Voltage measurement using multimeter

Discussion in 'Homework Help' started by lippyf, Dec 9, 2015.

  1. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    A circuit with DC voltage of 5 V with an output resistance of 0.1 MΩ. What voltage do you measure when you
    using a digital multimeter APPA62 having 10 MΩ input impedance? If the defect is small or large in relation to the error specified in the instrument's specification? See attached APPA62 electrical specifications.

    When using APPA62 to measure the output resistance, I assume the ciruit will be like a DC volage U=5V connected with two parallel resisitor R1=0.1MΩ and R2=10MΩ. So I just have to calculate the voltage over R2 which is the value the multimeter will be showing?
    Before I start to attempt to calculate, am I thinking this right?
     
  2. RBR1317

    Active Member

    Nov 13, 2010
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    Can you explain why you would need to 'calculate' the voltage across two parallel resistors?
     
  3. crutschow

    Expert

    Mar 14, 2008
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    The two resistors are in series, not parallel.
    They form a voltage divider. For 0.1MΩ and 10MΩ the measurement error will be about 1% low.
     
  4. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    Oh ok, so the multimeter will be showing:
    5*((0.1*10^6*10*10^6)/(0.1*10^6+10*10^6))=495049.505≈4950
    4950/495049.505=0.009999≈1%
    is this right?
     
  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I don't think so. Your question was:

    If a 5 volt source with the source resistance of 0.1 megaohm resistance has a a 10 megaohm resistor load, a 10 megaohm input resistance meter connected across the load will read _____ volts.
     
  6. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    Yes, you're right. How should I do then?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    When voltage source is modeled, it has two components, the source of voltage and a resistor, the resistor is in series with the source of voltage. Therefore when a voltmeter is introduced into the circuit, the resistor inside the voltmeter will be in series with the resistor inside the voltage source.
     
  8. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    Not sure if I understand it correct, but doesnt it form a circuit with a voltage source connected with two resisitor in series?
     
  9. shteii01

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    Feb 19, 2010
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  10. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    You need to DRAW the circuit as you see it.

    You have two questions to answer. What is the calculated voltage across the 10 megaohm resistor and what voltage will the meter read?
     
  11. Bordodynov

    Active Member

    May 20, 2015
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    Vmeas=5*(10*10^6/(0.1*10^6+10*10^6))=5*(10/(0.1+10))=5*(1/(0.01+1))=5/1.01=4.950495049505
    and another error of +/-1.5% --->Vmeas will be in the range [4.876 ; 5.025].
     
    Last edited: Dec 10, 2015
  12. crutschow

    Expert

    Mar 14, 2008
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    I thought the 10MegΩ resistance was the meter impedance, not a separate resistor. :confused:
     
  13. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    This is simply a voltage divider problem. The top resistor is 100k and the bottom resistor is 10Megs.
    The 5v ideal part of the supply is connected to the 100k, and the voltmeter sees the voltage only across the 10Meg resistor.
    An approximation is to say that the voltage will read 1 percent low because 100k is 1 percent of 10Megs, but there is a more exact calculation which you should do too which takes it down very slightly more than 1 percent.
     
  14. MikeML

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    Oct 2, 2009
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    What does the "+x digit" mean in the accuracy specification (x is 2, 3 or 5 depending which spec you look at)?

    157.gif
     
  15. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    The first part was measuring the power source.

    The second part I assumed was measuring the voltage across a 10 megaohm load resistor, R2.

    Anyways, the TS is to take into account the meter specifications. I'm waiting for him to return to answer Mike's question.
     
    Last edited: Dec 10, 2015
  16. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Which question did you answer, the first one or the second? And you shouldn't be taking it out so far because the meter doesn't support that many places to the right of the decimal place.

    Where did you get the error was +/- 1.5%?

    You will need to revisit your answer.
     
  17. Bordodynov

    Active Member

    May 20, 2015
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    Yes, I have to clarify my statement. It should read:

    Vmeas=5*(10*10^6/(0.1*10^6+10*10^6))=5*(10/(0.1+10))=5*(1/(0.01+1))=5/1.01=4.950495049505 (measure ideal voltmeter)
    and another error of +/-0.5% + +/-delta --->Vmeas will be in the range [4.9257-delta ; 4.9752+delta].

    If the display after the dividing point 3 digits, the delta = 2 * 0.001 = 0.002V -->
    Vmeas will be in the range [4.923 ; 4.977].
     
  18. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    4.950495049505-delta=4.923742574
    how do the multimeter digit rounding? will it not round up to 4.924?
     
  19. lippyf

    Thread Starter New Member

    Sep 26, 2015
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    "If the defect is small or large in relation to the error specified in the instrument's specification?"
    also can someone explain what do they mean with that?
     
  20. Bordodynov

    Active Member

    May 20, 2015
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    Lippyf. The It belongs to the interval error of the instrument 0.5% + 2digit. How many digits after the dividing point you see in the measurement?
    delta=2digit !
    Vmeas It belongs to the interval [Videal-0.5%-2digit; Videal+0.5%+2digit]
     
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