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