ANSWER/ERROR FOUND: resistance of the leads of the multi-meter is the culprit. The true resistance of the wrench cannot be measured using a basic multimeter.
The resistance of the wrench is more so roughly around 0.000016 ohms.
I searched quite a number of websites for an answer, but no joy as yet..(for this specific question).
If a car battery is short circuited with a wrench that has 0.5 ohms resistance, then theoretically using Ohm's law the current = V/R = 12.65 volts / 0.5 ohms = 25.3 amperes.
Many people (even on this site here) claim that 100's to 1000's of amperes will flow through the wrench and weld it to the terminals.
How is that possible when only max 25.3 amperes can theoretically flow through that wrench of 0.5 ohms using 12.65 volts?
NB: I measured the resistance of the wrench using a home "basic" use multimeter and it shows that best resistance of the wrench is 0.5 ohms. I hope that I measured this correctly. 🙂
Best Answer
A wrench does not have a resistance of 0.5 ohms, it's way lower.
Your basic multimeter cannot measure resistances to better than an ohm or so, the resistance of leads, and the unreliability of contact resistance make it impossible.
The way resistances as low as a wrench are measured is to use a 4-terminal Kelvin method. What you do here is to pass a current through the sample using two terminals, then measure the voltage across the sample using a different pair of terminals. With a wrench, if you used perhaps 1A from end to end, you would see a few mV or so voltage drop.
Let's put some numbers on your wrench. I don't like looking up resistivity, the large factors of 10 cause me concern whether I'm going to get them right on the back of an envelope, so I remember just one fact. A 1m length of 1mm\$^2\$ copper wire is about 17mohm, and then work from there.
Let's assume your wrench is 250mm long, and has a 10mm x 5mm shaft. It's 1/4 of 1m long, and 50mm\$^2\$, so is 1/200th of the resistance of my 1m x 1mm\$^2\$ wire. If it was made of copper, it would have a resistance of 17mohm/200, which is roughly 100μohm. But it's not copper, it's steel, and probably an alloy. After a quick rush around Wikipedia, let's assume it's 50x more resistive than copper, so has a resistance of about 5mohm.
12v dropped across 5mohm would give a current of 2400A. The CCA of the battery is way below that, so the wrench is not limiting the current, the battery is.
Contact resistance is a further complication. In the case of a battery shorted by a wrench, there's likely to be a plasma arc between the contacts, which can have a very low resistance indeed. The small contact area is also worth considering, though as that region is very short, it's often insignificant compared to the length of the conductor.
In practical terms the true resistance of the wrench is close to zero.The battery will deliver the maximum instantaneous current that can be extracted from its cells, which will be way lower than any calculations you make. The net effect is that the wrench will become essentially a fuse: it will burn through at its narrowest point. I have seen it happen to a crescent spanner, and it is spectacular, as it blew the head clean off. Fortunately, the person who did it was not hurt, but it was very dangerous and he was very lucky. It may well also explode the battery, particularly if the spanner is big enough to sustain the current for a little longer.
DO NOT RISK DOING THIS, IT MAY WELL KILL YOU OR AT LEAST GIVE YOU SEVERE ACID BURN DAMAGE. In short, don't be an idiot.