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Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which 0.200 mA flows.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

$6750 \textrm{ } \Omega$

Solution Video

OpenStax College Physics Solution, Chapter 20, Problem 19 (Problems & Exercises) (0:21)

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OpenStax College Physics, Chapter 20, Problem 19 (PE) calculator screenshot 1
Video Transcript

This is College Physics Answers with Shaun Dychko. The pocket calculator has a 1.35 volt battery and the current flowing through it is 0.2 times ten to the minus three amps. So resistance is voltage divided by current, so 1.35 volts divided by 0.2 times ten to the minus three amps gives us 6750 ohms.


Submitted by garrettasakawa on Sat, 07/13/2019 - 10:56

The video and answer don't match the question