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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. :)

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    \$\begingroup\$ Check the reistance of your multimeter test leads, its probably in the order of 0.5Ω \$\endgroup\$ – PlasmaHH Jul 2 '18 at 8:34
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    \$\begingroup\$ Some types of wrench will have a DC resistance of almost infinity when facing a powerful enough battery - the state where the metal still exists can be considered transient. \$\endgroup\$ – rackandboneman Jul 2 '18 at 9:42
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    \$\begingroup\$ Instead of editing the answer into your question, you might want to mark the answer of @Neil_UK as correct. Unlike a forum, the Stackexchange network will not lose a marked answer in the static of other content. \$\endgroup\$ – I'm with Monica Jul 2 '18 at 14:25
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    \$\begingroup\$ I have seen some questions where the OP answered and accepted his/her own question IN THE ANSWER SECTION. That works, too if no other answer is correct. But it is better to mark an answer if possible. \$\endgroup\$ – mkeith Jul 2 '18 at 15:13
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    \$\begingroup\$ @arioch, sure a lot can relate it out but they have resolution of. 1ohm oily anyways so it would show 0 \$\endgroup\$ – PlasmaHH Jul 3 '18 at 17:15
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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.

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  • \$\begingroup\$ and to get an idea of the resistance usually the resistivity of the material is used with the length and the area to calculate it... \$\endgroup\$ – Solar Mike Jul 2 '18 at 8:34
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    \$\begingroup\$ Thank you Neil. I appreciate your feedback. Seems like my error is that I am using the wrong measurement tool to establish the resistance of the metal in query. What would you advise is the true resistance of a standard wrench (steel/vanadium alloy) Thanks so much! Satz. \$\endgroup\$ – Satz T Jul 2 '18 at 8:38
  • \$\begingroup\$ @SatzT A little research on your part would have found you this : eddy-current.com/conductivity-of-metals-sorted-by-resistivity \$\endgroup\$ – Solar Mike Jul 2 '18 at 8:40
  • \$\begingroup\$ Thanks Solar Mike, I appreciate your feedback. Thanks so much for that reference as well. I really appreciate your time taken to send me that. Wishing you a pleasant day! Satz. \$\endgroup\$ – Satz T Jul 2 '18 at 8:51
  • \$\begingroup\$ You forgot to include resistance of contacts. It would be limiting factor too. \$\endgroup\$ – Chupacabras Jul 2 '18 at 9:36
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I think what all the other answers are leaving out is the internal resistance of the battery. When dealing with large currents and low resistance shorts, this becomes the significant factor limiting current.

An ideal battery can be modeled by a voltage source, but real batteries act more like a voltage source in series with a resistance.

$$V = IR$$

Let's imagine a hypothetical example. Say you have a 12 V battery. Now short the leads with a 0.1 ohm resistor. According to ohms law, you get 120 A.

$$V = I(R+R_{internal})$$

Now imagine that same battery, except with a 1 ohm internal resistance. With the same short, you get 1.1 ohms of resistance, or approximately 10.9 A. Big difference!

This should line up with everyday experience. When you directly short a battery, you don't get infinite current. You get it's voltage divided by it's internal resistance.

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    \$\begingroup\$ Car batteries, however, are designed to have surface area at the expense of pretty much everything else. They don't like deep discharge, but they have internal resistances in the milliohms so that they can (briefly) supply hundreds of amps to a starter motor. \$\endgroup\$ – hobbs Jul 2 '18 at 22:42
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    \$\begingroup\$ True, however, I've read that the internal resistance of a lead-acid batter is in the range of 20 miliohms, and other answers put the resistance of the wrench at 5 miliohms. This still makes the internal resistance of the battery the most significant factor limiting current. \$\endgroup\$ – Michael Molter Jul 2 '18 at 23:01
  • \$\begingroup\$ I'll accept that, but Neil_UK's answer also says "the CCA of the battery is way below that, so the wrench isn't limiting the current, the battery is", which amounts to the same thing (albeit expressed differently), and the example of a battery with a 1 ohm internal resistance makes the effect seem more drastic than it is in fact. \$\endgroup\$ – hobbs Jul 2 '18 at 23:32
  • \$\begingroup\$ @MichaelMolter I clearly mentioned the internal resistance of the battery in my answer - perhaps you missed that... \$\endgroup\$ – Solar Mike Jul 3 '18 at 7:25
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Your measurement of resistance is erroneous - The current flow will be limited by the resistance of the wrench and the internal resistance of the battery - both very low.

There will be a current flow in the region of, or more, than 1000A easily. If it were not so dangerous I would suggest that you try it, BUT I have seen batteries explode from this sort of thing...

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  • \$\begingroup\$ Agree. Don't try it. But if you do, you will never look at the humble 12V battery the same way again. \$\endgroup\$ – mkeith Jul 2 '18 at 8:57
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    \$\begingroup\$ @mkeith I have used long jump leads with two batteries to weld bits as a temporary repair... but one does need to know what one is doing... \$\endgroup\$ – Solar Mike Jul 2 '18 at 9:03
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The contact resistance between your multimeter wires and the wrench is 0.5 Ohm.

The wrench, when pushed to the soft metal connecting lugs of the battery, will have a much lower short circuit resistance.

Still, the contact resistance it the highest of the short, hence the sparking. Lots of heat is wasted at the contact area.

Excellent do not try this at home videos can be found online.

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For resistance of the wrench is so low that you will get more accurate results that are only lower each improvement. A battery is a beast, and should not be taken lightly. Its ability to conduct instantaneous current is somewhat indicated by CCA (Cold Cranking Amps, as it turns out) which I don't know the exact definition and conditions of the parameter. (edit, SAE has a standard for that.) But I think you can take the number as the peak current the battery can conduct. (I think still valid after the edit, the actual capacity of the battery conducting this current is much higher.)

(Source: A truck battery can conduct ~1000Amps and I once welded a cable easily on the copper plate on the testing table during an EMC test.)

First improvement will be a 4-point measurement. A multimeter can only measure the resistance between its terminals, which means that it will measure the cables, flimsy contacts that you make with the probes, and the wrench. This is because in a 2-point measurement the meter conducts a current through its terminals and measures the voltage between the terminals. On a 4-point measurement, the meter conducts current through a pair of cables, and measures via another pair of cables, which conducts no current but carries the voltage without any alteration to the terminals. This way one can actually measure the resistance between two point contacts.

Another limitation would be the fact that one can sustain only a limited amount of surface area with the probes tips. A wrench can have a larger contact with the terminal of the battery, since the terminals are also made to be soft. I think the surface area will increase after the contact, because the wrench actually might fuse with the terminal (although, in many accidental cases it does not fuse, I think the contact creates a gas which repels the wrench, at least this is what i felt once just before breaking into cold sweat.)

After the wrench fuses, we have nothing we can do but wait (dodge?) for it to become so hot that it glows red. Only then, its resistance might increase, but not so much I presume, and I think we have already damaged the battery irrecoverably, and some chemical process in the battery will deteriorate its performance and stop conducting that much. (It might also mean that the battery will create gas and explode. Haven't trotted those lands.)

I would prefer to calculate the resistance by checking the type of stainless steel it was manufactured from. Please stay safe. (:

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    \$\begingroup\$ A truck battery will be on the order of 10-15C. The "C" rating is how many amps the battery can provide, as a function of its storage capacity. A 10C battery with a capacity of 85Ah can supply 850 amps. A 1000C battery is more like a supercapacitor, in that the 1000C rating means the battery could be drained from full in just over 3 seconds. \$\endgroup\$ – Bryan Boettcher Jul 2 '18 at 18:40
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    \$\begingroup\$ Correct, I was mistaken. The value is named as CCA. \$\endgroup\$ – mehmet.ali.anil Jul 2 '18 at 18:57
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Get some terminals similar to those on your battery and rest the wrench on them. Use a current-regulated power supply to apply 1 amp between the terminals. Use your multimeter to separately measure the voltage drop across the wrench and across each contact point. This will tell you where the biggest spark will be and you will not be disappointed by failure of your wrench to properly vaporize.

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I grew up in North Dakota and once in the 1970s, when faced with a -38F morning in January and a car without a headbolt heater plugged in, i did the "wrench trick" (i remember i used a large kitchen knife blade) to warm up the lead-acid battery. i was told 4 seconds and that's what i did. little sparks when i connected, and a bit more sparks when i disconnected. there were two small beads at the points of contact, but i was still able to pull the blade off.

and it worked. before the engine would hardly turn. after the 4-second short circuit, the battery delivered sufficient current to turn the engine over, and in about 10 seconds of cranking, the car started hesitantly.

saved my ass.

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Yes, your resistance measurement is likely incorrect and the resistance is lower. A standard Lead acid battery can put out as much as 1,000 amps for a short while, and probably will in the case of the wrench, if the contact areas are clean.

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  • \$\begingroup\$ This is essentially the same answer as Solar Mike, on a question that already has way too many answers IMO. \$\endgroup\$ – dim lost faith in SE Feb 5 '19 at 22:21
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Using Ohm's law it would be 24 ampere maximum. This is approximately 288 watts of power. In reality, some wrenches have much lower resistance and the question given is calculated on this data.

With all that power dissipating within the battery and no external circuit to drive it would start to heat up and possibly explode or go on fire in a short time. The internal electrodes would buckle and the battery would be damaged if left in this short circuit state for 30 seconds or more.

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  • \$\begingroup\$ Oh heavens no! Please do not suggest that someone can short a battery with a wrench for up to 30 seconds. You're off by a factor of hundreds, and by 30 seconds, you are likely to have melted the wrench, exploded the battery, suffered severe burns, sung with the angels, or all of the above. An answer like this is dangerous. \$\endgroup\$ – Dawood ibn Kareem Jul 3 '18 at 7:28
  • \$\begingroup\$ It's worth noting that if 24A were the actual correct figure, this would be perfectly safe: car batteries are specified for substantially higher currents than 24A, so would not be damaged. Typically, they can sustain several hundred or even a thousand amps for short periods (minutes or longer) and maybe a few hundred amps for as long as their charge lasts. \$\endgroup\$ – Jules Jul 6 '18 at 10:39

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