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What makes the internal resistance of a battery at the microscopic level?

And why does voltage drop when in a circuit compared to the open circuit voltage?

I think that the field between the plates should change, lower voltage means weaker electric field.

And the electric field is generated by excess charges in the two plates. But I don't think that there is less excess charges, because even if one more charge leaves the electrode then it get's remplaced by an other, it will make ,after a long, time the electrodes neutral.

My suggestion is that there is some sort of other field inside the cell that counter acts the electrodes plates when there is a current flowing this field must get stronger.

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  • $\begingroup$ What do you think the internal resistance is due to? Then we will correct or adjust your answer... $\endgroup$ – user207455 Jul 5 at 18:25
  • $\begingroup$ @Solar Done .... $\endgroup$ – mohamed azaiez Jul 5 at 19:22
  • $\begingroup$ I posted my answer not seeing your latest edit so it does not specifically address the edits. Sorry $\endgroup$ – Bob D Jul 5 at 19:30
  • $\begingroup$ @Bob it's my fault i should've made my question clear from the beginning $\endgroup$ – mohamed azaiez Jul 5 at 20:21
  • $\begingroup$ I problem no fault $\endgroup$ – Bob D Jul 5 at 20:24
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What makes the internal resistance of a battery at the microscopic level?

There is no simple answer due to the many different types of batteries and technologies. So the following is only an overview.

There are generally two categories of internal resistance of a battery.

One has to do with the resistivity of the materials of the internal components (e.g electrodes) that conduct charge from and to the terminals.

The other has to do with the mobility of the ions moving internally between the electrodes through the electrolyte and the medium separating the electrodes.

Other factors include, but are not limited to, battery size, the particular battery chemistry, temperature and age.

And why does voltage drop when in a circuit compared to the open circuit voltage? My suggestion is that there is some sort of other field inside the cell that counter acts the electrodes plates when there is a current flowing this field must get stronger.

I think you are getting close, but I think that what you are referring to as "some sort of field inside the cell" is in reality what we call the emf (electromagnetic force) or voltage that the battery generates inside the battery.

The internal battery voltage is in series with its internal resistance. When current is delivered to the circuit there is a voltage drop across the internal resistance. The terminal voltage is then the emf minus the internal voltage drop. When no current is delivered (open circuit), the terminal voltage is the same as the emf.

The figure below illustrates the point. If $R_L$ is infinite (meaning we have an open circuit) the voltage across the battery terminals equals it internal voltage, or emf. For any finite value of $R_L$ current will occur. That means a voltage drop across its internal resistance $R_b$. The voltage available at the terminals will equal the emf minus the voltage drop across $R_b$.

Hope this helps

enter image description here

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  • $\begingroup$ Thanks for the effot you've put in to awnser my question but i would like to clarify something The There are generally two categories of internal resistance of a battery." is the paragraph that correspends to my question i wanted to know at the micro scopic level what makes this resistance and how does it affect the electric field of the battery if you could add more details about this part of a link to a subject / extern resource that would help because i didn't find any on the internet . Thanks :) $\endgroup$ – mohamed azaiez Jul 6 at 0:51

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