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In class we see how a current reduces the voltage across the battery if it has internal resistance. And we see that Vterminal=EMF-Ir. We don't really see the theory behind this, and I went on to investigate a little further.

What I found out was that when a current is flowing there is less time for a battery to create a potential difference. However, I wanted to know what is happening microscopically and was hoping someone could tell me.

Does the length between the positive and negative terminals become shorter such that the voltage becomes less? Or does the amount of accumulated charge on both sides decrease?

Also, and don't answer this if you consider this irrelevant, how is that a battery manages to establish a voltage in my wire?

Also I was wondering if the so-called "nonelectrostatic force" inside a battery is always equal to the electric force inside the battery?

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This is as much a chemistry question as it is an electricity question, because batteries are chemical devices. A battery is constructed such that there's a redox reaction split into two half-reactions which must move electrons through your circuit in order to complete. The reaction happens at a finite speed, and may also be limited by how fast various ions diffuse or drift through the cell.

The effect is that as you draw more current from a battery, the voltage on its terminals will droop; this can be approximated by imagining that you have an ideal battery with a resistor in series.

(There is also some actual electrical resistance in the battery's electrodes and internal wiring, of course, which can be significant sometimes.)

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  • $\begingroup$ Thanks. The "Ir" term bothered me, and it's good to know it's an approximation. You're answer has made me think of a battery as a "source" of electrons... Is this true? Is it in fact a source? $\endgroup$ – DLV Oct 4 '14 at 3:39
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    $\begingroup$ A better analogy is to think of it as a "pump"— it pumps electrons from one terminal to the other (since electrons are negative, it's pumping from the positive terminal to the negative), as long as the "back pressure" (the voltage gradient it's working against) isn't so high that it keeps the pump from running. $\endgroup$ – Wim Lewis Oct 4 '14 at 3:59
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The internal resistance of batteries is caused by a number of different mechanisms. A major contribution comes from the ionic conduction mechanisms in the electrolyte solution. Ions are large and can only move very slowly in electrolytes. Another source for the voltage drop is the concentration and current density dependent polarization of electrodes. To zeroth order these effects can be modeled by a second EMF source which reduces the battery voltage (i.e. a second ideal battery with a lower voltage) and an effective resistor, but these models are insufficient for most applications, though, because both the electrode polarization and internal resistance show memory effects, i.e. they depend on the past current flow trough the battery. There is, unfortunately, no simple explanation for what is happening inside of batteries and some of it is still part of active research.

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  • $\begingroup$ Thanks. So one source for the voltage drop is a decrease in the polarisation of the electrodes? $\endgroup$ – DLV Oct 3 '14 at 20:12
  • $\begingroup$ It's an increase in the polarization, because the electrode EMF is in the opposite direction of the battery EMF. One can see this effect after a short current burst: the battery voltage drops, then recovers slowly to its original value when the current goes back to its original value. $\endgroup$ – CuriousOne Oct 3 '14 at 20:25
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You ask 'how is that a battery manages to establish a voltage in my wire?'

This is conceptually incorrect - think of voltage or EMF as potential energy, i.e. that which causes the current to flow. We engineers say that we 'apply' a voltage 'across' a networks terminals, and indeed it is sometimes referred to as a potential difference (PD) in some older textbooks.

Sorry to be pedantic, btw ;-)

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