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

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Homework Help Overview

The problem involves determining the uncertainty in measurements taken with a digital ammeter, specifically focusing on a constant reading and fluctuating current values. The subject area includes measurement uncertainty and electrical current analysis.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the uncertainty for a constant reading and questions the rounding of the result. They also express uncertainty in approaching the second part of the problem regarding fluctuating current values.
  • Some participants question the clarity of the uncertainty description provided in the problem statement.
  • Others suggest looking into external resources to clarify the interpretation of the uncertainty quoted.

Discussion Status

Participants are actively discussing the interpretation of the uncertainty quoted in the problem. Some guidance has been offered regarding the calculation of uncertainty, particularly in relation to the least significant digits. The original poster has made progress on part (a) and is seeking further assistance for part (b).

Contextual Notes

The discussion includes a reference to a specific textbook and a datasheet for a digital multimeter, which may provide additional context for understanding the uncertainty calculations. There is an emphasis on the precision of measurements and the significance of rounding in reporting uncertainty.

aurao2003
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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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