Voltage query -- A power supply, resistors and a voltmeter....

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Discussion Overview

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.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants attempt to apply Ohm's law to calculate the voltage across the resistors and the percentage error of the voltmeter reading.
  • There is confusion regarding the relevant equations needed for the problem, with some participants suggesting formulas for combining resistors and voltage dividers.
  • Participants discuss the resistance of the voltmeter, with one suggesting it is calculated based on its sensitivity rating and the scale selected.
  • There is a debate about the total resistance in the circuit, with some participants initially neglecting the voltmeter's resistance when calculating the equivalent resistance.
  • Participants express uncertainty about the calculations and the meaning of the variables used, leading to requests for clearer explanations and circuit diagrams.
  • Disagreement arises over the correct calculation of the total resistance when the voltmeter is connected in parallel with one of the resistors.
  • Some participants propose different methods for calculating the voltage across the resistors, leading to varying results and interpretations of the circuit behavior.

Areas of Agreement / Disagreement

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.

Contextual Notes

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.

Who May Find This Useful

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.

  • #31
ok battery voltage/total resistance x R2= voltage across r2

12v / 150k ohms x 50k ohms = 4 V
 
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  • #32
alsy said:
ok battery voltage/total resistance x R2= voltage across r2

12v / 150k ohms x 50k ohms = 4 V
Yes! Well done.

So the meter will read 4 V. That answers part 2 of the question. Now, how will you address part 3?
 
  • #33
thank you.

ok part 3.

10,000ohms sensitivity x 4v= 40,000 /10v range = 4000 / 100= 40%
 
  • #34
I think what they're looking for is the percent error in the measured value when compared to what an ideal meter would read (that is a meter which would not affect the circuit operation at all, one with infinite ohms per volt sensitivity).

What would the voltage be across R2 if the meter was not attached?
 
  • #35
12v / 200k ohms x 100k ohms = 6v voltage across r2 without meter
 
  • #36
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?
 
  • #37
4v / 6v = 0.66 x 100 =66.6% error
 
  • #38
alsy said:
4v / 6v = 0.66 x 100 =66.6% error
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.

What's the formula for percent error (or percent difference)?
 
  • #39
percentage error formula

accepted - measured / accepted x 100
 
  • #40
alsy said:
percentage error formula

accepted - measured / accepted x 100
Yup. Again, get in the habit of using parentheses to clear up operation precedence vagueness in your formulas:

[(accepted - measured) / accepted] x 100
 
  • #41
sure

[( 6v -4v /6 x 100 = 33.33 %
 
  • #42
Looks good. Keep in mind that they may want to know the "direction" of the percentage error, indicating this with a sign. So if the measured value is 33.3% low, then it might be written as -33.3%. If this is a written hand-in assignment then you can use the sign method or make an appropriate annotation on the value to indicate your understanding that it is a "low reading" of the true value.
 
  • #43
ok but if they say percentage error then I just express as 33.33%
 
  • #44
alsy said:
ok but if they say percentage error then I just express as 33.33%
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.
 
  • #45
ok thanks.

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
 
  • #46
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
Always start a new thread for a new question/topic.
 
  • #47
ok thanks
 

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