What is the most likely value of the current, with its uncertainty?

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The discussion focuses on determining the uncertainty in current readings using a digital ammeter with a specified uncertainty of plus-minus 1.5% plus-minus 2 digits. For part (a), the calculation of uncertainty for a reading of 2.64A leads to a result of plus-minus 0.06A, which aligns with the significant figures rule. In part (b), the user seeks clarification on how to find the most likely value of fluctuating current readings between 1.58A and 2.04A. It is explained that the uncertainty includes adding two units of the least significant digit to the percentage uncertainty. The user is encouraged to apply this understanding to complete part (b) of the problem.
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Homework Statement


Hi
I need some help with this problem. It reads this way:
The manufacturer of a digital ammeter quotes its uncertainty as plus-minus 1.5% plus-minus 2 digits.

a) Determine the uncertainty in a constant reading of 2.64A
b) The meter is used to measure the current from a d.c power supply. The current is found to fluctuate randomly between 1.58A and 2.04A. Determine the most likely value of the current, with its uncertainty.



Homework Equations


uncertainty as plus-minus 1.5% plus-minus 2 digits.





The Attempt at a Solution


Solving (a)
2.64 x 1.5% = 0.0396
1.5% x 2 = 0.03
Adding the above =0.0696
The answer in my physics book is plus - minus 0.06A. Can anyone kindly clarify way? I know the number of significant figures determines the uncertainty. Can the answer not be 0.07A? (After rounding up)

Solving (b)
I am not sure how to being. Since uncertainty is the range of values, I decided to find the range
So, 2.04-1.58 = 0.46A

But not sure how to progress further. Any suggestions? (Please bear in mind I study independently)
 
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The uncertainty quoted doesn't make much sense to me. Can you indicate exactly what was written? Did it really say exactly this:

± 1.5% ± 2 digits

or did you modify it in some way?
 


cepheid said:
The uncertainty quoted doesn't make much sense to me. Can you indicate exactly what was written? Did it really say exactly this:

± 1.5% ± 2 digits

or did you modify it in some way?
Yes.Its correct (Advanced level Practical work for Physics, Chapter 1, p7- M Crundell & C Mee)
 
I couldn't find anywhere that had even a preview of the inside of that book, but Googling the phrase "± 1.5% ± 2 digits"

let me to this datasheet for an actual digital multimeter:

http://www.bstcaltek.com/catalog/download/manual/1801_M.pdf

If you look above the table of electrical specifications in section 3.2 on that PDF document, there is a note that says:

Accuracies are ± (% of reading + number of least significant digits)

So this explains what your book is talking about. "2 digits" actually means "2 of the least significant digits", which means, "add two units of whatever the smallest digit in your measurement is."

In this case, the precision of your measurement is to the nearest hundredth of an amp, which means that after calculating 1.5% of the measured value, you add 2*0.01 to that. The least significant digit is in the hundredth's place in this example. So you add two hundredths of an amp. If you do that, you'll get something close to 0.06 A.
 
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cepheid said:
I couldn't find anywhere that had even a preview of the inside of that book, but Googling the phrase "± 1.5% ± 2 digits"

let me to this datasheet for an actual digital multimeter:

http://www.bstcaltek.com/catalog/download/manual/1801_M.pdf

If you look above the table of electrical specifications in section 3.2 on that PDF document, there is a note that says:



So this explains what your book is talking about. "2 digits" actually means "2 of the least significant digits", which means, "add two units of whatever the smallest digit in your measurement is."

In this case, the precision of your measurement is to the nearest hundredth of an amp, which means that after calculating 1.5% of the measured value, you add 2*0.01 to that. The least significant digit is in the hundredth's place in this example. So you add two hundredths of an amp. If you do that, you'll get something close to 0.06 A.
Thanks a lot. I have done the part you suggested as my (a) solution. Its (b) that I will like further assistance.
 
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