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When a battery is being charged, let's say it has an emf $\mathcal{E}$ and internal resistance $r$. At first it's completely drained, and we use a battery to charge it back, and lets say the emf of battery we use for the charging is $\mathcal{E}'$; it has negligible internal resistance with an external resistance $R$ in series.

When asked to find the current in the circuit, the formula for this calculation is $$I=\frac{\mathcal{E'-E}}{R+r}.$$ Now I can understand why the net resistance in the circuit is going to be $R+r$, but why is the net voltage in this case considered to be $\mathcal{E'-E}$ ? The first battery is fully dried out, so it doesn't have its capacity to give power. Therefore, it must also not be able to resist external supply, but with this formula it resists the supply. How can I understand this intuitively?

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The batteries, and the external resistor are in series, but the batteries are oriented with opposite polarizations, right? The (+) of the charging battery is connected to the (+) side of the drained battery, so the Loop law says we should subtract the emfs as we go around the loop. Even if the battery is mostly drained, $\mathcal{E}$ might not be zero. Moreover, as it charges up, $\mathcal{E}$ definitely becomes nonzero.

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  • $\begingroup$ > Loop law KVL is about sums of potential differences, not emfs. If we base our thinking on battery emfs, it is more appropriate to apply the original Kirchhoff's second circuital law, stating that sum of emfs in a closed path equals sum of terms $R_kI_k$ in that path. $\endgroup$ Aug 10, 2023 at 21:22
  • $\begingroup$ I'm not sure what you are trying to say here. The Loop Law says in this case that $\varepsilon-Ir -\varepsilon'-iR=0$. When writing the Loop Law, one adds the EMF (which results in a voltage gain going from (-) to (+)) while subtracting terms $iR$ which are voltage drops as one goes around the loop. $\endgroup$ Aug 24, 2023 at 16:03
  • $\begingroup$ KVL is a modern formulation of a rule, it refers to potential drops in closed loop, not to emfs. The way you're describing the procedure using emfs and terms $R_k I_k$ for all items in the loop, you're using the original formulation of the Kirchhoff 2nd circuital law, not KVL. It's a terminology detail, the resulting equations are the same. $\endgroup$ Aug 24, 2023 at 16:10
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If the battery emf $\mathcal{E}$ and battery internal resistance $r$ in the equation are for a fully discharged battery then $\mathcal{E}$ will generally be less than the fully charged battery emf, and $r$ will generally be greater than the fully charged battery resistance.

So $\mathcal{E}$ an $r$ are both variables, with $\mathcal{E}$ increasing and $r$ decreasing during charging.

Hope this helps.

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  • $\begingroup$ $r$ can also slightly increase in the later stages of charging. $\endgroup$ Aug 10, 2023 at 21:15
  • $\begingroup$ @JánLalinský yes, just wanted to keep it simple. $\endgroup$
    – Bob D
    Aug 10, 2023 at 21:28
  • $\begingroup$ Of course, I think the answer is good, just wanted to improve on that detail. $\endgroup$ Aug 10, 2023 at 21:30
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The first battery is fully dried out, so it doesn't have its capacity to give power. Therefore, it must also not be able to resist external supply, but with this formula it resists the supply.

Imagine instead of chemical batteries we have gravity batteries: masses that can be raised and lowered by receiving or producing power.

If we allow the mass to drop all the way to the floor, we cannot get any more energy from it. But if we want to recharge/lift it again, we have to overcome the full weight. It can resist something lifting it even when it is completely on the floor.

An ideal chemical battery might have exhausted 100% of the products that produce power, so that no energy can be gleaned. But it will take a specific voltage to drive that reaction in reverse.

In both cases it's the process that creates the resistance (raising the mass, driving the chemical reaction), not the quantity of energy available within the system.

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