Trying to understand math for Battery resistance ?

Jan 24, 2023 Last reply: 3 years ago 2 Replies

I’m working though understanding this description in this book. see attachment link below. I have tried this method for testing household batteries. I know there are other ways of testing batteries, but really I'm just trying to work my way through this book as a personal exercise. Please refer to link below for the book example Im trying to follow. The hand written notes are mine based on what things the terms mean. Hopefully I have them right. Can you tell me If I am performing the example properly. Here is the math using the equations in the document, step by step broken down based on the formula given in this book.



  1. VL/RL = IL
  2. VNL – VL = NewVL
  3. NewVL/IL = RI

My Tests


1.5 Volt Battery (Good Battery) I have an (A) Size 1.5 Volt battery, the following are my results. VNL 1.569 No load. VL 1.548 Load tested with R100 Resistor. RL R100 Ohm Resistor


  1. VL/RL= IL 1.548 / R100 = 0.1548 (IL)



  1. VNL-VL= (NewVL) 1.569-1.548= 0.021 (NewVL)
  2. VLNew/IL=RI 0.021/.1.548=0.1356589147286822 (RI)

The final result and internal resistance is 0.13 Ohms. However when I go to the following web site

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punch in the numbers I come to 1.357. Yes it’s the same number rounded up, but why is my number 0.13 and the web site moving the decimal place.1.3 Where am I wrong ?? They do seem to use a different math I think (This type of math is not my strong suit so Im not that clear on it.) Who is right the web site my book or is my understanding / math wrong?



Here are the pages from the book that Im using. I hope they come through, I don't post many pictures and things. thank you very much.



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Here are the numbers I have entered into the
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web site calculator. [img]
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1.548 / 100 = 0._0_1548

0.021 / 0.01548 = 1.357

Human error, Thank you, Carless of me. Regards

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