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Thevenins problem

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data

    vuZohAF.png
    2. Relevant equations
    V=IR
    P=I^2R
    maximum power when load resistance = thevenin's resistance

    3. The attempt at a solution
    I have been able to work out the new load resistance but none of the other values correctly.
     
  2. jcsd
  3. May 2, 2015 #2

    Svein

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    Science Advisor

    1. Remove RL. Now the voltage at the junction between 20Ω and 30Ω is [itex]\frac{30\Omega}{(20+30)\Omega}\cdot 15V = 9V [/itex] with a source impedance of [itex] 20\Omega \left\lvert \right\rvert 30\Omega =\frac{20\cdot 30}{20 + 30}\Omega[/itex]...
     
  4. May 2, 2015 #3
    Thank you for your reply, also how do you calculate the other values
     
  5. May 2, 2015 #4

    Svein

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    Science Advisor

    I have given you the first steps. Now try the next: With RL still disconnected, find the voltage and the source impedance at point A.
     
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