, it's not fun), you will be met with a huge current arc that will likely leave a burn mark on whatever was used to short it. This is because the car battery is capable of discharging a large amount of current in a very short period of time.

I'm not sure how this could work given Ohm's law V=IR. If we assume the resistance of toucher is constant, then we'd expect the current to be the same as well.

Could it be that a car battery has less internal resistance than a garage battery?
Does it have anything to do with contact area? If it does, then how would you model it? Generally resistances add in series, but if I only half touch a contact, then neither the battery nor I change, so you'd expect our resistances to stay the same, but somehow our total resistance changes. So which objects resistance would change - mine or the batteries?

Explanation / Answer

The answers of Martin and Edward are quite right, but the

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