Voltmeter reading and error with 12V supply across two 100k resistors

[alsy]

ok battery voltage/total resistance x R2= voltage across r2

12v / 150k ohms x 50k ohms = 4 V

 

[gneill]

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?

 

[alsy]

thank you.

ok part 3.

10,000ohms sensitivity x 4v= 40,000 /10v range = 4000 / 100= 40%

 

[gneill]

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?

 

[alsy]

12v / 200k ohms x 100k ohms = 6v voltage across r2 without meter

 

[gneill]

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]

4v / 6v = 0.66 x 100 =66.6% error

 

[gneill]

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)?

 

[alsy]

percentage error formula

accepted - measured / accepted x 100

 

[gneill]

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

 

[alsy]

sure

[( 6v -4v /6 x 100 = 33.33 %

 

[gneill]

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.

 

[alsy]

ok but if they say percentage error then I just express as 33.33%

 

[gneill]

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.

 

[alsy]

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

 

[gneill]

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.

 

[alsy]

ok thanks