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## Homework Statement

Find the Uncertainty of the Total Resistance of two (2) parallel-connected resistors:

[tex]R1 = 2.2 kΩ ± 5\% \ ;\ R2 = 1.8 kΩ ± 20\%[/tex].

## Homework Equations

We know that resistors in parallel are:

[tex]\frac{1}{R_t}={\frac{1}{R_1±\delta_{R_1}}+\frac{1}{R_2±\delta_{R_2}}}[/tex]

This implies: (Eq. 1)

[tex]R_t=\frac{1}{\frac{1}{R_1±\delta_{R_1}}+\frac{1}{R_2±\delta_{R_2}}}[/tex]

Which is also equal to: (Eq. 2)

[tex]R_t=\frac{(R_1±\delta_{R_1})(R_2±\delta _{R_2})}{( R_1±\delta_{R_1}) + (R_2±\delta _{R_2)}}[/tex]

## The Attempt at a Solution

According to rules established by our teacher:

In addtion: if given [tex] f±\delta_f = (x±\delta_x)+(y±\delta_y)[/tex].

it implies:

[tex]f±\delta_f = (x+y)±(\delta_x+\delta_y)[/tex]

This means that in addition, δf is equal to the sum of the absolute errors.

In multiplication: if given [tex] f±\delta_f = (x±\delta_x)(y±\delta_y)[/tex].

it implies:

[tex]f±\frac{\delta_f}{f} = (x*y)±(\frac{\delta_x}{x}+\frac{\delta_y}{y})[/tex]

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

[tex]\frac{\delta_f}{f} = \%Err_x[/tex]

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

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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:

[tex] 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})} [/tex]

[tex]\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\%}[/tex]

[tex]\Rightarrow \frac{3.96MΩ}{4kΩ }±(25\%+11.75\%) 990Ω±36.75\%[/tex]

[tex]\Rightarrow990Ω±36.75\% [/tex]

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:

[tex]if \quad r_x = \frac{1}{x} ,~ then\quad \frac{\delta_{r_x}}{r_x} = \frac{\delta_{x}}{x}[/tex]

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:

[tex]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}}[/tex]

[tex]\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\%}[/tex]

[tex]\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\%}[/tex]

[tex]\Rightarrow990Ω±13.25\% [/tex]

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...

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

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

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