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Suppose a cell having emf, V and terminal voltage V' having internal resistance, r connected with an external resistance, R in the circuit in which current, I flows. We know that

V=V'+Ir

Can anyone tell why is that the emf is independent of the external resistance and internal resistance both conceptually ?

Thanks

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The ideal cell is modelled by a fixed emf and fixed internal resistance. By definition the emf is independent of both internal and external resistance, temperature, current, etc.

Real cells do not necessarily behave like this. They get depleted, so the emf decreases over time. You have to recharge or replace them. If you draw too much current the emf also gets depleted quickly. The emf arises from a chemical reaction, and if the temperature rises the reaction goes faster so the emf increases.

The emf of a real cell is only fixed and independent of current for times which are short compared with the cell lifetime and for currents which are not too large.

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  • $\begingroup$ Sorry for this very late reply.@sammy why emf decreases for a real cell. Is it due to internal resistance.(This part is optional) If yes, can you please tell me why internal resistance decreases the emf of a cell when used for longer period of time. Please reply $\endgroup$ – Perspicacious Nov 25 '16 at 9:32
  • $\begingroup$ I do not know enough about cell chemistry to explain reliably. You can read about it at Battery University and the wiki article on Electrochemistry. The main reason why terminal PD declines is the increase in internal resistance due to the build-up of deposits on or around the electrodes which inhibit ion flow. Also cell emf depends on the concentration of the reactants, which decreases as charge is drawn from the cell, but the relation is not linear and the emf is fairly constant for much of the battery life. $\endgroup$ – sammy gerbil Nov 25 '16 at 13:50
  • $\begingroup$ Another article about a more realistic battery model is How Batteries Discharge. $\endgroup$ – sammy gerbil Nov 25 '16 at 13:51

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