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Why the terminal potential difference greater than the emf while charging the battery by passing current $I$? When the current is drawn from the circuit terminal potential is less than the emf?

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  • $\begingroup$ How would you expect charging to occur if the applied charging voltage were not greater than the battery emf? $\endgroup$
    – Bob D
    Apr 23, 2020 at 14:45
  • $\begingroup$ If the current is flowing from the plus poles to the minus pole of a battery the potential difference on the battery terminals is greater than emf. Because the external field is forcing to flow the current in the opposite direction than when that is absent. $\endgroup$
    – Constantin
    Apr 23, 2020 at 15:36
  • $\begingroup$ When charging it is like 2 batteries placed in parallel with unequal voltages. See: youtube.com/watch?v=XoLzn49IZjM with relvant links in about section for theory. Also when charging overpotential exists pushing the voltage higher, when discharging overpotential push the voltage lower. Overpotential is energy loss (mainly as heat) in the battery with the 3 main losses being Activation loss (slow reaction kinetics), mass transport/concentration loss (slow diffusion) and ohmic losses $\endgroup$
    – ChemEng
    Apr 23, 2020 at 20:35

1 Answer 1

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A battery acts like an emf in series with a resistor. When supplying current, the resistor causes a loss of output voltage. When being charged, the voltage drop on the resistor requires that the termnal voltage be larger than the emf. (The emf can depend on the level of charge.)

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