A cauterizer, used to stop bleeding in surgery, puts out 1.98 mA at 16.5 kV. (a) What is its power output? (b) What is the resistance of the path? Ω

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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A cauterizer, used to stop bleeding in surgery, puts out 1.98 mA at 16.5 kV.
(a) What is its power output?
(b) What is the resistance of the path?
Ω
Transcribed Image Text:A cauterizer, used to stop bleeding in surgery, puts out 1.98 mA at 16.5 kV. (a) What is its power output? (b) What is the resistance of the path? Ω
Expert Solution
Step 1

Given that the cauterizer used to stop bleeding in surgery puts out current (I)=1.98mA=1.98×10-3A at Voltage (V)=16.5kv=16.5×103

We know that, 

Power=(voltage)×(current) 

or, power (p)=VI 

Putting the values of voltage and current we get 

Power (p)=16.5×103×1.98×10-3  

Power (p)=32.67 w. 

So, the power output of the cauterizer is 32.67 watt.

 

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