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gneill
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Great. So given that the meter reported the voltage as 4 V, what's the percent error in the reading?alsy said:12v / 200k ohms x 100k ohms = 6v voltage across r2 without meter
Great. So given that the meter reported the voltage as 4 V, what's the percent error in the reading?alsy said:12v / 200k ohms x 100k ohms = 6v voltage across r2 without meter
No, that's the percentage of the "true" voltage that's seen. What you want is the percentage that the error in the reading represents.alsy said:4v / 6v = 0.66 x 100 =66.6% error
Yup. Again, get in the habit of using parentheses to clear up operation precedence vagueness in your formulas:alsy said:percentage error formula
accepted - measured / accepted x 100
As I said, it depends upon who's looking at the result and how you've been taught to present a percent error. Certainly the magnitude of the error is 33.3%. If that's all they want to see, then that's fine.alsy said:ok but if they say percentage error then I just express as 33.33%
Always start a new thread for a new question/topic.alsy said:I have a shorter engineering science related question,regarding linear velocity.not sure if that's your area or if it can be answered here