Uncertainty error of resistors in parallel

[valiantt]

Homework Statement

Find the Uncertainty of the Total Resistance of two (2) parallel-connected resistors:
$$R1 = 2.2 kΩ ± 5\% \ ;\ R2 = 1.8 kΩ ± 20\%$$.

Homework Equations

We know that resistors in parallel are:

$$\frac{1}{R_t}={\frac{1}{R_1±\delta_{R_1}}+\frac{1}{R_2±\delta_{R_2}}}$$

This implies: (Eq. 1)
$$R_t=\frac{1}{\frac{1}{R_1±\delta_{R_1}}+\frac{1}{R_2±\delta_{R_2}}}$$

Which is also equal to: (Eq. 2)
$$R_t=\frac{(R_1±\delta_{R_1})(R_2±\delta _{R_2})}{( R_1±\delta_{R_1}) + (R_2±\delta _{R_2)}}$$

The Attempt at a Solution

According to rules established by our teacher:

In addtion: if given $$f±\delta_f = (x±\delta_x)+(y±\delta_y)$$.
it implies:
$$f±\delta_f = (x+y)±(\delta_x+\delta_y)$$
This means that in addition, δf is equal to the sum of the absolute errors.

In multiplication: if given $$f±\delta_f = (x±\delta_x)(y±\delta_y)$$.
it implies:
$$f±\frac{\delta_f}{f} = (x*y)±(\frac{\delta_x}{x}+\frac{\delta_y}{y})$$

From here on, let's established for simplification that:

$$\frac{\delta_f}{f} = \%Err_x$$

This means that in multiplication, δf is equal to the sum of the relative errors. (Note that this also applies to division)-----------------------------------------------------------------------

So given those above, when should those rules be applied? My teacher insists that you should apply those rules at Eq. 2. This means his method yields:

$$R_t = \frac{(R_1*R_2) ± (\%Err_{R_1}+\%Err_{R_2})}{(R_1+R_2) ± (R_1 * \%Err_{R_1}+R_2*\%Err_{R_2})}$$

$$\Rightarrow \frac{2.2kΩ * 1.8kΩ ± (5\% + 20\%)} {2.2kΩ + 1.8kΩ ± (110Ω + 360Ω)} = \frac{3.96MΩ±25\%}{4000 ±(470Ω)} = \frac{3.96MΩ±25\%}{4000Ω ±11.75\%}$$

$$\Rightarrow \frac{3.96MΩ}{4kΩ }±(25\%+11.75\%) 990Ω±36.75\%$$

$$\Rightarrow990Ω±36.75\%$$But according to this link, http://uregina.ca/~szymanss/uglabs/companion/Ch3_Error_Prop.pdf , Eq. 2 is not in a appropriate form for error calculation. It then shows that:

$$if \quad r_x = \frac{1}{x} ,~ then\quad \frac{\delta_{r_x}}{r_x} = \frac{\delta_{x}}{x}$$

In this case, it says that the relative error is unchanged if you take the reciprocal of a quantity. This is also supported by this link: http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf

Now using Eq. 1 and the reciprocality rule:

$$R_t=\frac{1}{\frac{1}{R_1±\delta_{R_1}}+\frac{1}{R_2±\delta_{R_2}}} = \frac{1}{\frac{1}{R_1}±\%Err_{R_1}+\frac{1}{R_2}±\%Err_{R_2}}$$

$$\Rightarrow \frac{1}{\frac{1}{2.2kΩ±5\%}+\frac{1}{1.8kΩ±20\%}} = \frac{1}{\frac{1}{2.2kΩ}±5\%+\frac{1}{1.8kΩ}±20\%}$$

$$\Rightarrow \frac{1}{\frac{1}{2.2kΩ}+\frac{1}{1.8kΩ} ± (\frac{1}{44kΩ}+\frac{1}{9kΩ})} = \frac{1}{\frac{1}{990}±\frac{53}{39600}} = \frac{1}{\frac{1}{990}±13.25\%}$$

$$\Rightarrow990Ω±13.25\%$$That's pretty much it. But I do remember someone telling me that uncertainty error should always increase so I don't know...

And also, if assuming that the second case is correct, would that mean given a bunch of resistors, say for example, 1 pc. of a 1kΩ ±20% and 99 pcs. of 1kΩ ±5% resistors, all connected in parallel, what would happen? I tried making an equation but I'm not sure if I got the math right...

$$R_t = \frac{1}{\frac{1}{1kΩ}*100} ± \frac{\frac{1}{1kΩ}*(0.20+(0.05*99)}{\frac{1}{1kΩ}*100} = 10Ω±5.15\%$$

lastly, given enough 1kΩ± 5% resistors to parallel, would that mean that the %error value will approach 5%?
Not really part of the homework but just a thought experiment. =D

 

[vela]

Your teacher is wrong. To demonstrate this, try calculating $R_t$ using the extreme values for $R_1$ and $R_2$. I found a range for $R_t$ from 853 ohms to 1116 ohms, which is consistent with the 13% error you obtained using the second method. Your teacher's method grossly overestimates the error.

 

[MC Wong]

Both calculations are corrected. The worst possible combination is the key to report the worst possible range to others so that your teacher has no mistake. However, when you stand on manufacturer side, you may report the lowest one.

 

[haruspex]

$\Rightarrow \frac{1}{\frac{1}{2.2kΩ}+\frac{1}{1.8kΩ} ± (\frac{1}{44kΩ}+\frac{1}{9kΩ})} = \frac{1}{\frac{1}{990}±\frac{53}{39600}} = \frac{1}{\frac{1}{990}±13.25\%}$

Your notation is rather confusing. Sometimes your r±x means r(1±x) and at other times literally r±x.
This item is particularly puzzling: $\frac{1}{\frac{1}{990}±\frac{53}{39600}}$.
I think you mean $\frac{1}{\frac{1}{990}±\frac{53\%}{39600}}$.
However, 13.25% is so large that you can't really use $\frac{\delta_{r_x}}{r_x} = \frac{\delta_{x}}{x}$. If you don't use that you get a touch over 15%.

As @vela notes, your teacher's method is unnecessarily pessimistic. It overlooks that the errors that make the numerator largest also make the denominator largest, so they partly cancel.

 

[haruspex]

Both calculations are corrected. The worst possible combination is the key to report the worst possible range to others so that your teacher has no mistake. However, when you stand on manufacturer side, you may report the lowest one.

No, the teacher's calculation is wrong, as demonstrated.