How to Calculate Resistors in Parallel Uncertainty

AI Thread Summary
The discussion centers on calculating the equivalent resistance of two resistors in parallel, each with specified uncertainties. The calculated series resistance is 11,500Ω with an uncertainty of 575Ω, while the parallel resistance is initially calculated as approximately 2,779Ω with an uncertainty of around 241Ω. However, further analysis suggests that the uncertainty should be closer to 139Ω, prompting a reevaluation of the method used for uncertainty calculations. The participants recommend using a specific formula found online for better accuracy, as the original method may not adequately account for the complexities of combining products and quotients in uncertainty. Ultimately, refining the approach to uncertainty is essential for accurate results in resistor calculations.
LCHL
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Homework Statement



We are told that resistor 1 has a resistance of 4700Ω ± 5% and resistor 2 has a resistance of 6800Ω ± 5% and the equivalent series and parallel resistances are to be calculated with an estimate of uncertainty.

Homework Equations



RSeries = R1 + R2
RParallel = R1 × R2 / R1 + R2

Uncertainties can simply be added if their measured numbers are added or subtracted.

The formula we were given for multiplying or dividing uncertainty is the following:

If z = x ⋅ y or z = x / y
Then

Δz=z(√(Δx/x)2+(Δy/y)2) (The square root covers everything in the equation except for z)

The Attempt at a Solution



I believe that I am doing the math correctly, but my the method doesn't really appear to correspond to anything else I have seen.

I have done the following:

RSeries = 4700 + 6800 =11500Ω
Uncertainty = 235 + 340 = 575Ω (That's just 5% of each number)

I feel that my method might be going wrong here:

R1 × R2 = 4700 × 6800 = 31,960,000
Uncertainty ≅ 2,259,913.273 (Using the uncertainty multiplication/division formula)

The bottom half of the parallel equation has already been calculated.

To get the equivalent resistance and uncertainty, I used the multiplication/division formula again.

RParallel = 31,960,000 / 11500 ≅ 2,779Ω
Uncertainty = 2779(√(2,259,913.273 / 31,960,000 )2 + (575 / 11500)2)

Which gives ~241Ω.

Is 2779 ± 241Ω a correct and acceptable way to display this answer or have I done something wrong here? Any feedback would be appreciated. :smile:
 
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LCHL said:

Homework Statement



We are told that resistor 1 has a resistance of 4700Ω ± 5% and resistor 2 has a resistance of 6800Ω ± 5% and the equivalent series and parallel resistances are to be calculated with an estimate of uncertainty.

Homework Equations



RSeries = R1 + R2
RParallel = R1 × R2 / R1 + R2

Uncertainties can simply be added if their measured numbers are added or subtracted.

The formula we were given for multiplying or dividing uncertainty is the following:

If z = x ⋅ y or z = x / y
Then

Δz=z(√(Δx/x)2+(Δy/y)2) (The square root covers everything in the equation except for z)

The Attempt at a Solution



I believe that I am doing the math correctly, but my the method doesn't really appear to correspond to anything else I have seen.

I have done the following:

RSeries = 4700 + 6800 =11500Ω
Uncertainty = 235 + 340 = 575Ω (That's just 5% of each number)

I feel that my method might be going wrong here:

R1 × R2 = 4700 × 6800 = 31,960,000
Uncertainty ≅ 2,259,913.273 (Using the uncertainty multiplication/division formula)

The bottom half of the parallel equation has already been calculated.

To get the equivalent resistance and uncertainty, I used the multiplication/division formula again.

RParallel = 31,960,000 / 11500 ≅ 2,779Ω
Uncertainty = 2779(√(2,259,913.273 / 31,960,000 )2 + (575 / 11500)2)

Which gives ~241Ω.

Is 2779 ± 241Ω a correct and acceptable way to display this answer or have I done something wrong here? Any feedback would be appreciated. :smile:
Welcome to the PF.

You can easily check your answers by substituting the max values for each resistor in the equations for the series and parallel combinations, and then substitute in the minimum values and re-calculate. Then you can see what the % difference is (+/-%) between the max and min combinations...
 
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Hi. Thanks for the reply. Doing the following suggests that the uncertainty is ~139Ω.

Looking online, I found a formula that corresponds to this answer.

http://physics.stackexchange.com/qu...quivalent-resistance-of-resistors-in-parallel

I would therefore have to think that I should apply this formula instead. Thank you for guiding me to the answer. :oldsmile: Having said that, I don't know what's wrong with what I did above. My two best guesses are that you can't use it for both a product and a quotient, because the uncertainty gets compounded somehow, or alternatively, the formula is wrong.
 
LCHL said:
Hi. Thanks for the reply. Doing the following suggests that the uncertainty is ~139Ω.

Looking online, I found a formula that corresponds to this answer.

http://physics.stackexchange.com/qu...quivalent-resistance-of-resistors-in-parallel

I would therefore have to think that I should apply this formula instead. Thank you for guiding me to the answer. :oldsmile: Having said that, I don't know what's wrong with what I did above. My two best guesses are that you can't use it for both a product and a quotient, because the uncertainty gets compounded somehow, or alternatively, the formula is wrong.
Yeah, I'd go with the formula that you found online. The parallel combination calculation involves both products and quotients. I'm no expert in uncertainty, but my simple method usually works for me. :smile:
 
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I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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