Potential difference across a resistor

AI Thread Summary
The discussion revolves around determining the external resistance R for which the potential difference across the first cell in a series circuit becomes zero. Participants analyze the relationship between the internal resistances r1 and r2 of two cells with the same emf E. The calculations involve the total current in the circuit and the potential difference across the internal resistor r1. A key point is that for the potential difference to be zero, the derived equation suggests R should equal r1 minus r2. The conversation emphasizes the importance of using parentheses correctly in equations to avoid confusion.
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
 
palkia said:
the total current in the circuit which came to be-E/(R+r1+r2)
No, you forgot something.
 
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2E/R+r1+r2
 
palkia said:
2E/R+r1+r2
Right, but please use parentheses correctly.
If the PD across the first battery is zero, what current would you expect?
 
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E=2E/(R+r1+r2) r1

But why are solving for the terminal voltage of the battery?
 
palkia said:
E=2E/(R+r1+r2) r1

But why are solving for the terminal voltage of the battery?
I believe that equation gives the right answer.
 
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E-Ir1
 
  • #10
palkia said:
E-Ir1
I do not see how you got that from the equation in post #7. Please show your working.
 
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  • #11
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)
 
  • #12
So the potential difference across the interal resistor issame as terminal voltage of the battery?
 
  • #13
palkia said:
So the potential difference across the interal resistor issame as terminal voltage of the battery?
Right.
 
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  • #14
palkia said:
E-2 E r1 /(R+r1+r2)
That is not an equation. You had E=2 E r1 /(R+r1+r2). What does that give you for r1 in terms of R and r2?
 
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  • #15
I am getting R=r1-r2
 
  • #16
palkia said:
I am getting R=r1-r2
Right.
 
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