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Working out EMF and internal resistance

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Power supply of EMF = E and internal resistance of R. Need E and R.
    This is connected to resistor of 10'000 Ohm. The current drawn from the power supply is 0.91A.
    A voltmeter of internal resistance 10'000 Ohm is connected in parallel to the resistor. The voltmeter reads 8.3 Volts.

    Original question (number 2) can be found here:

    https://www.dropbox.com/s/n842i3ojnanhtxc/2013-01-13 14.37.43.jpg

    2. Relevant equations

    V = IR and Terminal PD = E - IR, presumably.

    3. The attempt at a solution

    Overall plan: Use the two situations to set up two simultaneous equations, then solve for E and R.
    I'm assuming the questions is saying that the drawing of the current was independent of connecting the voltmeter.

    From 1st part:

    V = IR,
    R = 10'000 Ohm
    I = 0.91A
    So V = 9'100 V
    V = terminal PD so:
    9'100 = E - 0.91×R (Call this 1)

    From 2nd part:

    Total resistance of Voltmeter and resistor is:

    1/10000 + 1/10000 = 1/5000 so R_tot = 5000 Ohm.

    Voltage measured is terminal voltage (8.3V)

    Current is I = V/R = 8.3/5000 = (C 0.00166 A

    So:

    8.3 = E - 0.00166×R (Call this 2).

    HOWEVER: combining 1 and 2 gives a negative internal resistance!

    Any ideas what I've done wrong?



    Thanks guys!
     
    Last edited: Jan 13, 2013
  2. jcsd
  3. Jan 13, 2013 #2

    gneill

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    Staff: Mentor

    I note that they refer to it as "A power supply" and not a battery. Depending upon the complexity of the power supply and its ultimate purpose, it is possible that it will exhibit a negative internal resistance. This sort of thing is achievable with active components (amplifiers).
     
  4. Jan 13, 2013 #3

    TSny

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    Homework Helper
    Gold Member

    The potential difference across a 10 kΩ resistor carrying 0.91A would be 9100 V. I wonder if there's a typo in the book. Maybe the current was meant to be stated as .91 mA. Hard to say.
     
  5. Jan 14, 2013 #4
    Ok thanks both of you.
    I wanted to check whether I was missing something but I wasn't. It was a typo - setting the first current to 0.91mA gives the answer in the back of the book.
     
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