How to find the relative error in the voltage reading

[Brittany King]

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!

 

[mfb]

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?

 

[Brittany King]

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!

 

[gneill]

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.

 

[Brittany King]

Yup, i'll post a picture of the question and the formulas below

 

[gneill]

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?

 

[Brittany King]

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.

 

[gneill]

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.

 

[Brittany King]

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.

 

[gneill]

FYI, the Thevenin method will yield: