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enter image description here

Everything is clear and understood, I only don't understand why he added enter image description here only in the second question and did not add it in the first question

Also, what he meant by "If the meter resistance is much greater than the element's resistance, the current through the meter will be negligible"

Source:

The Electrical Engineer’s Guide to Passing The Power PE Exam

by A. S. Graffeo, P.E.

PowerPE, LLC.

July 2016 Printing

ISBN: 978-0-9881876-1-0

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  • \$\begingroup\$ It's in the first question too; in the second equation of the solution. \$\endgroup\$
    – user16324
    Jul 12, 2021 at 11:20

2 Answers 2

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Think about what 100mV/30mA means physically. Think about Ohm's law.

Now see if you can find 100mV/30mA in the first question. Maybe it's multiplied by a current such as 30mA.

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  • \$\begingroup\$ OK. So, Im*Rm is added but as 100 mV good now I got it. But, what does he mean by "If the meter resistance is much greater than the element's resistance, the current through the meter will be negligible" it doesn't make sense! \$\endgroup\$
    – OMAR
    Jul 12, 2021 at 7:26
  • \$\begingroup\$ If you think about what 100mV/30mA means physically, imagine it in parallel with an element of, say, 0.1 ohm. Most of the current will flow through the 0.1 ohm element. At about 1A the meter will read full scale. If you don't ignore the meter current it will take 1.03A to make the meter read full scale, a difference of 3%. \$\endgroup\$ Jul 12, 2021 at 9:38
  • \$\begingroup\$ but From the Figure Rm (meter resistance) and Rs (series resistance) are in series not paralleled! \$\endgroup\$
    – OMAR
    Jul 12, 2021 at 9:41
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    \$\begingroup\$ The text is not talking about the figure, it's asking you to imagine something else- specifically an ammeter. \$\endgroup\$ Jul 12, 2021 at 9:42
  • \$\begingroup\$ I think you meant Voltmeter, that what is mentioned in the text not an ammeter \$\endgroup\$
    – OMAR
    Jul 12, 2021 at 9:45
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Also, what he meant by "If the meter resistance is much greater than the element's resistance, the current through the meter will be negligible"

If you put a very high resistance in parallel with a low resistance, you get a low resistance which is almost exactly the same as the small resistance alone.

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  • \$\begingroup\$ See the Figure Rm (meter resistance) and Rs (series resistance) are in series not paralleled! \$\endgroup\$
    – OMAR
    Jul 12, 2021 at 9:39
  • \$\begingroup\$ @OMAR You need to read what it says just before that: "When a voltmeter is connected across an element...". Voltmeters are normally connected in parallel with things, not in series. \$\endgroup\$
    – Simon B
    Jul 12, 2021 at 11:02

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