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Internal Resistances

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Calculate internal resistance of meter. There is a 10V power supply with a variable resistor connected in series. The meter will be connected across the resistor. I adjust the resistor until the reading in the meter read about 5.37V. The value of resistance is 15 ohms. How do I calculate the internal resistance? This is the equation I used

    5.37V=10V((Rm*15)/(Rm+15))

    I solve for Rm and get 559kOhm. Does this look correct?
     
  2. jcsd
  3. Sep 11, 2008 #2

    dlgoff

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    Looks correct to me. Just curious, how did you measure the resistance being 15 ohm? With the same meter?
     
  4. Sep 12, 2008 #3

    alphysicist

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    I don't see how you are getting 559kOhm from that equation; is there a calculation error?

    Also, perhaps I'm visualizing the circuit incorrectly, but could you show how you got that equation?
     
  5. Sep 12, 2008 #4
    That 15 ohms was what it took to make the meter reading drop to half of 10V. It is a variable resistor. I tried many ohmage settings and 15Ohms dropped the reading to half of 10V to 5.??.

    alphysicist: As for a the miscalculation, you may be right. I calculated that quickly in lab and was unsure if I was even approaching the problem in the correct manner. I will check that but can't do it at the moment.
     
  6. Sep 12, 2008 #5

    alphysicist

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    What bothered me was that the equation you had in your post was not dimensionally consistent (volts on the left, volts*ohms on the right); I was wondering if there was more to the equation that you had already dealt with that took care of that. Unfortunately I think I am not visualizing your circuit correctly, so there could something that I'm not thinking about.
     
  7. Sep 12, 2008 #6

    dlgoff

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    So you were using a decade box? There may be tolerances of the resistors to consider when determining the internal resistance of your meter.
     
  8. Sep 12, 2008 #7

    Redbelly98

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    Are you supposed to calculated the internal resistance of the power supply or of the meter?
     
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