Can single DC 1.5V battery generate 1500 watts of power? 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?
 A: 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.
A: 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.
A: 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.
A: 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.
