alsy
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ok battery voltage/total resistance x R2= voltage across r2
12v / 150k ohms x 50k ohms = 4 V
12v / 150k ohms x 50k ohms = 4 V
The discussion revolves around a circuit involving a 12V DC power supply connected across two 100k ohm resistors in series, with a voltmeter measuring the voltage across one of the resistors. Participants explore the calculations for voltage readings, the impact of the voltmeter's resistance, and the overall circuit analysis.
Participants do not reach a consensus on the calculations and interpretations of the circuit. Multiple competing views and methods are presented, and the discussion remains unresolved regarding the correct approach to the problem.
Limitations include unclear definitions of variables, unresolved mathematical steps, and varying interpretations of circuit configurations. The discussion reflects a range of assumptions about the behavior of the circuit components.
This discussion may be useful for students studying circuit analysis, particularly those interested in understanding the effects of measurement devices on circuit behavior and the application of Ohm's law in practical scenarios.
Yes! Well done.alsy said:ok battery voltage/total resistance x R2= voltage across r2
12v / 150k ohms x 50k ohms = 4 V
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