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# (IMAGE PROVIDED)uncertainty help (as level coursework) Watch

1. hello there

I am doing my ASlevel coursework piece on the EMF and internal resistance of a cell.

I have some results which I have got from another source which include the uncertainty in volts and the voltage measurement. I have been told the ABSOLUTE uncertainty in the volt meter is+/- 0.01v and the % uncertainty is 2%

How do I get the uncertainty values in that column? (B) thanks

PICTURE:

EDIT: In column B all the values should have "+/- " in front of them
2. Okay lets say I am using a voltmeter which has an absolute uncertainty of +/-0.01v and a percentage uncertainty of +/-2%

I then take a reading using that voltmeter and get the reading to be 1.36v

What would the uncertainty be in the reading? thanks. Would it be +/- 0.03? The answer book says it is but I don't know how to get that. Thanks
3. (Original post by upthegunners)
hello there

I am doing my ASlevel coursework piece on the EMF and internal resistance of a cell.

I have some results which I have got from another source which include the uncertainty in volts and the voltage measurement. I have been told the ABSOLUTE uncertainty in the volt meter is+/- 0.01v and the % uncertainty is 2%

How do I get the uncertainty values in that column? (B) thanks

PICTURE:

EDIT: In column B all the values should have "+/- " in front of them
The values in that column look like the absolute uncertainty corresponding to a percentage uncertainty of 2%
Without any further details about how these measurements were performed it's not possible to say much more.
4. (Original post by upthegunners)
Okay lets say I am using a voltmeter which has an absolute uncertainty of +/-0.01v and a percentage uncertainty of +/-2%

I then take a reading using that voltmeter and get the reading to be 1.36v

What would the uncertainty be in the reading? thanks. Would it be +/- 0.03? The answer book says it is but I don't know how to get that. Thanks
This question is related to the one you already have in the other thread on uncertainty.
I've moved it there.
5. (Original post by Stonebridge)
The values in that column look like the absolute uncertainty corresponding to a percentage uncertainty of 2%
Without any further details about how these measurements were performed it's not possible to say much more.
thanks for moving it

I basically just had a series circuit connect up to a 1.5v cell, rheostat and ammeter and took 10 voltage readings a difference resistances and that's it
6. (Original post by upthegunners)
Okay lets say I am using a voltmeter which has an absolute uncertainty of +/-0.01v and a percentage uncertainty of +/-2%

I then take a reading using that voltmeter and get the reading to be 1.36v

What would the uncertainty be in the reading? thanks. Would it be +/- 0.03? The answer book says it is but I don't know how to get that. Thanks
There are two things going on here.
Firstly the ± 0.01V is the precision of the instrument. This is the smallest division on its scale. This value is not necessarily the uncertainty in a reading taken with that instrument.

% uncertainty is the actual uncertainty divided by the value of the reading and times 100
So if you have 1.00 ±0.01V that is a % uncertainty of 1% for example.

(0.01/1.00) x 100

If, on the other hand, you are told that an instrument has a % uncertainty in all readings of, say, 2% then each reading will be ± 2%
so if you have a reading of 2.00V the uncertainty will be ± 0.04V
This is because 0.04 is 2% of 2.00
7. (Original post by Stonebridge)
There are two things going on here.
Firstly the ± 0.01V is the precision of the instrument. This is the smallest division on its scale. This value is not necessarily the uncertainty in a reading taken with that instrument.

% uncertainty is the actual uncertainty divided by the value of the reading and times 100
So if you have 1.00 ±0.01V that is a % uncertainty of 1% for example.

(0.01/1.00) x 100

If, on the other hand, you are told that an instrument has a % uncertainty in all readings of, say, 2% then each reading will be ± 2%
so if you have a reading of 2.00V the uncertainty will be ± 0.04V
This is because 0.04 is 2% of 2.00
thanks sooo much dude! You're explained that better than my teacher

And he also did say that the +/-1.00 was in fact the uncertainty
8. (Original post by Stonebridge)
There are two things going on here.
Firstly the ± 0.01V is the precision of the instrument. This is the smallest division on its scale. This value is not necessarily the uncertainty in a reading taken with that instrument.

% uncertainty is the actual uncertainty divided by the value of the reading and times 100
So if you have 1.00 ±0.01V that is a % uncertainty of 1% for example.

(0.01/1.00) x 100

If, on the other hand, you are told that an instrument has a % uncertainty in all readings of, say, 2% then each reading will be ± 2%
so if you have a reading of 2.00V the uncertainty will be ± 0.04V
This is because 0.04 is 2% of 2.00
sorry to hassle you again but how would I calibrate the voltmeter?
9. (Original post by upthegunners)
sorry to hassle you again but how would I calibrate the voltmeter?
There are a number of ways.
It depends on
- what type of voltmeter it is
- what equipment you have at your disposal

Try Googling
"Weston Cell" for a start.

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