Potential difference across a resistor

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palkia
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Homework Statement


Two cells of same emf E and different internal resistance r1 and r2 are connected in series to an external resistance R.The value of R for which the potential difference across the first cell is zero is given by

(A) R=r1/r2
(B)R=r1+r2
(C)R=r1-r2
(D)R=r1r2

Homework Equations



E=IR[/B]

The Attempt at a Solution



I first founded the total current in the circuit which came to be-E/R+r1+r2 then used the equation Ir1 to get the potential difference but I am not getting my answer[/B]
 
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What did you get as answer?

What must be the voltage drop at r1 to make this scenario possible?

By the way: It is useful to put brackets around denominators to make clear what is in the denominator. is x/y+z =(x/y)+z (as it would be usually read) or x/(y+z) (what you mean here)?
 
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Potential difference across r1is Er1/(R+r1+r2) so for zero potential difference r1=0 which looks weird
 
E=2E/(R+r1+r2) r1

But why are solving for the terminal voltage of the battery?
 
haruspex said:
I do not see how you got that from the equation in post #7. Please show your working.
E-2 E r1 /(R+r1+r2)
 
So the potential difference across the interal resistor issame as terminal voltage of the battery?
 
palkia said:
So the potential difference across the interal resistor issame as terminal voltage of the battery?
Right.
 
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