Equation For Charging Current In Accumulator

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Homework Statement



Suppose we have a circuit represented, which is used for charging an accumulator from a.d.c supply of e.m.f, E1 and internal resistance, r1. If the e.m.f of the accumulator is E2 with an internal resistance of r2, then the equation for the charging current is what?
In this question, I want to know how to derive the equation for charging current in an accumulator.


Homework Equations



I know that there is equation that goes like this:
E = V + v
E = IR + ir
E = I(R + r)
I = E/(R + r)
But am not sure the above equation relates to my question in any way. Am optimistic that there are other equation that will be fit enough for this my questions.

The Attempt at a Solution



E = V + v
E = IR + ir
E = I(R + r)
I = E/(R + r)
but I don't think that this my working is correct or is fit for the question. Any help?
 
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I would say that the overall emf = E1-E2 and the total resistance - r1+r2.
So the current will be (E1-E2)/(r1+r2)
This checks out because when the battery is charged E2 = E1 and the current will then be zero
 
Emilyjoint said:
I would say that the overall emf = E1-E2 and the total resistance - r1+r2.
So the current will be (E1-E2)/(r1+r2)
This checks out because when the battery is charged E2 = E1 and the current will then be zero

Can you make the current, I the subject of the formula in any of the equation?
 
Emilyjoint said:
I would say that the overall emf = E1-E2 and the total resistance - r1+r2.
So the current will be (E1-E2)/(r1+r2)
This checks out because when the battery is charged E2 = E1 and the current will then be zero

Thank you so much. I understand from your reply that the current, I = (E1-E2)/(r1+r2)
I would say that the overall emf = E1-E2
But why is the overall emf = E1-E2? Can you give a principle that supports that?