thevenin and norton rlc circuits

Discussion in 'Homework Help' started by rents89, Feb 5, 2009.

  1. rents89

    Thread Starter Member

    Feb 5, 2009
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    any idea on how to solve these??


    cheers
     
  2. The Electrician

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    Did you read the sticky post? You must show you're own work as far as you've been able to get before anyone will help you.
     
  3. rents89

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    Feb 5, 2009
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    yeh sorry didnt read it, i have done some calculations, but i am unsure about the method.. was wonderin if anyone could point me in the right direction...? thanks alot
     
  4. The Electrician

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    Tell us the method you've been trying to use.
     
  5. mik3

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  6. rents89

    Thread Starter Member

    Feb 5, 2009
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    ok
    for q1:

    - tried to find thevenin's impedance (Z_Th)
    -then calculated V_Th
    -finally V_Th/(Z_Th - j40) to find the current in the capacitor.

    q2:

    -converted voltage source to current and got 0.375<0° A
    -rearranged the circuit and trying to find Z_Norton

    pretty stuck basically...
    cheers guys
     
  7. The Electrician

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    Start with the very first part. If you'll show your calculations for finding Zth, then we can show you where you went wrong.
     
  8. rents89

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    Feb 5, 2009
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    Z_th= ((50+j50)(30))/((50+j50)+(30))

    but im pretty sure its wrong
     
  9. rents89

    Thread Starter Member

    Feb 5, 2009
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    do i need to add anything else? cheers
     
  10. mik3

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    No, you are correct.
     
  11. The Electrician

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    So far, so good. Now show the calculation for Vth. Show the final result.

    Then show the final number you get for the current in the cap as a single complex number, but also show the steps to get that result.
     
  12. rents89

    Thread Starter Member

    Feb 5, 2009
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    really?? ok, so how to calculate V_Th?
    I got,

    ((50+j50)/(80+j70))*200<0°
     
  13. rents89

    Thread Starter Member

    Feb 5, 2009
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    sorry just saw the post,

    V_Th= 133.04< -141.70°

    I= ((V_Th/(Z_Th - j40))= 3.31<-88.49°

    =2.86+1.66i

    ?!?!?
     
  14. rents89

    Thread Starter Member

    Feb 5, 2009
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    helppppp meeee
     
  15. The Electrician

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    I think this should be Vth = ((50+j50)/(80+j50))*200<0°
     
  16. rents89

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    Feb 5, 2009
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    safe cheers, is the method for calculating the current in the cap correct?

    also any help on q2? cos i converted the voltage source to current, is that right?

    thanks alot man
     
  17. The Electrician

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    You have Vth wrong; I showed the mistake in the previous post.

    Vth = 149.91<12.995°

    Then I = ((Vth/(Zth-j40)) = ((149.91<12.995°/(22.486<12.995°-j40) =
    3.6346<70.907° = 1.1889 + j3.4346

    I think your only mistake was getting Vth wrong.

    Have you attempted the second problem yet?
     
  18. The Electrician

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    Once again, it's difficult to help if you don't show all your steps; that's the only way I can see if you made a mistake somewhere, and just where you made it.
     
  19. rents89

    Thread Starter Member

    Feb 5, 2009
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    yeh sorry once again, my attempt at converting the voltage source is on the diagram... am i on the right track?
     
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  20. mik3

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    Feb 4, 2008
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    Yes you are. ;)
     
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