13
$\begingroup$

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.

  1. Could it be that a car battery has less internal resistance than a garage battery?
  2. 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?
Improve this question
$\endgroup$
6
  • $\begingroup$ You could check this article in wikipedia en.wikipedia.org/wiki/Battery_%28electricity%29 , "battery capacity and discharging". $\endgroup$
    – anna v
    Commented Apr 2, 2011 at 7:30
  • $\begingroup$ And beyond what everyone else said, it should be explicitly stated that Ohm's law isn't a universal law of nature, like Newton's law of gravity, or the Maxwell equations. It is an empirical observation about how some materials behave in some situations. $\endgroup$
    – Zo the Relativist
    Commented Apr 2, 2011 at 14:14
  • $\begingroup$ And you can always decrease the effective internal resistance by ganging several similar batteries together in parallel. Put twenty car batteries together in parallel, and the short circuit voltage can become frighteningly high. $\endgroup$
    – Omega Centauri
    Commented Apr 2, 2011 at 14:45
  • 4
    $\begingroup$ There's a usually unspoken rule of thumb when designing stuff with batteries: The people designing the battery promise they will keep the voltage within a certain range as long as the the current pulled from the battery is less than some maximum current. The people designing other stuff promise to pull less than some maximum current from the battery as long as the voltage applied is in the normal operating range. When these promises are violated, the convenient simple rules of thumb we use ("batteries put out a constant voltage") don't work anymore. $\endgroup$
    – David Cary
    Commented Apr 2, 2011 at 17:15
  • 1
    $\begingroup$ You can get an arc from a 9V "flashlight" battery if you use a NiCd or Li-ion cell. Been there. done that. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented May 26, 2011 at 22:00

5 Answers 5

Reset to default
9
$\begingroup$

The answers of Martin and Edward are quite right, but theý lack a componet which plays a role in the case of that "garage opener battery".

If current is not limited in the circuit outside the battery, first internal resistance can be the limiting factor. But, often in such small (the measure for "small" is the area/volume of active material at the battery poles) batteries the current is limited by the speed of chemical reaction at the electrodes. This is not a linear function of current, thus it cannot viewed as a part of internal resistance.

Does it have anything to do with contact area? If it does, then how would you model it?

Yes, contact area is important, but this is a problem at high current densities mostly. Depending on voltage in such a circuit, small contact area (= high current density) can result in contact welding, sparking, or ignition of an arc, or in some strange semiconductor effects of the contact point. (metal surfaces are nearly always coated by some oxide) Did You ever watch when an electric arc is started by a welder? This is main business when designing contacts in relays, swiches in home, or the switches in 230 kV lines.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ Ah, so it does have something to do with area -- not the contact area between the wire and the battery terminal, but the internal "wetted area" at the boundary of the solid plates and the liquid electrolyte. I guess this is why "deep cycle" batteries have relatively smooth solid plates (wetted area approximately LxW of the plate), while "starter" batteries have plates "like a sponge" (wetted area many times the LxW of the plate). $\endgroup$
    – David Cary
    Commented Apr 2, 2011 at 13:57
  • $\begingroup$ Okay, so what you are saying is similar to David's comment. Once the current gets above a certain range, the chemical reaction can't maintain the voltage $\endgroup$
    – Casebash
    Commented Apr 3, 2011 at 1:02
  • $\begingroup$ @Casebash, No, I see a lot of difference. Especially because Davids comments are somewhat erratic. The one above where he jumps from my explanation of small contact areas to the inside of batteries is very strange. This was reason for me not to respond. $\endgroup$
    – Georg
    Commented Apr 3, 2011 at 11:44
3
$\begingroup$

First an aside, before getting directly to the question.

There are many standard battery types that output 1.5V, with some example common ones being AA,AAA,C,D. You can even buy one of each type with the same exact chemistry inside. So what is the difference between these, and why would a designer require a couple big D cell batteries for a flashlight when a couple AAA have the same voltage rating? There are two main reasons. One is the total energy contained (a D cell is larger in volume, and thus has more room to store chemical energy, and thus can run a device longer than a AAA could). The second is what you are getting at here, and that is the short circuit current. A D cell battery should be able to output more current than a AAA.

This leads us to your question, of how this can possibly fit with Ohm's law since the battery voltage rating is the same. From an electrical engineer's standpoint, the open circuit voltage of a battery is only half the story. One needs to know the open circuit voltage as well as the internal resistance rating, to be able to predict the actual output of a battery into some device. You could consider the "equivalent circuit" of a real voltage source as an ideal voltage source in series with a resistor.

For example, you can look at the technical documents for typical batteries here:
http://www1.duracell.com/oem/productdata/default.asp
and here's one for a D cell alkaline
http://www1.duracell.com/oem/Pdf/new/MN1300_US_CT.pdf
notice it lists the "nominal voltage" and the "impedance". For a D cell the impedance is listed as 136 mOhm. If we look at a AAA using the same chemistry we see
http://www1.duracell.com/oem/Pdf/new/MN2400_US_CT.pdf
the impedance is about 250 mOhm.

While this two parameter description of power sources is intuitive and the first order approach to describing a device, or course the real life devices can be a bit trickier (the impedance can depend on frequency as batteries can only respond so quickly, draining the battery quickly is usually less efficient, etc.). So the technical documents often provide real test curves if you want to see even more detail.

So a little 12V battery in a remote control will not be designed to have the low impedance required to start a car engine. The 12V car battery should definitely be able to put out more current when shorted.

Improve this answer
$\endgroup$
3
$\begingroup$

"Could it be that a car battery has less internal resistance than a garage battery?"

Yes! Let's denote a car battery internal resistance by $r_c$ , a garage battery internal resistance by $r_g$ , resistance of a toucher by $R$ ,voltage of a battery by $U$. Then current through the toucher:

$$I=\dfrac{U}{R+r}$$ The toucher is releasing power: $$P=I^2R= \dfrac{U^2R}{(R+r)^2}$$ Let's introduce the ratio:

$$\mu=\dfrac{P_c}{P_g}= \left(\dfrac{R+r_g}{R+r_c}\right)^2$$ $P_c$ is the power the toucher is releasing on a car battery and $P_g$ is the power the toucher is releasing on a garage battery. Internal resistance of a car battery is estimated about $r_c=0.001$ ohms and internal resistance of a garage battery around $r_g=0.1$ ohms.

For simplicity let's suppose that $R<<r_c$. Then:

$$\mu=\dfrac{P_c}{P_g}= \left(\dfrac{r_g}{r_c}\right)^2=100^2$$

Improve this answer
$\endgroup$
1
$\begingroup$

If you are going to short out a battery (generally a bad idea), you are completing a circuit in which the total resistance is that of the wire plus the internal resistance of the battery (all batteries have some amount of internal resistance due to a variety of factors).

Small batteries are generally designed to drive small current loads, so their internal resistance (perhaps an ohm or two) could be fairly large in comparison to a piece of wire (which might be milliohms or microohms). So, the short circuit current might be just a couple of amps, and doing the calculations, you'd get a few watts of power dissipated inside the battery as heat due to its internal resistance, with a little dissipated as heat in the wire.

Car batteries are designed to efficiently deliver hundreds of amps at close to their open-circuit voltage, because their biggest job is to turn a hefty electric motor to start your car. A car battery will have internal resistance in the fractions of an ohm. In a short-circuit condition, the total resistance will be very low, allowing many hundreds of amps to flow into a "short circuit". Multiplied by the open-circuit voltage, that's a lot of power dissipated within the battery and/or the load. Guaranteed something spectacularly bad will result.

Improve this answer
$\endgroup$
-5
$\begingroup$

It's all about capacity, not internal resistance. Compare a shotgun shell to a .22.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.