Question

The percentage uncertainty in capacitance C of a capacitor is 2.5 +/- 10%. What is the absolute uncertainty in 1/C? Please help!

Answer 1

your value is C=2.5 ±10% , you have to find 1/C ?
simply 1/C=1/2.5=0.4 means 1/C=0.4
now about ±10% so it is tolerance of capacitor so a capacitor with tolerance 10% means it tolerate 10F capacitance more from it own value ,on solving 10%=10, so your value lie -10<2.5>+10 so your capacitor capacitance power tolerant will lie between -12.5 to +12.5...after this value your capacitor will burnout. so 1/C=(1/ ±10%)= ±10 so final answer is 1/C=0.4 ± 10 [means capacitance lie between -10.4< C > +10.4]... i hope you will understand what i want to explain..Thanks

Answer 2

That is what I did but the mark scheme gives this answer as the absolute uncertainty From ± 0.4 (or ± 0.5) to ± 0.1
(or ± 0.2). there are other values of C. It's physics paper5 9702/52/oct/nov/14. I'm confused or maybe I'm getting the question wrong. Any ideas?

Answer 3

From a mathematical point of view 2.5 ±10% should mean 2.5 ± 0.25
So C between 2.25 and 2.75

Answer 4

yeah, but what's the absolute uncertainty in 1/C. Can you check the attachment, question 2 please.

Answer 5

I agree with the way skybrel approached the solution but I'm not sure how he got those numbers, if we assume that the percentage variation here was gotten over just the basic standard deviation then what has been done to get it is the following:

\[\sigma _N = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2} \]

As a sidenote here, standard deviation isn't the only or the best deviation known to man there are others too such as relative and absolute deviation. From the context of that exam I don't assume they actually mean they used these deviations but the following:

Absolute:
\[dev = \frac{\sigma}{\overline x}\]
Percentage:
\[dev = \frac{\sigma}{\overline x}*100\]

Which seems to assume that the average value is the correct value? Anyhow from that point on it's easy to see that:

\[\frac{dev*\overline x}{100} = \sigma \]
\[\sigma = \frac{10*2.5}{100} = 0.25 \]
then again we have the requirement to find:
\[\frac{1}{C} = ( C \pm \sigma )^n =( C^n \pm n\sigma ) = ( 2.5^n \pm (-1)*10\% ) = (0.4 \pm 10\%) = (0.4 \pm 0.25) \]

The basic rules for uncertainty operations are easy to find:
http://web.uvic.ca/~jalexndr/192UncertRules.pdf
http://spiff.rit.edu/classes/phys273/uncert/uncert.html#fraction
http://www.rit.edu/~w-uphysi/uncertainties/Uncertaintiespart2.html
http://virgo-physics.sas.upenn.edu/uglabs/lab_manual/Error_Analysis.pdf

Answer 6

this no edit button is annoying, if you pause your answer and continue later you're bound to have to fix half of it and can't.
The error is in the last "=" of 1/C calculation where since I just copy-paste most of the time left in the old result of 0.25 and not 10% of 0.4, also there's an exponent "n" on 2.5 which should be -1.

In any case this power calc. seems to be applicable to percentage uncertainties only, which computing compound errors including powers seems to work with "absolute" uncertainties as well.

In any case I've not done stats. in a while and don't guarantee correctness.

Answer 7

The last equation is what my teacher thought us but I don't see it fitting here. The answer from the mark scheme gives a range of +/- 0.4 (or +/- 0.5) to +/- 0.1 (or +/- 0.2). I've solved errors from previous papers and they were correct. I'm having trouble with this paper in particular but thanks.

Answer 8

are you sure you're looking at the right column, doing the reciprocal of 0.2-8.8 gives 0.4-0.1 range, but the errors are not that.It seems weird to have errors of the same magnitude as 1/C.

If this same procedure worked on the last exam/paper with different numbers (but same principle idea) could it be possible that someone marked 1/C as errors by accident?

Answer 9

No. I meant 0.4 to 0.1 are the errors.

Answer 10

you can look this stuff up on the BBC website.

Answer 11

look at rule #3, with n = -1

from here

http://web.uvic.ca/~jalexndr/192UncertRules.pdf