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Suppose there is a DC $1.5\mathrm{V}$ battery connected with $1.5\times10^{-3}\mathrm{\Omega}$ resistor.

Then the amount of circuit current is $I=V/R=10^3\mathrm{A}$. (according to Ohm's law)

I know the formula $P=IV$ which is electrical power.

Finally I got the electrical power of $1500 \mathrm{W}$ in this case.

I can't believe this! How can single DC 1.5V battery generate 1500 watts of power?

What am I missing?

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    $\begingroup$ How long does it generate that power? $\endgroup$ – Jon Custer Sep 3 '18 at 14:01
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Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 and 0.1 ohms, limiting its power output to a few watts.

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  • $\begingroup$ I'm sorry but I don't understand. I have a multimeter right next to me and I've measured the voltage of 1.5V energizer battery. It was 1.552[V]. And also I've measured the amount of current of the battery which is 4.21[A]. And the resistance of my multimeter's measuring tip is 0.4Ω. Those are all measured value, and it follows Ohm's law well. The power of this battery is 6.53W(according to the formula). If the resistance of the tip goes weaker, the amount of current rises, also power goes stronger. In this reason, there is no power limit. $\endgroup$ – user28936 Aug 29 '13 at 2:31
  • $\begingroup$ I find it hard to discern your precise point, but if you reduce the resistance of the circuit more and more, the battery's internal resistance will become increasingly important. $\endgroup$ – Mark Eichenlaub Aug 29 '13 at 2:54
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    $\begingroup$ No, you can't just make some sort of measurement and extrapolate down to 10^-3 ohms. As the external resistance decreases, the voltage across the external resistor will decrease as well. $\endgroup$ – Mark Eichenlaub Aug 29 '13 at 3:11
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    $\begingroup$ Damn, I was so stupid. I've solved the power function with variables, but I didn't noticed. You were right. The internal resistance limits the power. So sorry. s8.postimg.org/mjvexxwp1/K_20130829_447352.png $\endgroup$ – user28936 Aug 29 '13 at 3:25
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    $\begingroup$ Note that dissipating several Watts in internal resistance will heat up the battery, which is just one reason why the "fixed internal resistance" model is flawed. $\endgroup$ – MSalters Aug 29 '13 at 7:41
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Unless a battery is made using superconductors, which hasn't happened yet, at least in commercial batteries, there is a resistance to the materials that conduct the electricity inside the battery, usually only a small fraction of an ohm. For example, on the Duracell Ultra AA (1.5V) battery's datasheet, it says that the internal resistance is approximately 81 milliohms. So the battery itself, because it is not perfectly conductive, provides resistance to the circuit. If you do the math, 1.552V / 81milliohms = 19.16 Amps, the theoretical maximum current output of the Duracell Ultra AA battery. Multiplied by 1.552V, that gives you 29.74 Watts, the maximum power output of the battery.

Of course, different batteries have different internal resistances, but all commercial batteries have an internal resistance which limits the current and power output, preventing a single AA battery from outputting 1500 Watts. If you want to calculate the actual maximum power output of your battery, look up the battery's datasheet, which lists technical information about the battery, and is sure to have the internal resistance listed, though possibly not by that name.

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    $\begingroup$ ... except that 29.74 W would all be dissipated inside the battery, and the actual power output would be 0. Max power output is obtained by matching the load to the source resistance, where the output power is 1/4 of what you calculated. $\endgroup$ – Art Brown Aug 29 '13 at 3:54
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Try measuring your battery voltage with and without the load connected. There will be a (small) reduction in voltage when current is flowing. That difference, divided by the current, is the internal battery resistance; it may be small but it's non-zero.

This internal resistance becomes increasingly important as you scale up the current (by reducing the load resistance value). Even 1 milli-ohm at your extrapolated load has a major effect, and your battery should have considerably more than that.

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  • $\begingroup$ Let me summarize this. Let the internal resistance of the battery be $r$ and electromotive force be $\epsilon$. If the current of $I$ flows through the circuit which resistance is $R$, then the voltage loaded on the resistor $R$ will be $\epsilon-Ir(=IR)$. And the current $I=\frac{\epsilon}{r+R}$, the voltage $V=\frac{\epsilon R}{r+R}$. Finally the power will be $P=\frac{\epsilon^2 R}{(r+R)^2}$. $\endgroup$ – user28936 Aug 29 '13 at 2:43
  • $\begingroup$ @user28936: Yes. $\endgroup$ – Art Brown Aug 29 '13 at 2:47
  • $\begingroup$ $R$ is a constant(suppose the temperature not changes) and $\epsilon$ goes lower as long as we use battery(consuming energy). BTW why does $r$ exist? Is it a constant or a variable? Anyway I think I've solve the current problem. $\endgroup$ – user28936 Aug 29 '13 at 2:49
  • $\begingroup$ @user28936: r is an inevitable feature of real batteries. It's roughly constant, but electrical models of batteries can easily get very complicated. For instance, see my answer to this question: physics.stackexchange.com/questions/62258/… $\endgroup$ – Art Brown Aug 29 '13 at 2:59
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In this article one can see that an alkaline, $1.5(V)$ battery contains an energy of $9.36(kJ)$. If we assume that the battery has no internal resistance and connect the anode and cathode by a thick metal wire then almost all the contained energy will be released in a very short time because of the very small resistance of the connecting metal. So it could be possible that the $9.36(kJ)$ energy is set free in a short time, like in the case of connecting the two poles by a thick piece of metal. The battery's energy is released in a fraction of a second, which is to say, $9.36(kJ)$ is released in such a way that the generated power is more than $1500(W)$. In this case, the piece of metal connecting the poles heats up fast. The internal resistance though takes care that the energy that starts to flow when we connect the anode and cathode by the thick piece of metal can't flow that fast so it takes longer before the energetic electrons have moved from the minus side of the battery to the positive side of the battery (unless the internal resistance is relatively small).In this case, the battery is heating up on the inside.

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