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Source Voltage and Max Power

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A practical voltage source has a voltage effieciency of 90% when connected to a load of 9 Ohms. When connected to a load of 4 Ohms, the load voltage was 160V. Determine the source voltage and the max power.

    Can anyone help me with this question?


    2. Relevant equations
    P = (I^2)*(R)


    3. The attempt at a solution
    I'm not getting the correct answer.
     
  2. jcsd
  3. Nov 19, 2007 #2

    berkeman

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    Staff: Mentor

    Show us a little more work. A "practical" or real voltage source has an internal resistance (call it Ri) in series with an "ideal" voltage source (that has an open-circuit voltage of Vo). Write the two equations for the two load sitations given, and see if that helps you to determine what Vo is.

    And once you know Vo and Ri, do you know how to choose the load resistor value to get the maximum power transfer to it?
     
  4. Nov 19, 2007 #3
    Hi, if there is a voltage effieciency of 90%: Voltage effieciency = R / R+r

    Does this mean the internal resistance is equal to 1 Ohm.
     
  5. Nov 19, 2007 #4

    berkeman

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    Staff: Mentor

    I honestly have no idea what "voltage efficiency" is. Is this your translation from another language? There is power efficiency, but efficiency does not pertain to voltage alone.

    Maybe they mean that 0.9 * Vo = Vload?
     
  6. Nov 19, 2007 #5
    Is 1 Ohm incorrect then? How else can internal resistance be calculated in this case? That's the only way I can figure out.
     
  7. Nov 19, 2007 #6

    berkeman

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    Staff: Mentor

    Well if they mean voltage divider instead of voltage efficiency, write the equation for a voltage divider to figure out what Ri is.
     
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