In Section 10.1 I said that the primary function of a battery is to maintain a constant potential difference, but in practice batteries can't always manage to do that. For example, suppose I connect a standard AA battery (1.5 volts) to the ends of a wire with resistance 0.001Ω. How much power will the wire emit? $$P=\frac{(\Dl V)^2}{R}=\frac{(1.5)^2}{0.001}=2250\u{W}$$ That's enough to light up more than 50 large light bulbs! Experience tells us that a real AA battery isn't able to supply that much power

$$\def\Rint{R_{\rm int}}$$A simple way to model this limitation is to suppose that real batteries are made up of an ideal battery (which always maintains a constant emf \({\cal E}\)) and some internal resistance \(\Rint\):

The potential difference across the real battery isn't a constant, but depends on the current that flows through the battery: the more current you try to draw, the lower the potential difference will be.

What is the maximum current that a real battery can provide, as a
function of its emf and internal resistance?

The battery is only going to provide current if the potential
difference across its terminals is greater than zero. If \(\Dl V>0\), then
$${\cal E}-I\Rint > 0 \implies \boxed{\displaystyle{I<{{\cal E}\over \Rint}}}$$

What is the maximum power that a real battery can provide to the wire
or devices connected to it?

The battery produces \(I{\cal E}\) of power, but some of that power is
dissipated by the internal resistance as heat (which is why batteries
get warm, especially when under a heavy load). The power provided by
the battery is
$$\begin{eqnarray}
P&=& I\Dl V\\
&=& I({\cal E}-I\Rint)\\
P&=&I{\cal E}-I^2\Rint\\
\end{eqnarray}$$
The graph shows this power as a function of current: we see that when
the current reaches its maximum value \({\cal E}/\Rint\), the power
actually drops to zero. The power is maximal when
$$0={dP\over dI}={\cal E}-2I\Rint \implies I_{@max}=\frac{\cal E}{2\Rint}$$
where the power is equal to
$$\begin{eqnarray}
P_{max}&=&I_{@max}{\cal E}-I_{@max}^2\Rint\\
&=& \frac{\cal E}{2\Rint}{\cal E}-\left(\frac{\cal E}{2\Rint}\right)^2\Rint\\
&=& \frac{\cal E^2}{2\Rint}-\frac{\cal E^2}{4\Rint}\\
&=& \boxed{\displaystyle\frac{\cal E^2}{4\Rint}}\\
\end{eqnarray}$$

What is the maximum amount of power that a AA battery can provide if
its internal resistance is 0.1Ω?

Using the formula we just derived, and remembering that a AA battery
has an emf of \({\cal E}=1.5\u{V}\):
$$P_{\max}=\frac{(1.5\u{V})^2}{4(0.1\u{\Omega})}=5.6\u{W}$$
which is much smaller than the 2250W above! :)