How to Calculate Resistors in Parallel Uncertainty

In summary, Homework Statement indicates that two resistors with known resistance values are to be combined to create equivalent resistances with an unknown amount of uncertainty. The uncertainty can be expressed as a percentage of the max and min values of the resistors. The unknown equivalent resistances are calculated using a multiplication/division formula that takes into account the uncertainty values.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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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Related to How to Calculate Resistors in Parallel Uncertainty

1. What is the formula for calculating resistors in parallel with uncertainty?

The formula for calculating resistors in parallel with uncertainty is:
Req = 1/[(1/R1) + (1/R2) + (1/R3) + ... + (1/Rn)]
where Req is the equivalent resistance and R1, R2, R3, ... , Rn are the individual resistances in parallel.

2. How do you calculate the uncertainty in resistors in parallel?

To calculate the uncertainty in resistors in parallel, you need to first find the uncertainty in each individual resistance. This can be done by multiplying the percentage uncertainty by the numerical value of the resistance. Next, you need to use the formula for combining uncertainties in parallel, which is:
ΔReq = √[(ΔR1/R1)² + (ΔR2/R2)² + (ΔR3/R3)² + ... + (ΔRn/Rn)²]
where ΔReq is the uncertainty in the equivalent resistance and ΔR1, ΔR2, ΔR3, ... , ΔRn are the uncertainties in the individual resistances.

3. Can you calculate the resistance when there is no uncertainty?

Yes, the formula for calculating resistors in parallel without uncertainty is the same as the formula with uncertainty. However, in this case, the individual resistances will have an uncertainty of 0, so the final result will not have an uncertainty either.

4. How do you interpret the uncertainty in resistors in parallel?

The uncertainty in resistors in parallel represents the range of values that the equivalent resistance could potentially fall within. This means that the actual value of the equivalent resistance could be higher or lower than the calculated value by the amount of uncertainty. It is important to take this into consideration when using the calculated value in further calculations or experiments.

5. Are there any assumptions made when calculating resistors in parallel with uncertainty?

Yes, the formula for calculating resistors in parallel with uncertainty assumes that the uncertainties in each individual resistance are independent of each other. It also assumes that the resistors are connected in a parallel circuit, with no other components affecting the overall resistance. These assumptions may not hold true in all cases, so it is important to carefully consider the situation before using this formula.

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