Where "Rdg"is for reading and "dgt"is for digits. So, 15,23 * 0,015 = 0,22845. Rounding to most significant digit it's 0,2. So is this correct? $$15,23 \pm 0,2$$

Since 4dgt is smaller we ignore it. Or am i doing this wrong? I need to find uncertainty in the way I did for a physics lab.

  • \$\begingroup\$ What do "rdg" and "dgt" mean? I can guess, but I shouldn't have to. Are you using comma or point as a decimal mark? \$\endgroup\$
    – The Photon
    Commented Jun 3, 2020 at 17:59
  • \$\begingroup\$ @ThePhoton "Rdg"is for reading and "dgt"is for digits. \$\endgroup\$ Commented Jun 3, 2020 at 18:08
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    \$\begingroup\$ Please edit the question to make it clear. Comments are to help you improve the question, but the question should be understandable by new readers without reading comments. \$\endgroup\$
    – The Photon
    Commented Jun 3, 2020 at 18:14

1 Answer 1


Why did you round to 0.2 for just one significant digit? Nothing anywhere in the question has only one significant digit. The operand with lowest number of digits in the question is two so why did you round to just one significant digit?

\$ 15.23 \times 0.015 = 0.23\$ rounded to the same number of significant digits as the operand with the lowest number of significant digits (which is the 1.5%)

Then \$\pm4\$ digits on top of that means \$\pm4\$ of the least significant digit of your reading which was 15.23. The least significant digit of that reading is 0.01. You use the digit from the 15.23 and not the 1.5% because the digit error was quantified relative to your reading. It makes more sense if you just think about how your meter must have an finite minimum absolute error. If it was purely percentage error then it would imply the absolute error of your meter becomes infinitesimal as you approach a reading of zero.

Therefore \$ \pm0.23 \pm0.04 = \pm0.27\$

  • \$\begingroup\$ Thank you, so it should be added. Answer here says otherwise, that why I wasn't sure electronics.stackexchange.com/questions/316345/… \$\endgroup\$ Commented Jun 3, 2020 at 18:11
  • \$\begingroup\$ @MuradDavudov It's more like you can ignore the counts if it is much smaller than the percent error since it gets swamped out. WIth small readings the counts dominates and with large readings the percent error dominates. But your count error and percentage error are fairly similar in magnitude so you can't just neglect one. \$\endgroup\$
    – DKNguyen
    Commented Jun 3, 2020 at 18:13
  • \$\begingroup\$ I, see. It's all clear then. \$\endgroup\$ Commented Jun 3, 2020 at 18:14

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