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Electrical Circuit Problem

  1. Aug 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A real voltage source is of the Thevenin Type. The open circuit voltage of the source is 12 volts. If a 100 ohm load is added to the circuit, the voltage at the same terminals is 11.2 volts. What is the internal resistance of the source.


    3. The attempt at a solution


    I found the total current in the circuit to be 12/100 = 120mA

    If there is .8 volts missing, the internal resistance should be: .8/120mA????
     
  2. jcsd
  3. Aug 27, 2007 #2

    berkeman

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    Staff: Mentor

    When they say "real voltage source", that means that you should not ignore the built-in output resistance of the source. So the total current will not be 12V/100 Ohms.... you need to include the Rout in that equation as well....
     
  4. Aug 27, 2007 #3
    I need to find the internal resistance before I can include it in the equation right? After doing .8/120mA=6.66 Ohms. The new total current could be 12V/(100 + 6.66)
     
  5. Aug 27, 2007 #4

    berkeman

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    Staff: Mentor

    I'm not sure if that's right, but try it this way -- use the voltage divider equation, since you know the voltage source is 12V, and the divided down voltage across Rout and 100 Ohms is 11.2V.....
     
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