# Uncertainty error of resistors in parallel

1. Jul 27, 2014

### valiantt

1. The problem statement, all variables and given/known data

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

2. Relevant 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)
$$R_t=\frac{(R_1±\delta_{R_1})(R_2±\delta _{R_2})}{( R_1±\delta_{R_1}) + (R_2±\delta _{R_2)}}$$

3. 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 [Broken], 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

Last edited by a moderator: May 6, 2017
2. Jul 27, 2014

### vela

Staff Emeritus
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.