Question

What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is $2.00\textrm{ }\Omega$?

Final Answer

$2.9994\textrm{ V}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 21, Problem 16 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A lithium-ion battery in a wrist watch has an

*emf*of 3.0000 volts and an internal resistance of 2.00 ohms and we are told that the current being drawn from this battery by the watch circuit is 0.300 milliamps, which I write as 0.300 times 10 to the minus 3 amps. So the terminal voltage is the question: what is this voltage between the terminals of the battery? And that's going to equal the*emf*minus the voltage drop across the internal resistance, which is the current multiplied by*r*. So that's 3.0000 volts minus 0.300 times 10 to the minus 3 amps times 2.00 ohms and this is 2.9994 volts.