2
$\begingroup$

Let's assume that a positive charge passes from the negative pole to the positive pole , hence it will gain potential energy since the electrochemical reactions in the battery does work on the charge.

  • Then how is it that a battery can also simultaneously cause a drop in the voltage of a charge due to internal resistance .
  • Let's now assume that we have a simple circuit consisting of a single cell and a single resistor , let's take internal resistance out the picture according to kirchhofs law the voltage drop across the resistor would equal to the voltage gain in the battery , now let's take internal resistance into account , now the voltage drop across the resistor will decrease , but how does the charge know how much voltage to lose at the resistor ?, I know according to ohms law the current will decrease but I want more of a conceptual understanding to this situation .
$\endgroup$
  • $\begingroup$ If it helps, the current through the cell is through the electrolyte and the simple fact is that the electrolyte has low but non-zero resistance. $\endgroup$ – Alfred Centauri Jul 1 '17 at 16:00
3
$\begingroup$

Realistically cells always have some internal resistance, even if it is small. Now consider the circuit diagram bellow for a cell with internal resistance $r$ connected to a resistor of resistance $R$.

When we deal with this scenario we consider this as a series circuit, with suppose the original e.m.f. of the cell of 10 V then as the whole dotted box is now the cell's circuit, due to the internal resistance $r$ some of the voltage from the 10 V e.m.f. is lost there( You can use $V=Ir$ to find how much). The remaining potential difference will be across the external resistor $R$( according to Kirchhoff's law the e.m.f. is equal to the voltage drops in a closed loop) thus the situation is just like a potential divider where voltage is divided across a series circuit proportional to the resistance of the resistors( if you don't know exactly how then it's useful to derive it yourself with total resistance $R+r$ to find current and then the voltage across each resistor). Thus, the potential difference across $R$ is a bit less than 10 V as some electric potential energy is lost by the flowing charge across the internal resistance as the charge can finally make it out of the cell.

For extra information, voltage across $R$,$V_R$ is :

$$V_R=\frac{R}{R+r}(V_{e.m.f.})$$

$\endgroup$
  • 1
    $\begingroup$ "Realistically cells almost always have some internal resistance" - drop the almost; physical cells always have non-zero internal resistance. A cell with zero internal resistance could provide unlimited current while any physical cell has finite short-circuit current. $\endgroup$ – Alfred Centauri Jul 1 '17 at 15:54
  • $\begingroup$ @AlfredCentauri, I'll be a bit speculative on this subject, but I strongly suspect that this internal resistance occurs due to the fact that ions must migrate through an electrolyte inside the battery in order to provide current to the circuit. As the ions migrate, I am certain that there is some type of flow resistance, as the bulk of the electrolyte is not moving, and the ions will undoubtedly be "bumping into" molecules of the electrolyte as they move. This naturally means that the current inside the battery will become limited at some point. $\endgroup$ – David White Jul 1 '17 at 16:03
  • $\begingroup$ How does the voltage in the resistor know that it should drop when internal resistance is taken into account? $\endgroup$ – LM26 Jul 1 '17 at 18:23
  • $\begingroup$ It 'knows' because there IS internal resistance r, the voltage drops as per internal resistance. ( V drop= Ir) $\endgroup$ – Tausif Hossain Jul 2 '17 at 5:18
0
$\begingroup$

The circuit diagram that you have drawn is a model of the system (battery and external resistor $R$) that you are considering.

If you did an experiment to measure the potential difference across the terminals of the battery $V$ (which is 10 V in this case) against the current passing through the battery $I$ your results would produce a graph something like this.

enter image description here

To model this arrangement you assume that you have a perfect cell of emf $10\,\rm V$ and internal resistance $r$.
Such a model will give you the same VI characteristics for the battery.

In terms of energy the internal resistance represents a loss of electrical energy produced by the electrochemical reaction within the battery when a current passes through the battery.

The workings of a battery are exceeding complex as can be judged from the fact that the model described above can only be used if the characteristics of the battery do not change with time.
The reason for this is that the "internal resistance" is not a resistor placed within a cell but rather convenient way of accounting for the complex processes occurring within the battery.
Contributions to the internal resistance will certainly come from the resistance of the metal conductors (which might change due to corrosion or formation of crystals/chemicals on them), the resistance of the electrolytes (which includes ion mobility which may change due to loss or change of composition of the electrolyte) and the resistance of any separators used within the battery (which are supposed to be porous but the pores may be clogged up).
Then there are changes due to the rate of the electrochemical reactions within the battery. So as a battery delivers current many things happen within the battery and these may or may not be reversible and in the real world the internal resistance of a battery depends on many factors including time and temperature.

If you want to know more as to the workings of a battery and hence what might contribute to its internal resistance you need to start looking at specific types of batteries.
As a start you may find this link useful or this one?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.