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Internal resistance of a digital multimeter

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Provided that the input resistance of the DMM is
    much larger than the resistance values in the circuit to
    which it is connected, the DMM will not draw a
    significant current. Estimate the input resistance of the
    DMM by connecting the 10 M resistor supplied in
    series with it and noting the voltage reading when
    this series combination is connected to your DC black
    box. Explain clearly the logic of doing this.

    Basically I have a simple circuit with a DC black box which is a box with a 1.5V battery and three resistors (1 in series and 2 in parallel) and I have a digital multimeter that I have to estimate its internal resistance by connecting a 10 MΩ resistor to the circuit in series with the multimeter. I can't really think of why connecting such a large resistor would help determine the internal resistance.

    3. The attempt at a solution

    I haven't been able to come up with an answer because first I need to understand the role of the 10 MΩ resistor.
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2


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    Science Advisor
    Homework Helper

    Draw the meter as a meter + a 1M resistor in series.
    Now connect a 10M resistor accross the meter inputs.

    What do you know about resitances in parallel
  4. Oct 7, 2008 #3
    First of all, I appreciate your post.

    Well, for resistances in parallel we know that the overall resistance will be smaller than any individual resistance. so in this case 1 / R = 1 / R_s + 1 / R_p where R_s is the series resistor and R_p is the parallel resistor. so R = 0.9M, roughly. I still don't get why the 10M resistor is needed.
    Last edited: Oct 7, 2008
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