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Internal Resistance of a Battery 
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#1
Oct2507, 11:39 PM

P: 90

1. The problem statement, all variables and given/known data
The potential difference across the terminals of a battery is 8.20 V when there is a current of 1.54 A in the battery from the negative to the positive terminal. When the current is 3.56 A in the reverse direction, the potential difference becomes 9.40 V. What is the internal resistance r of the battery? 2. Relevant equations V_ab = E  Ir 3. The attempt at a solution So I thought that it would make no difference if the charge was reversed. So I just plugged the values into the equations, so since I have two equations and two variables I could solve for r. 8.56 = E  1.54r > E = 8.56 + 1.54r 9.40 = E  3.56r > E = 9.40 + 3.56r  0 = .84  2.02r r = .416 Obviously the resistance can't be negative. What am I doing wrong, does it have something to do with the reversing part of the problem? 


#2
Oct2507, 11:51 PM

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#3
Oct2507, 11:54 PM

P: 90

I don't understand why that makes a difference, because won't the change in potential be the same? So what's negative the potential or the amps??? I'm confused.



#4
Oct2607, 12:08 AM

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P: 1,662

Internal Resistance of a Battery
8.20 V = E (in V)  (1.54 A) · r (in ohms) , as you did. But in the second part, we are told that the 3.56 A current is being applied in the opposite direction (positive to negative terminal inside the battery), so the sign of the current must be switched. 


#5
Oct2607, 12:16 AM

P: 90

Thanks. I guess I need to go back and read up on charges deff. of a Coulomb and such. I knew the potential would be the same.



#6
Oct2607, 12:20 AM

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