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A common real battery is often modelled (to a good approximation) as a perfect voltage source having emf $\varepsilon$ with a series internal resistance $r$. So, when a current $I$ is drained from it, the voltage across the terminals is $\varepsilon - I r$.

Now, does it follow that necessarily the power dissipated as heat equals $I^2 r$? If so, is the internal resistance due to some actual component within a battery?

(This would certainly be the case if the chemical processes inside the battery actually generated an emf $\varepsilon$ and something inside actually worked as a resistance. However, it need not be true if for some reason the "actual" emf is actually current-dependent and so there wouldn't be an exact $I r$ voltage drop across any part of the battery as to dissipate $I^2r$ power.)

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2 Answers 2

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Results for an experiment to measure the voltage across the terminals of a battery$v$ as a function of the current passing through the battery $i$ many look something like this.

enter image description here

Where $v_{oc}$ is the open circuit terminal voltage and $i_{sc}$ is the current when the terminals are short circuited.

This is a straight line graph of the form $v = v_{oc} - ki$

When the terminals are short circuited then $0=v_{oc} - k i_{sc} \Rightarrow k = \dfrac {v_{oc}}{i_{sc}}$

So the abstraction which is used is to say that the $v_{oc}$ is the emf $\mathcal{E}$ of the battery and as $k$ has the dimensions of a resistor make $k = \dfrac {v_{oc}}{i_{sc}} = r$ the internal resistance of the battery.
Note that it is minus the gradient of the battery's $v-i$ characteristic.

The power dissipated in the battery is $i^2r$.

There is no actual resistor in the battery (unless your teacher or lecturer has put one into a black box) rather the internal resistance represents as a lumped component the battery's resistance to the flow of current.
It is not a constant for a battery and changes with the battery's age and the current passing through the battery.

It is a useful approximation to a real voltage source.

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  • $\begingroup$ This is a nice result but doesn't really address the heat question. $\endgroup$ Feb 19, 2016 at 22:42
  • $\begingroup$ I was trying to show that there is no actual resistor in the battery rather it behaves as though there was a resistor $r$ in the battery. A current, $I$, passing through the battery will generate heat and to a reasonable approximation the heta generated per second is $i^2r$. $\endgroup$
    – Farcher
    Feb 19, 2016 at 23:04
  • $\begingroup$ I understood what you were up to, I was suggesting that without actually measuring the heat you haven't excluded the possibility that the voltage sag is due to some feature of the chemistry in the cell. (Not that I believe for a minute that it is, I'm just picking nits.) Of course, that is a harder experiment because you have to address the question of the heat released or absorbed by the chemical reaction in the first place. $\endgroup$ Feb 19, 2016 at 23:08
  • $\begingroup$ from the standpoint of actually using a battery to drive a load, the measured internal resistance of the battery will generate heat, as pointed out above, no matter how the resistance is distributed inside the battery nor whether that resistance is, for instance, in the electrodes or is part of the electrochemistry occurring at the electrode surfaces. The key point is that the battery then must be able to rid itself of that internally-generated heat; if it cannot, it may catch fire or explode. This is a known problem with lithium cells. $\endgroup$ Oct 19, 2017 at 4:56
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An argument I found is actually the energy conservation.

Let's model your idea with an idealized battery with the following features:

  • internal resistance $r_0$
  • current-dependent emf during discharging: $\varepsilon(I) = \varepsilon_0 - k I$
  • charge capacity $Q$
  • no drop of emf nor rise of internal resistance after a partial discharge (not as essential assumption but simplifying the calculations)

From the first two points it follows that the voltage across the terminals is $$U(I) = \varepsilon(I) - r_0 I = (\varepsilon_0 - kI) -r_0 I = \varepsilon_0 - r I.$$ Here $r = k+r_0$ is the apparent inner resistance but the heat produced inside the battery is actually given only by $r_0 I^2$. The emf has an apparently constant value $\varepsilon_0$.

Now connect the battery to a circuit with a resistor with the resistance $R$ until it totally discharges. Then, the total produced heat is $$W = (r_0 + R) I^2 t = (r_0 + R) I Q = (r_0 + R) \frac{\varepsilon_0}{r + R} Q = \frac {r_0 + R}{k + r_0 + R} \,\varepsilon_0 Q.$$ Notice that the amount of this heat depends on what external resistor $R$ you used: it ranges from $W_\text{min}=\frac {r_0}{k + r_0} \,\varepsilon_0 Q$ for a curcuit shorted by an ideal wire up to $W_\text{max}=\varepsilon_0 Q$ for $R\gg r$. However, the total produced heat should be independent of $R$ since it is given by the total chemical energy that was stored in the battery and which we wasted. This paradox does not happen if and only if $k=0$, i.e., if the emf does not decrease with the current.

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  • $\begingroup$ Yep. I've done the experiment (discharge a battery in a calorimeter). Your model worked. I was checking for extra heat generated by possible side reactions, but could not detect any: The current times the difference of the open circuit and under load voltages accurately predicted the dissipation. $\endgroup$
    – John Doty
    Jul 12, 2022 at 23:22

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