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We all know the usual (simplistic) model for batteries, that is, representing it by a pure constant voltage source in series with its "internal resistance". This is probably sufficiently precise for many applications during a relatively short time. I think this model (implicitly ? / not at all?) assumes that the battery is discharging into some load, or what is the same, that the current is flowing from the + terminal to the - terminal. But let us assume that we have a rechargeable battery (say a lead acid battery) and, for the sake of simplicity, that is it is fully charged (so that the pure voltage source of the model cannot much increase its voltage). To put flesh on bone, let us assume its voltage is 12 V.

If this battery is submitted to a somewhat higher voltage in the reverse direction, say +17V at its + terminal, is the above model still approximately valid ?

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  • \$\begingroup\$ Driving current the wrong way through a battery is also known as charging it. Some batteries don't like to be charged, and no type of battery likes to be charged if it already has a full charge. Also, as Tony's answer points out, batteries only behave like voltage sources within some range of currents. Exceed their limits, and they can behave in other ways (e.g., like fireworks). \$\endgroup\$ – Solomon Slow Apr 8 '18 at 21:02
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$$Yes$$ it is still valid.

schematic

simulate this circuit – Schematic created using CircuitLab

Quiz. If you understand now, what is the new battery voltage when these two batteries are connected? (expect sparks to fly)

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  • \$\begingroup\$ But if battery was ideal with ESR or Rs=0 then infinite current. \$\endgroup\$ – Sunnyskyguy EE75 Apr 8 '18 at 18:54

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