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Maximum Power in Circuit

  1. Feb 15, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the difference between maximum power in load and maximum power by load and circuit?

    3. The attempt at a solution

    I know max power in load is when Rth (Thevenin) = RL(load). And the equation to find power in load is (Vth^2)/(4*Rth). Not sure how to find max power by load and circuit, is it just P=I*V?
     
  2. jcsd
  3. Feb 15, 2014 #2

    gneill

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    Staff: Mentor

    Suppose you consider the simple case where there's a voltage supply with some internal resistance and a load resistance:

    attachment.php?attachmentid=66654&stc=1&d=1392497874.gif

    The only source of power is the voltage supply. But that power can be dissipated in two places.

    Can you determine an expression for:
    1) the power produced by the voltage supply
    2) the power dissipated by ("Delivered to") the load
    3) the power lost in the source

    Suppose that the source resistance and voltage are both of fixed value. What value of RL will draw the most power from the voltage supply? What value of RL will deliver the most power to the load?
     

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  4. Feb 15, 2014 #3
    Below
     
    Last edited: Feb 15, 2014
  5. Feb 15, 2014 #4
    1) P=I*(Rs+RL)
    2) P=I*RL
    3) P=I*Rs

    Since P=V^2/R, The lower the RL the higher the power from the voltage supply
    Rs=RL will deliver the most power to the load.
     
  6. Feb 15, 2014 #5

    gneill

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    Staff: Mentor

    You'd need to square the current in those expressions. ##P = I^2 R## ; ##P = V^2 / R## ; ##P = I V## .

    You might want to write them in terms of the given fixed values V and Rs.

    So, if the load RL heads toward zero the power in the (overall) circuit gets larger. Will this be larger or smaller than the maximum power deliverable to the load?
     
    Last edited: Feb 15, 2014
  7. Feb 15, 2014 #6

    Yeah forgot to square the current.
    The power should be larger than the maximum power deliverable to the load.
     
  8. Feb 15, 2014 #7

    gneill

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    Staff: Mentor

    Right. So in general, when there are unavoidable losses (such as the resistance associated with a power supply), the total power in a circuit will be larger than that delivered to the load. That's in a DC circuit.

    When you start looking at AC circuits, you'll find that there are other ways that power delivered to a load can differ from the power generated by the power supply.
     
  9. Feb 15, 2014 #8
    Thank you very much for your help! :)
     
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