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# Uncertainty Question. watch

1. As the percentage uncertainty in each value is of 1% and average value of current is 2.00 A, so I solved it like this:

And as I = 2.00 A

So according to me the answer should be B, but the mark scheme gives the answer as D. I just can't believe that how can be equal to 0.04 A?
2. (Original post by Zishi)
As the percentage uncertainty in each value is of 1% and average value of current is 2.00 A, so I solved it like this:

And as I = 2.00 A

So according to me the answer should be B, but the mark scheme gives the answer as D. I just can't believe that how can be equal to 0.04 A?

What is the maximum and minimum possible value of current?
As the reading fluctuates between 1.98 and 2.02, the actual value of current may be 1.98 as well as 2.02.
Now, the reading has a systematic error of 1%. That means that minimum possible value is . The maximum value is .
The mean reading is 2.00, therefore the maximum error is 0.04.
3. (Original post by jaroc)

What is the maximum and minimum possible value of current?
As the reading fluctuates between 1.98 and 2.02, the actual value of current may be 1.98 as well as 2.02.
Now, the reading has a systematic error of 1%. That means that minimum possible value is . The maximum value is .
The mean reading is 2.00, therefore the maximum error is 0.04.
Oh well, that's really a helpful way. But why can't it be solved using that fractional uncertainty way? i.e , etc?
4. (Original post by Zishi)
Oh well, that's really a helpful way. But why can't it be solved using that fractional uncertainty way? i.e , etc?
You can, but as the manufacturer has used a % you might as well use a %.

The point here is that the sytematic error is added to the random error.
The sytematic error is caused by something within the device. The random error is caused by the act of measurement.
Both are 1% or 0.02 in 2.00 which equates to 0.01 in 1 or 1/100
It doesn't matter which you use, you will still arrive at the same answer that the sum is 2% or 2/100 or 0.02 in 1.00 or 0.04 in 2.00
5. (Original post by Stonebridge)
You can, but as the manufacturer has used a % you might as well use a %.

The point here is that the sytematic error is added to the random error.
The sytematic error is caused by something within the device. The random error is caused by the act of measurement.
Both are 1% or 0.02 in 2.00 which equates to 0.01 in 1 or 1/100
It doesn't matter which you use, you will still arrive at the same answer that the sum is 2% or 2/100 or 0.02 in 1.00 or 0.04 in 2.00
Hmm, I also thought that their percentage uncertainties would add up. But as we take the mean of those two currents, we add them up and divide them by 2. My teacher gave me the concept that the same happens with their uncertainties, they'll also add up and will also be divided by two.

As 1% of 1.98 = 0.02 and 1% of 2.02 = 0.02

So uncertainty in the mean value = (0.02+0.02)/2 = 0.02

So is that concept wrong? Do they just simply add up? Or does this happens in other cases?
6. It's not the mean of 2 currents.
It's the mean of an unknown number of currents that fluctuate between those two values.
It doesn't say in the question that there were 2 measurements.
However, that isn't really the point here.
The point is that there is a random fluctuation in the readings of ± 0.02 or 1%
Added to this there is an error in the instrument of 1% (equivalent to ± 0.02)
This means that the true reading could be anywhere between 1.96 and 2.04 in the worst case where both errors combine to reinforce each other.
In an experiment with these known errors you would quote the maximum (possible) error.
7. (Original post by Stonebridge)
It's not the mean of 2 currents.
It's the mean of an unknown number of currents that fluctuate between those two values.
It doesn't say in the question that there were 2 measurements.
However, that isn't really the point here.
The point is that there is a random fluctuation in the readings of ± 0.02 or 1%
Added to this there is an error in the instrument of 1% (equivalent to ± 0.02)
This means that the true reading could be anywhere between 1.96 and 2.04 in the worst case where both errors combine to reinforce each other.
In an experiment with these known errors you would quote the maximum (possible) error.
Oh! I get it now. But if there were two readings, then the error in mean of those two values would've been ±0.02, right?
8. If you only took 2 readings of a current and obtained the values 1.98A and 2.02A you would not really have sufficient data to be able to estimate the error in the readings. The idea is to take at least 6 or 7, preferably more, and take the mean value of those.
Once you have a mean current value from this you can find the mean value of the difference between that and the readings to give you the mean error.

I paste below an answer I gave to this some time ago

"For example, you get these values
9.8mm
9.9mm
10.0 mm
10.0 mm
10.1 mm
10.2 mm
giving a mean value 10.0mm

There are a number of ways of dealing with the error/uncertainty.

These readings deviate from the mean by
-0.2, -0.1, 0, 0, 0.1, 0.2 respectively.

Change the - signs to a plus and add all these and you get 0.6
(if you just add them as they are you get zero)
divide by 6, the number of readings, and you get 0.1mm
You could express the average uncertainty of the readings as ± 0.1mm

A very rough and ready method of expressing the uncertainty in the repeated readings is to just take the maximum range of ±0.2 (from 9.8 to 10.2 with mean of 10.0)
This would be the maximum uncertainty. The probable uncertainty is lower than that, but is a statistical calculation. (The average uncertainty was less as we saw above.)"
9. (Original post by Stonebridge)
If you only took 2 readings of a current and obtained the values 1.98A and 2.02A you would not really have sufficient data to be able to estimate the error in the readings. The idea is to take at least 6 or 7, preferably more, and take the mean value of those.
Once you have a mean current value from this you can find the mean value of the difference between that and the readings to give you the mean error.

I paste below an answer I gave to this some time ago

&quot;For example, you get these values
9.8mm
9.9mm
10.0 mm
10.0 mm
10.1 mm
10.2 mm
giving a mean value 10.0mm

There are a number of ways of dealing with the error/uncertainty.

These readings deviate from the mean by
-0.2, -0.1, 0, 0, 0.1, 0.2 respectively.

Change the - signs to a plus and add all these and you get 0.6
(if you just add them as they are you get zero)
divide by 6, the number of readings, and you get 0.1mm
You could express the average uncertainty of the readings as ± 0.1mm

A very rough and ready method of expressing the uncertainty in the repeated readings is to just take the maximum range of ±0.2 (from 9.8 to 10.2 with mean of 10.0)
This would be the maximum uncertainty. The probable uncertainty is lower than that, but is a statistical calculation. (The average uncertainty was less as we saw above.)&quot;
Awesome, thanks a lot for the explanation.

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