This is the question:

There are two resistors with resistance values $R_1=100\pm3$ ohm and $R_2=200\pm4$ ohm. Find the equivalent resistance of parallel combination.

According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if $$y=\frac{\text {AB}}{\text C}$$ then $$\%\;\text{error in y}=\%\;\text{error in A}+\%\;\text{error in B}+\%\;\text{error in C}$$

For the above problem, let $R_s$ denote series combination. Then $R_s=300\pm7$ ohm.

Let $R_p$ denote parallel combination.

$$\therefore R_p=\frac{R_1R_2}{R_1+R_2}=\frac{R_1R_2}{R_s}$$

Ignoring errors, we get $R_p=\frac{200}{3}$ ohm $=66.67$ ohm

$\%\;\text{error in R}_1=3$, $\%\;\text{error in R}_2=2$, $\%\;\text{error in R}_s=\frac73$

Hence, $\%\;\text{error in R}_p= 3+2+\frac73=\frac{22}{3}$

So, error in $R_p$ will be $\frac{22}{3}\%$ of $\frac{200}{3}$, which is approximately $4.89$.

Hence, I got $R_p=66.67\pm4.89$ ohm.

However, the book used the formula described and proved here and arrived at the answer $R_p=66.67\pm1.8$ ohm.

So, is the percentage error method wrong?

  • 3
  • How did you determine the error in $R_{s}$? Regardless, if your addition of errors is the correct way to find the % error in y, then it can never be 1.8 Ohms, since that is only ~2.7% of 66.67 Ohms (i.e., it is smaller than the sum of the errors in your A and B terms). – honeste_vivere Jul 17 '17 at 17:32
  • I'm voting against closing this as a duplicate; the error made in this question is rather different from the error made in the proposed "duplicate." – Michael Seifert Jul 18 '17 at 2:58

You should not simply add the errors, you should sum them squared in case of $y=AB/C$.

$dy/y=\sqrt{ (dA/A)^2 + (dB/B)^2 + (dC/C)^2 }$

This comes from the partial derivation of the function $y$ by all the components and weighting them by the uncertainty: $dy^2=(dA\frac{\partial y}{ \partial A})^2 + ... $ The square is there because you treat the different components as orthogonal - Pythagoras is 2 or more dimensions. Correct shape should include also correlations.

Hint: do the two partial derivations of $R=\frac{R_1R_2}{R_1+R_2}$ and input the numbers.

$dR = \sqrt{ (dR_1 \frac{\partial R}{\partial R_1})^2 + (dR_2 \frac{\partial R}{\partial R_2})^2 }$

However, your number 1.78$\Omega$ I get when I do

$dR= dR_1\frac{\partial R}{\partial R_1} + dR_2\frac{\partial R}{\partial R_2}$

  • That's interesting that the result you need to have 2.7% error, while the input has 2% and 3%. In case of something like $R_1/R_1$, your error would be canceled out. Let us see if 2.7% is possible... – jaromrax Mar 30 '15 at 11:18
  • So, on my opinion, the result 1.78$\Omega$ is not quite correct, I would rather prefer 1.40$\Omega$. I cannot find my Hudson's book to look at, there should be a matrix notation that naturally includes sum of squares, if I remember correctly. – jaromrax Mar 30 '15 at 13:35

I think the approach given by jaromax is correct (+1, I also get 1.4$\Omega$), whereas the formula quoted in the linked question should not be used if the measurements of $R_1$ and $R_2$ are independent and (slightly) overestimates the total uncertainty.

However, I am adding this answer because the approach you adopted based on percentage errors is definitely incorrect.

In order to use error combination formulae of the type $$dy/y=\sqrt{ (dA/A)^2 + (dB/B)^2 + (dC/C)^2 },$$ it is implicitly assumed that the uncertainties in $A$, $B$ and $C$ are independent.

If $A=R_1$, $B=R_2$ and $C= R_1+R_2$, then this is clearly not the case -- a positive deviation in either of $R_1$ or $R_2$ will result in an increase in $R_1+R_2$ too.

  • Yes, I also remarked that a correct formula includes correlations, so just the formula itself does not really answer the question. – jaromrax Mar 31 '15 at 12:05

I would like to share one more simple method to compute the error:

  1. Minimal Parallel Resistance => 97 ohm || 196 ohm = 64.887 ohm
  2. Maximal Parallel Resistance => 103 ohm || 204 ohm = 68.442 ohm

So, Measured Parallel Resistance will be from 64.887 ohm to 68.442 ohm.

Writing in terms of Average Value ± Error ==> 66.665 ± 1.778 ohm.

Hope it helps !

Just to see another way of doing this: Consider the quantities $x_1 = 1/R_1$ and $x_2 = 1/R_2$. We can calculate the uncertainties in these quantities easily. We can then define $x_p = 1/R_p$; we can find the uncertainty in this quantity, since $x_p = x_1 + x_2$; and finally we can use the uncertainty in $x_p$ to find the uncertainty in $R_p$.

This method avoids the error of blindly applying the product/quotient rule when the products being multiplied/divided are correlated with each other; since the errors in $R_1$ and $R_2$ are assumed to be uncorrelated, the errors in $x_1$ and $x_2$ will be as well, and so the standard "addition of error" formula applies.

Be careful when applying the division rule: $$R_p = \frac{R_1R_2}{R_1 + R_2} = \frac{A}{B}$$

You can only apply this rule to $A$ and B $if$ they are independent. In the case of parallel resistance, A and B both contains the terms $X$ and $Y$ and, hence , are not independent. Therefore, this rule cannot be applied. You can use the general addition/subtraction rule on

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = A + B$$ since $A$ and $B$ are independent.

This is the path to use the rules to find the uncertainty error in $R_p$ instead of going through the max and min method. Please see Donald Simanek's wehsite article on Propagation of Errors.

Basically what you are doing for $y=\frac{AB}{C}$ is adding up all the percentage errors, which is wrong.

Take log function on both side, so you get $\log(y)=\log(A)+\log(B)-\log(C)$ so for percentage error becomes: $\frac{dy}{y} = \frac{dA}{A} + \frac{dB}{B} - \frac{dC}{C}$ so if you actually follow your step, you get: $ %error= 3 + 2 - 7/3 = 2.6%$ so 2.6% of 200/3 gives you 1.8%

My calculation dR/R = dR1/R1 + dR2/R2 -dR1/(R1+R2) -dR2/(R1+R2) (Come from the derivative of R = (R1R2)/(R1+R2))

Do manipulation dR/R = [dR1/R1 -dR1/(R1+R2)] + [dR2/R2 -dR2/(R1+R2)]

dR/R = (dR1/R1)[R2/(R1+R2)] + (dR2/R2)[R1/(R1+R2)]

dR/R = (3/100)[200/300]+(4/200)[100/300] = 0.0267

dR=0.0267R= 0.0267(66.7) = 1.78

The methods you were trying is for finding error when the resultant resistance is to the power one. Like in series. But here it is of power -1. So differentiating we get $$\frac{dR}{R^2} = \frac{dr}{(r1)^2} + \frac{dr}{(r_2)^2}$$ This implies: $$dR = \bigg[\frac{3}{10000} + \frac{4}{40000}\bigg] \times 4444.9$$. $$dR=1.78\ \rm ohm$$ The answer you required is obtained.

protected by AccidentalFourierTransform Aug 1 at 14:03

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