A battery produces 40.8V when 7.10 A are drawn from it and 47.3V when 2.20 A are drawn. What is the emf and internal resistance of the battery?

Question

anonymous · Answer

not sure how to solve this without knowing the open circuit voltage(ocv).  A battery's IR is equal to the difference between the OCV and the CCV, divided by the applied load.  This formula is derived from ohm's law E/IxR.  For example:  If the OCV=14V and the CCV=12V with a 100A load applied, the internal resistance is 0.02ohms.  14-12=2     2/100=.02.  
   Without having the OCV, i aint sure how to do it.  sorry.

stormfire1 · Answer

You need to write out the two equations with two unknowns and solve it:

E - 7.2 R = 40.8V
E - 2.5 R = 47.0V

E =~ 50.3 and the internal resistance =~ 1.32 ohms