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Let's take, for example, a 9 V battery. Forgetting about internal resistance or any temperature restrictions, what is the maximum current I can draw from this?

Using Ohm's law with a 1 Ω load, this should give us:

  • V = I/R
  • I = 9 V * 1 Ω
  • Current = 9 A

According to my calculations, this would give us ≈3.5 min of battery life

I also thought of it like this:

  • 9 V battery, 550 mAh battery life
  • 550 mA for 1 hour
  • 550 mA/h * 3600 secs = 1980 A for 1 sec

Drawing this much current at 9 V would require around 5 milliohms according to my calculations. I know this isn't possible in the real world, but theoretically maybe?

Even theoretically ignoring the temperature restrictions and internal resistance, this seems obviously impossible. Where am I going wrong?

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    \$\begingroup\$ Where you're going wrong is ignoring the device's thermal limits and internal resistance. \$\endgroup\$ – The Photon Jun 5 at 18:50
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    \$\begingroup\$ So those are the only issues with this? If we theoretically ignored the internal resistance and temp restrictions, this would be possible? Obviously it's not, just wanting to see if I had my math right. \$\endgroup\$ – redstone851 Jun 5 at 18:55
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    \$\begingroup\$ Yes, theoretically anything is possible if you just ignore physical reality. What is your point? \$\endgroup\$ – StarCat Jun 5 at 19:07
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    \$\begingroup\$ No point, just wanted to see if I was doing the math correctly and that those were the only two things that I will have to factor into my math. (Thermal limits and internal resistance) \$\endgroup\$ – redstone851 Jun 5 at 19:09
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    \$\begingroup\$ While everyone else has pointed out how real life is important here, I just want to add that your equations are the wrong way around. V = I*R, not the other way around. \$\endgroup\$ – usernumber Jun 6 at 14:49
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If you "forget about" internal resistance, then the maximum current is infinite. An "ideal" component, non-existent in the real world, can provide mathematically "pure" infinite or zero amounts of resistance, voltage, current, and all the rest.

Different battery compositions will have different amounts of real-world "impure" limitations. Internal resistance, temperature versus performance characteristics, "memory" and recovery effects, and so on.

One of the difficult times I had learning about electronics was doing calculations and then wondering why the physical components on the breadboard were different. The figures on paper say I should measure 9 volts. I'm actually measuring 8.654 volts. What gives?

A short length of wire might well be only 5 mΩ, but when you connect the battery using only the wire, it doesn't vaporize the wire with a massive surge of almost 2000 amperes. Why? Because the battery is limited by real-world physics.

Some batteries are capable of some extremely high current. Consider automotive "wet cell" lead batteries. You'll find that they're capable of 1000 amperes or more, especially for turning over huge engines during start. In electronics and physics, many things are a trade off. If you want super high current, you may have to accept lower voltage, lower battery life, or extremely high cost.

A capacitor, as another example, can supply extremely high currents (compared to batteries), but they store charge, and are not a charge pump, as a battery is. As such, they're sort of like super-high-speed batteries with extremely limited capacity.

It was the biggest eye-opener for me as a kid in school to realize that applying Ohm's law to components was not exactly straightforward. You have to take the physics into consideration, and it's messy. A capacitor isn't just a capacitor: it has some resistance and inductance as well. The best way to think about components and batteries, I think, is that any component is a mixture of a bunch of other components, but imagine a control panel with sliders. A resistor might have its "resistance" slider at a large amount, but the "capacitance" and "inductance" sliders can't be at zero. A wirewound resistor, for example, will have more inductance than say a carbon composition resistor.

Your math isn't wrong, but it's for ideal components. Check out a battery datasheet; it'll provide you with some figures that show where it isn't exactly ideal.

(If you happen to have a 2000A-capable 9V battery, I know some electric vehicle engineers that would like to chat with you!)

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    \$\begingroup\$ Hmm, wire up about 3000 of them in parallel? \$\endgroup\$ – Michael Jordan Jun 6 at 3:14
  • \$\begingroup\$ A car battery does nicely, though! \$\endgroup\$ – MicroservicesOnDDD Jun 6 at 8:18
  • \$\begingroup\$ A component being a combination of "resistance", "capacitance" and "inductance" is still only a model (possibly a useful model). \$\endgroup\$ – Peter Mortensen Jun 7 at 0:27
  • \$\begingroup\$ @Peter Agreed. One step at a time. :) \$\endgroup\$ – JYelton Jun 7 at 0:29
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Even theoretically ignoring the temperature restrictions and internal resistance, this seems obviously impossible. Where am I going wrong?

The answer is right there in your question summary. You're overlooking the fact that while in theory, theory and practice are the same, in actual practice, they're not.

You can't ignore the internal resistance (actually, the real-world electrochemical processes that we model as resistance), because it directly limits the short-circuit current, and for high-performance batteries it can lead to battery damage (sometimes spectacular) due to overtemperature.

A really simple theoretical approach treats a battery as a constant voltage source -- but this only works for applications where the combination of current draw, run time, and sensitivity to voltage drop is low. Where the current draw is high enough that either run time is too low or voltage drop too high, then you need to plug more complexity into your theory.

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When we consider a battery to be a voltage source, that assumes that you are using it in the regime where that is a good approximation, because an approximation it is. Over a range of discharge currents it will be a good approximation, but at very low or very high currents it will fail. Similarly, when we draw a wire in a schematic we consider it to be zero resistance most of the time, assuming that the voltage drop will be small enough not to matter.

I typed "9V battery discharge curve" into a famous search engine and one of the results was this page. It shows results at 100 mA and 500 mA, commenting that 500 mA is an unreasonably high current for such a battery and even 100 mA is rather high. If you want to use batteries outside the usual range, I would look at specific manufacturers to see if they have data. It will probably vary more than capacity at nominal current, because capacity is what people usually buy for.

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Of course there are other physical effects that prevent arbitrarily high currents in a short time:

  • Inductance prevents the curent through the wires to change too fast (e.g. you will have difficulty to empty the theoretical battery in 1ns)

  • In physics, similiar to the maximum speed of light, there is a maximum power through a surface of any size; it is c^5/(4G) or 9.1*10^51 W

I am sure there are many more effects that prevent a perfect battery from being emptied in arbitrary short durations.

(the physical power limit corresponds to annihilating the matter of the sun into pure energy and outputting it through a surface within 20µs)

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  • \$\begingroup\$ What's the derivation of that maximum power, just out of curiosity? \$\endgroup\$ – Hearth Jun 7 at 18:33
  • \$\begingroup\$ The gist is that the highest possible energy concentration is a black hole. Time stands still, if you enter the event horizon (Schwarzschild radius). If the sun was a black hole, it would have a diameter of 6 km; the energy could leave with a maximum speed of c (300.000km/s), taking 20µs. This paper (arxiv.org/pdf/physics/0309118.pdf) has some other interesting units with absolute limits on pages 9 and 11. Including maximum voltage, minimum inductance, minimum capacitance, minimum resistance "per conductance channel". \$\endgroup\$ – Sebastian Jun 8 at 10:54
  • \$\begingroup\$ The same is true for smaller black holes; if you have one the size or the mass of a 9V battery, the energy could escape more quickly, but would be proportionally less. Interestingly, the maximum reachable power stays the same value. You cannot put many of those batteries too near to each other and multiply the output power as the spacetime would curve and form a larger black hole at the point, where you combine that power, or you (as observer) would be inside an event horizon and could not receive that power. \$\endgroup\$ – Sebastian Jun 8 at 11:12
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The battery has its internal resistance that is not only non-zero, but also non-linear and also depends on temperature and the state of charge of the battery. For a typical 6f22-form factor battery it is something 2-20 ohm for a new battery at room temperature. It gets higher as the battery gets discharged, rises with discharge current and gets a bit lower for moderately elevated temperature (say, ~50C).

The initial short-circuit current for such a battery is ~1 Ampere.

The dependance between the useful capacity and the discharge current is approximated by https://en.wikipedia.org/wiki/Peukert%27s_law . The capacity is linear only for small currents and drops at higher discharge currents.

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