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

In summary: No, the voltmeter's 100k is in parallel with...?R2, so the total resistance becomes 55.56k ohms and the voltage drop becomes 8.64vIn summary, a 12V DC supply is connected across 2 100k resistors in series. A voltmeter with a sensitivity of 10,000 ohms per volt is switched to its 10V range and connected to measure the voltage of 1 resistor. The total resistance in the circuit is 150k ohms, and the voltage drop across R1 is 8 volts. By applying Kirchhoff's Voltage Law, it can be determined that the voltage drop across the R2||Rm
  • #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?
 
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  • #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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