I know the relation between the terminal voltage, V and emf of the cell, V' while charging given by

V = V' + Ir

where I = Current in the circuit and r = Internal Resistance of Cell

Can anyone please explain why in charging a cell, the sign changes from - (which is when the cell discharges) to + in detail by deriving the formula ?

Thanx in advance

@Farcher Can u explain/deduce that formula(V = V' + Ir) on the basis of following diagram:-

enter image description here

The diagram shows a cell of emf, V' and internal resistance, r being charged up by a charger.

  • $\begingroup$ The left hand side of the charger is positive and the right hand is negative. Starting at the top right hand corner and applying Kirchhoff's voltage law one gets $V-V'-Ir=0$ Which when rearranged gives you the required result. $\endgroup$
    – Farcher
    Nov 17, 2016 at 16:29

1 Answer 1


To drive a current through a ideal cell (no internal resistance) from its positive terminal to its negative terminal requires there to be an external voltage source which exceeds the emf of the cell $\mathcal E$ by an infinitesimal amount.
However with a cell that has internal resistance $R$ an extra potential difference must be applied across the cell terminals to drive the current $I$ through the internal resistance $=IR$.

So the total applied potential difference across the cell must be $V = \mathcal E + IR$.

It might be clearer if both sides of the equation are multiplied by the current $VI = \mathcal E I +I^2R$.

Now $VI$ is the power being delivered by an external source to the cell.
$I^2R$ is the rate at which heat is produced due to the cell having an internal resistance.
$\mathcal EI$ is the rate at which energy is being supplied to the cell to reverse the chemical change which the cell uses when it is discharging (converting chemical energy into electrical energy.

Now doing the same when the terminal potential difference is less than the emf of the cell.

$V= \mathcal E - IR \Rightarrow VI = \mathcal E I - I^2R \Rightarrow \mathcal E I = VI + I^2R$

$\mathcal E I$ is the electrical power supplied by the cell from the chemical reaction within the cell.
$I^2R$ is again the rate at which heat is produced due to the cell having an internal resistance.
$VI$ is the power delivered to the external circuit by the cell.

So in one case (discharging) the cell is producing the electrical energy from chemical energy and in the other case (recharging) the cell is consuming electrical energy which is being converted into chemical energy.

  • $\begingroup$ I have added some more things in the post. Please Check it out. Hoping you will answer. Thanx and sorry for any inconvenience $\endgroup$ Nov 17, 2016 at 16:20
  • $\begingroup$ can i consider terminal voltage to be equal to EMF of a NON-IDEAL CELL when it is kept in an open circuit? $\endgroup$ Nov 18, 2016 at 5:47
  • $\begingroup$ KES because the current is then zero and so the voltage drop across the internal resistance $IR$ is also zero. $\endgroup$
    – Farcher
    Nov 18, 2016 at 5:50
  • $\begingroup$ I am really very sorry but there's one more doubt. The formula ,V=V'+Ir is valid for an empty cell or for an fully charged-up cell. Can u please explain. I am asking this as there is one question related to it in which a cell having emf of 12V is connected to dc charger having emf of 110V and it's asking about the cell terminal voltage. So what is not seeming right to me in this question is that if the cell has 12V emf(which i think it means it is fullcharged) across it, how will it be charged more? $\endgroup$ Nov 18, 2016 at 6:11
  • $\begingroup$ In a problem such as this you have to assume that the emf of the cell is independent of the state of charge of the cell. In other words there are still chemicals in the cell which when they react will produce an emf of 12 V. The behaviour of real cells as compared with an ideal imperfect cell (constant emf, unlimited chemicals, constant internal resistance) is very much more complex as other chemical reactions within the cell start playing a much more important role. $\endgroup$
    – Farcher
    Nov 18, 2016 at 6:25

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