How to Calculate Final Uncertainty in Quotient Calculations?

[roam]

Homework Statement

This is about experimental physics. I have a value of resistance R=9000 \pm 1000. Then I had to do the following calculation:

\frac{40000}{9000}=4.44 \ \Omega

So, I'm not quite sure how to calculate the final uncertainty from my uncertainty \sigma = \pm 1 \ k\Omega

The Attempt at a Solution

Since 9000 \times (4.8 \times 10^{-4}) = 4.44

I used the rule that says y=na \implies \sigma_{y}=n \sigma_a

so:

\sigma = 1000 \times (4.8 \times 10^{-4}) = 0.48 \ \Omega.

Is this correct? Would the final uncertainty be ±0.48?

 

[phinds]

I don't know any formal math on this, but if you start off with an uncertainty of +/- 11%, how can you not end up with an uncertainty of +/- 11% ?

 

[I like Serena]

Hi roam!

Yes, this is correct.

When multiplying or dividing the relative error remains the same.Generally we have for a quotient q = x/y
that the contribution in its uncertainty by y is:

\Delta q = | \frac {\partial q} {\partial y} | \Delta y = |\frac {-x} {y^2}| \Delta y = |q| \cdot |\frac {\Delta y} y|Note however that in your answer there should also be a contribution of the uncertainty of 40000. Or doesn't it have an uncertainty?

 

[roam]

Hi roam!

Yes, this is correct.

When multiplying or dividing the relative error remains the same.


Generally we have for a quotient q = x/y
that the contribution in its uncertainty by y is:

\Delta q = | \frac {\partial q} {\partial y} | \Delta y = |\frac {-x} {y^2}| \Delta y = |q| \cdot |\frac {\Delta y} y|


Note however that in your answer there should also be a contribution of the uncertainty of 40000. Or doesn't it have an uncertainty?

I think there is always uncertainty associated with a physical measurement. But we were not told what the uncertainty of 40000 was, and we didn't have any means of finding it out during the experiment. So in this case what would be the best thing to do?

 

[I like Serena]

I think there is always uncertainty associated with a physical measurement. But we were not told what the uncertainty of 40000 was, and we didn't have any means of finding it out during the experiment. So in this case what would be the best thing to do?

For a homework question I think it is ok.In practice you'd have to apply a rule of thumb.

If it's uncertainty is much lower, which would be typical for a physical constant, you can ignore it.

A measurement should always be given with its uncertainty.
If it isn't, the rule of thumb is that its uncertainty is 0.5 of the last digit.
In your case: 40000 +/- 5000.
Note that 40000 would usually be written as 4x10⁴ or 4.0x10⁴ to make the number of significant digits clear.EDIT: What do you mean that there were no means to find out during the experiment? There should always be a way.

 

[roam]

For a homework question I think it is ok.


In practice you'd have to apply a rule of thumb.

If it's uncertainty is much lower, which would be typical for a physical constant, you can ignore it.

A measurement should always be given with its uncertainty.
If it isn't, the rule of thumb is that its uncertainty is 0.5 of the last digit.
In your case: 40000 +/- 5000.
Note that 40000 would usually be written as 4x10⁴ or 4.0x10⁴ to make the number of significant digits clear.


EDIT: What do you mean that there were no means to find out during the experiment? There should always be a way.

I'm sure there is a way to find out. The experiment was about AC bridge measurements. What I posted above is from an equation

R=\frac{R_2 \times R_3}{R_4}

R₄ was a variable resistor and I found its value to be 9 +/- 1 k Ohms. R₂ and R₃ were stable resistors each of them were 200 \Omega. But they didn't tell us the uncertainty associated with them and we didn't investigate it during the course of the experiment.

So how would you find (using your rule) the uncertainty for each of the 200 Ohm resistors (0.5 of the last digit)?

 

[I like Serena]

I'm sure there is a way to find out. The experiment was about AC bridge measurements. What I posted above is from an equation

R=\frac{R_2 \times R_3}{R_4}

R₄ was a variable resistor and I found its value to be 9 +/- 1 k Ohms. R₂ and R₃ were stable resistors each of them were 200 \Omega. But they didn't tell us the uncertainty associated with them and we didn't investigate it during the course of the experiment.

So how would you find (using your rule) the uncertainty for each of the 200 Ohm resistors (0.5 of the last digit)?

On each resistor the uncertainty is specified.
The resistor would have 4 colored bands, the first 3 specify the ohms, the fourth specifies the uncertainty (usually 5% or 10%).Opposed to this, the rule of thumb basically says that all the numbers that are given are assumed to be correct, but to assume no more digits. The uncertainty is the corresponding possible rounding error.

Here it would be 200 +/- 50 ohms.

 

[roam]

On each resistor the uncertainty is specified.
The resistor would have 4 colored bands, the first 3 specify the ohms, the fourth specifies the uncertainty (usually 5% or 10%).


Opposed to this, the rule of thumb basically says that all the numbers that are given are assumed to be correct, but to assume no more digits. The uncertainty is the corresponding possible rounding error.

Here it would be 200 +/- 50 ohms.

Yes, I knew that the fourth band corresponds to the tolerance of the resistor. But we were using decade resistance boxes, so I couldn't see any colour bands. And I don't know the specifications of those resistance boxes. So I'm just going to use the rule of thumb that you mentioned.

I also have one last question: if we have the value x=147 ± 10 (absolute uncertainty), and we calculate

\frac{1}{x}=\frac{1}{147} = 6.8 \times 10^{-3}

How would you calculate the uncertainty here? The "1" here doesn't have any associated uncertainty, it's from the formula.

 

[roam]

I mean, in general when you have 1 over some number, what happens to the uncertainty of that number?

 

[I like Serena]

I mean, in general when you have 1 over some number, what happens to the uncertainty of that number?

"1" is not a measurement, it's a mathematical constant, and a whole number to boot.
Its uncertainty is zero.

Same for any mathematical constant (2, 3, pi, e, i, etcetera).

 

[roam]

"1" is not a measurement, it's a mathematical constant, and a whole number to boot.
Its uncertainty is zero.

Same for any mathematical constant (2, 3, pi, e, i, etcetera).

Right. So does this mean that for the example I gave where x=147 ± 10, for 1/x the error will be

\sqrt{p_1+p_x^2}=\sqrt{0+p_x^2}

where p_x is the percent error of x (i.e 10/147 * 100 = 6.8%). Is that right?

 

[I like Serena]

Right. So does this mean that for the example I gave where x=147 ± 10, for 1/x the error will be

\sqrt{p_1+p_x^2}=\sqrt{0+p_x^2}

where p_x is the percent error of x (i.e 10/147 * 100 = 6.8%). Is that right?

I assume you meant to use the square of p1, instead of just p1.

Apparently you're already aware that you should add uncertainties squared and then draw the square root.
Good!

However, you have to do this with the absolute uncertainties and not with the relative uncertainties.
These are standard deviations, and the typical symbol (from statistics) to use would be σ instead of p.
In physical error analysis the typical symbol would be Δx.

EDIT: as a coincidence, in this particular example the results will come out the same!

 

[roam]

I assume you meant to use the square of p1, instead of just p1.

Apparently you're already aware that you should add uncertainties squared and then draw the square root.
Good!

However, you have to do this with the absolute uncertainties and not with the relative uncertainties.
These are standard deviations, and the typical symbol (from statistics) to use would be σ instead of p.
In physical error analysis the typical symbol would be Δx.

EDIT: as a coincidence, in this particular example the results will come out the same!

Well, 0²=0 (since the uncertainty of a constant is 0 as you pointed out). Personally by the sigma (σ) I denote the absolute uncertainty whereas by "p" I mean the percentage uncertainty. For addition or subtraction of x and y, we have to add their absolute uncertainties in quadrature, i.e

√(σ_x+σ_y)

But for multipication or division we have to use the percent uncertainties:

√(p_x+p_y).

So, in my case I think I should use the percent uncertainty since there is a division! :)

Anyway, thank you very much for all the help, I really appreciate it.

 

[I like Serena]

Well, 0²=0 (since the uncertainty of a constant is 0 as you pointed out). Personally by the sigma (σ) I denote the absolute uncertainty whereas by "p" I mean the percentage uncertainty. For addition or subtraction of x and y, we have to add their absolute uncertainties in quadrature, i.e

√(σ_x²+σ_y²)

But for multipication or division we have to use the percent uncertainties:

√(p_x²+p_y²).

So, in my case I think I should use the percent uncertainty since there is a division! :)

Anyway, thank you very much for all the help, I really appreciate it.

Yes, sorry, you're right.
In this case you need to square the relative uncertainties, add, and take the square root.

I meant that in the general case, you'll get contributions to the total uncertainty from different sources.
This is applicable for any formula.

To compare, in the case

q= \frac x y

Your uncertainty is

\Delta q = \sqrt { (\Delta q_x)^2 + (\Delta q_y)^2 }

where \Delta q_x = |\frac 1 y| \cdot \Delta x and \Delta q_y = |\frac {-x } {y^2}| \cdot \Delta y.





Sorry if I'm overdoing it.
I guess you don't need this if you only do sums and quotients. ;)
(But what if you have square roots, or even worse, a sine? )

 

[lackos]

(But what if you have square roots, or even worse, a sine? )[/QUOTE]

RUN AWAY, well that's what i do anyway

 

[I like Serena]

(But what if you have square roots, or even worse, a sine? )

RUN AWAY, well that's what i do anyway

So what about KE =½mv²?

You may want to know the uncertainty in that one, before it overtakes you and hits you while you're running away!

 

[Ray Vickson]

Hi roam!

Yes, this is correct.

When multiplying or dividing the relative error remains the same.


Generally we have for a quotient q = x/y
that the contribution in its uncertainty by y is:

\Delta q = | \frac {\partial q} {\partial y} | \Delta y = |\frac {-x} {y^2}| \Delta y = |q| \cdot |\frac {\Delta y} y|


Note however that in your answer there should also be a contribution of the uncertainty of 40000. Or doesn't it have an uncertainty?

The statement "When multiplying or dividing the relative error remains the same" is only true for small changes. In this case, 1000 is not particularly small compared with 9000, so the percentages when going up from 9000 to 10000 or down from 9000 to 8000 are different. Just calculate them to see.

RGV