- #1

MrNewton

- 42

- 3

i am currently working on a project where i have to calculate the internal resistance of a battery based on the voltage drop caused by placing a resistor over the battery. I am an EE, and i understand how i can calculate the IR by schematic, but i found this equation:

((Open circuit voltage / load circuit voltage) -1) * R [1]

So the Open circuit voltage (OCV) is the voltage of the battery when nothing is connected to it. The load circuit voltage (LCV) is the voltage on the battery when a resistor (R) is placed.

So let's say OCV = 3.2V

LCD = 2.8V

R = 100 Ohm

So UR = 2.8V

IR = UR/R = 2.8/100 = 0.028A

Ur = OCV-UR = 0.4V

r = Ur/I = 0.4/0.028 = 14.2857 Ohms

If you do this by the first equation you get:

((3.2/2.8)-1)*100 = 14.2857

MY QUESTION: How did the get the first equation [1]? I can't simplify my equations and get the first equation.