Analogies for internal resistance and emf

[The_Lobster]

Hi!

I'm trying to fully grasp why the potential drops when an internal resistance is present in a source of emf. At first I thought that yes it would be harder to "push" the charges to the higher potential, but once there, why shouldn't the higher potential be the same as before? Just as if a water pump was to push water to the top of a fountain, if there was resistance in this path, it would go slower, but the top of the fountain would still be at he same height, and the potential energy of the water the same once it reaches the top?

Please help me in constructing a proper analogy for this...

J

edit: could one think of this way: the internal resistance decreases current, thus in the water fountain analogy, the water doesn't reach the same height it would do had it not been for the internal resistance, and since the height is lower, the potential is lower.

 

[tiny-tim]

Hi The_Lobster!

edit: could one think of this way: the internal resistance decreases current, thus in the water fountain analogy, the water doesn't reach the same height it would do had it not been for the internal resistance, and since the height is lower, the potential is lower.

Yes, that's correct …

in the water analogy, you could fix the height, but in emf there is no height (or anything else) to fix, so you end up with a lower potential (which in the case of water means lower height).

(ok, now try using a water analogy to explain energy storage in a capacitor! )

 

[The_Lobster]

Great! Thanks! :D

On a molecular level, could one say that since the current decreases (due to the internal resistance), the electric field on which the potential difference is based becomes weaker?

 

[tiny-tim]

hmm … not sure what the molecular level has to do with it (and there's a potential difference whether there's a current flowing or not) …

if the potential difference is lower, then the field must be lower.

 

[The_Lobster]

Right! Thank you again! :)