# Real Batteries

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! :)
A typical new AA battery has an internal resistance of 0.1Ω, but its internal resistance increases with age and use. As the internal resistance rises, the battery's potential difference becomes more sensitive to the "load" (the amount of current) on the battery. You may have had the experience of testing an older battery in a battery tester, finding it was good, but then having the battery not work in an electronic device. Battery testers put very little load on a battery, so as not to drain them, so their reading is pretty close to the emf of the battery $$\Dl V\approx {\cal E}$$. Put the battery in a high-load device, however, the potential difference across its terminals drops off quickly if it has a high internal resistance.