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Internal resistance of multimeter

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    I am taking a class in circuit analysis and am trying to measure the internal resistance of the Radio Shack multimeter. The internal resistance (unknown) is R_x. A battery (with voltage V_s) is available along with a resistor R that has the same order of magnitude of R_x.

    2. Relevant equations

    Kirchhoff's voltage law (KVL)

    3. The attempt at a solution

    I tried using the circuit shown in the attached file and connected it in the way shown below the circuit. Before making any connections, I tried to figure out what the voltages would be.

    I think the voltage across R should be V_s because of KVL around the loop containing the battery and the resistor.

    By applying KVL around the loop containing the multimeter, I get the following equation:

    V_x + V_s = v

    (V_x is the voltage across the internal resistance of the meter.)

    But when I hook up the circuit, the voltage v is less than V_s, which would mean that V_x is negative, right? This does not make sense to me. It looks like R_x and R are in series, which, if V_x was positive, would mean that I could write something like

    [tex]
    v \frac{R}{R+R_x} = V_s
    [/tex]

    which would mean

    [tex]
    R\left( \frac{v}{V_s} - 1 \right) = R_x
    [/tex]

    But because v is less than V_s, the parenthesized expression is negative and R_x is negative, which can't be true. Any advice?
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    You seem to be thinking of the meter resistance as being in series with a meter device which has infinite resistance. Instead, consider the meter resistance as a resistance value that is in parallel with ("shunts") a perfect meter. The perfect (or "ideal") meter tells you the potential across this meter resistance.

    attachment.php?attachmentid=52186&stc=1&d=1350874556.gif

    You might want to reconsider the configuration of your test circuit (hint: think voltage divider).
     

    Attached Files:

    • Fig1.gif
      Fig1.gif
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  4. Oct 22, 2012 #3
    Your equation:
    V_x + V_s = v
    should rather read
    V_s - V_x = V
    The emf of the battery is positve, but the voltage drop through Rx is negative.
     
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