Question
A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path?

a) $30.0 \textrm{ W}$

b) $7.50 \textrm{ M}\Omega$

Solution Video

# OpenStax College Physics Solution, Chapter 20, Problem 55 (Problems & Exercises) (0:43)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The power output of the cauterizer is the current through it multiplied by its voltage. So that's two times ten to the minus three amps and multiplied by 15 times ten to the three volts. This gives 30.0 watts. To find the resistance of the body tissue that's -- that the current of the cauterizer is going through, we'll use <i>V</i> equals <i>I R</i> and divide both sides by <i>I</i> and get the resistance is voltage divided by current. So that's 15 kilovolts divided by two milliamps. This is 7.50 megaohms. Mega is times ten to the six.