How to find the relative error in the voltage reading

In summary: Er= Vm-Vx/Vx *100% -the percentage relative loading error associated with the measured voltageVm=Vx(Rm/Rm+Rs) - from voltage divider equationRs= Source ResistanceRm= Meter ResistanceIn summary, the relative error in voltage measurement is -56.5%.
  • #1
Brittany King
12
0

Homework Statement


For a circuit with 3 resistors in series (R1=1.00 kOhms, R2=2.50 kOhms, R3= 4.00 kOhms and Va=12.0 V). A voltmeter was placed across R2 and R3. Calculate the relative error in the voltage reading if the internal resistance of the voltmeter was a) 5000 ohms, b) 50 kOhms, C) 500 kOhms.

Homework Equations



Er = - (Rs / Rs+Rm) * 100%

The Attempt at a Solution



Rs= R3+ R4 = 4000 ohms + 2500 ohms = 65000 ohms

Er= - (6500 ohms/( 6500 ohms + 5000 ohms) * 100% = -56.5 %

The answer is a_) 15% b) -1.7% c) -0.7%

I'm guessing how I'm calculating Rs wrong but I'm not sure what else to do. Any help would be appreciated. Thanks!
 
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  • #2
Brittany King said:
Er = - (Rs / Rs+Rm) * 100%
First: I guess you mean Rs/(Rs+Rm), otherwise your fraction is just Rs/Rs which is 1.

Unrelated to that: What are Rs and Rm, and where does this formula apply?

The problem statement asks for the relative error - what is compared to what here?
 
  • #3
The textbook I got this equation from doesn't have a bracket around (Rs+Rm), the equation is derived from:

Er= Vm-Vx/Vx *100% -the percentage relative loading error associated with the measured voltage
Vm=Vx(Rm/Rm+Rs) - from voltage divider equation

Rs= Source Resistance
Rm= Meter Resistance

The question is looking for the percentage relative loading error of the voltage measurement between the source and the voltmeter.

Relative loading error (Er) = -( Rs/ Rs+Rm) * 100%

The circuit has 3 resistors in series with the voltmeter attached to contacts 2 and 4 which span resistor 2 and resistor 3.

This question is from Skoog 6th ed Instrumental Analysis textbook question 2-3.

Thanks!
 
  • #4
Brittany King said:
The textbook I got this equation from doesn't have a bracket around (Rs+Rm), the equation is derived from:

Er= Vm-Vx/Vx *100% -the percentage relative loading error associated with the measured voltage
Vm=Vx(Rm/Rm+Rs) - from voltage divider equation

Rs= Source Resistance
Rm= Meter Resistance
Can you snap picture of the page in the textbook that presents this formula? As @mfb mentioned, any ratio of the form X/X reduces to 1. So applying the standard precedence of mathematical operations, the formulas as presented don't make sense.
 
  • #5
Yup, i'll post a picture of the question and the formulas below
 

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  • #6
Okay, so they are using a horizontal line to represent the division operation, which clearly separates the numerator and denominator. Effectively, this provides "implied" parentheses to group the quantities appropriately. When you write such an equation in a linear fashion, it's up to you to add actual parentheses to retain the required groupings. Otherwise the standard order of operations applies, often leading to unintended results.

It looks to me like the source resistance Rs that you want is the so-called "output resistance" of the network that the meter "sees" when it's attached to its connection points on the network. By any chance have you studied Thevenin Equivalent Circuits yet?
 
  • #7
gneill said:
Thevenin Equivalent Circuits

Okay thanks got it!

We haven't touched on Thevenin Equivalent Circuits, we were just going through a section with series and parallel circuits and learning about the DC current, voltage and resistance measurements with digital voltmeters.
 
  • #8
Brittany King said:
Okay thanks got it!

We haven't touched on Thevenin Equivalent Circuits, we were just going through a section with series and parallel circuits and learning about the DC current, voltage and resistance measurements with digital voltmeters.
I see. Well that means you'll have to read ahead a bit (the easy way!), or do a bit more work to find the result using the tools you have at hand now:

You can use your basic circuit analysis methods (KVL, KCL, etc.) to find the unloaded (open circuit) voltage across the R2+R3 combination, then the loaded (meter attached) voltage for the same. Use those values to determine the percentage change.
 
  • #9
gneill said:
I see. Well that means you'll have to read ahead a bit (the easy way!), or do a bit more work to find the result using the tools you have at hand now:

You can use your basic circuit analysis methods (KVL, KCL, etc.) to find the unloaded (open circuit) voltage across the R2+R3 combination, then the loaded (meter attached) voltage for the same. Use those values to determine the percentage change.

Ok awesome ill give it a read. Thanks for your help.
 
  • #10
FYI, the Thevenin method will yield:
upload_2018-2-18_20-10-58.png
 

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1. What is relative error in a voltage reading?

Relative error in a voltage reading is a measure of the difference between the measured voltage and the true or expected voltage. It is expressed as a percentage or decimal value and indicates the accuracy of the measured voltage.

2. Why is it important to calculate relative error in voltage readings?

Calculating relative error in voltage readings allows us to assess the accuracy and precision of our measurements. It helps us to identify and correct any sources of error and improve the overall quality of our data.

3. How do you calculate relative error in a voltage reading?

To calculate relative error in a voltage reading, you need to subtract the true or expected voltage from the measured voltage, divide that value by the true or expected voltage, and then multiply by 100 to convert it into a percentage. The formula for relative error in voltage readings is: (Measured Voltage - Expected Voltage) / Expected Voltage x 100%

4. What is an acceptable range for relative error in voltage readings?

An acceptable range for relative error in voltage readings varies depending on the specific context and application. In general, a relative error of less than 5% is considered good, while a relative error of 10% or more may indicate a need for further investigation or improvement of measurement techniques.

5. Are there any limitations to using relative error in voltage readings?

Yes, there are certain limitations to using relative error in voltage readings. It assumes a linear relationship between the measured and expected values, and it does not take into account any systematic errors that may be present. Additionally, it is only useful when comparing two measurements of the same quantity and does not provide information on the accuracy of individual measurements.

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