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feezul_93

Complex number in circuit

  • one year ago
  • one year ago

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  1. feezul_93
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    How to find the thevenin for c-d terminal ?

    • one year ago
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  2. feezul_93
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    I mean, thevenin voltage :)

    • one year ago
  3. ghazi
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    |dw:1356752136345:dw| open the current source and short the voltage source you will get Rth

    • one year ago
  4. feezul_93
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    I did, but I didn't get the right answer :(

    • one year ago
  5. ghazi
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    hold on let me solve

    • one year ago
  6. feezul_93
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    I got 2.667 – j4 Ω for Rth, n the answer for Vth is 9.614∠56.31° V

    • one year ago
  7. ghazi
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    did you convert 2.667 -j4 into polar form , you've got right answer

    • one year ago
  8. feezul_93
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    i got the Rth, but Vth ... I get it wrong :( wondering where's my mistake, Im using mesh analysis to find current flowing through capacitor, n times the capacitor to get Vth ... am I wrong ?

    • one year ago
  9. ghazi
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    i think you should be using superposition principle

    • one year ago
  10. feezul_93
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    ehh ehh, isn't it suppose to be supermesh ? D: there's current source within the loop

    • one year ago
  11. ghazi
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    there will be nothing messy , give a try by that first eliminate current source and then eliminate voltage one

    • one year ago
  12. feezul_93
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    ouhh, okayy, but the concept Im using just now is right right ?

    • one year ago
  13. ghazi
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    yep, its right i am sure you'll get your thevnin voltage by that

    • one year ago
  14. feezul_93
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    but I didn't T^T nevermind, i try to use superposition now, thanks :) btw, did u get the Vth ? cause I got the feeling that the answer given is wrong, hahaha

    • one year ago
  15. ghazi
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    well, Vth, you can use voltage division rule

    • one year ago
  16. feezul_93
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    hold on, currently using superposition to solve it :)

    • one year ago
  17. ghazi
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    |dw:1356753391535:dw| calculate voltage drop at point a and b

    • one year ago
  18. feezul_93
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    |dw:1356753632924:dw|

    • one year ago
  19. feezul_93
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    oh, some text missing ><

    • one year ago
  20. ghazi
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    actually you are supposed to calculate thevenin voltage across the capacitor that is i guess your load here, so you have to make it open terminal to calculate the voltage drop at load

    • one year ago
  21. feezul_93
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    yeah yeahh, that's exactly what I did just now ... n base on your drawing ... we're using nodal analysis ? cause I prefer to find current through capacitor n simply times it, but I don't really know if my concept is right ... there are too many ways to solve this @.@

    • one year ago
  22. feezul_93
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    can I just ignore the terminal 1st, n solve it like a normal question ?

    • one year ago
  23. ghazi
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    you dont need to go through all that stuff , just use superposition and you are done

    • one year ago
  24. feezul_93
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    T^T, i get the same answer like my previous method ... I guess the answer given is wrong ... but still ... I can't clarify it if Im the only 1 who get this answer ... did you get the same answer as given ?

    • one year ago
  25. ghazi
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    actually i hate these complex calculation lol so i can write the method you can match it , that is voltage drop due to 20 V source and due to current source

    • one year ago
  26. ghazi
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    i'll upload the solution and if you are getting same answer then i guess your book has got some mistake, i am not sure but i guess its 4 angle 56 V

    • one year ago
  27. feezul_93
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    hahahahaha, yeaappp, same goes to me, and I wonder who create this stuff, what kind of brain he had -___- but i guess i understand the flow of answering this kind of question, ouhh really ? thanks a lot :)

    • one year ago
  28. feezul_93
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    D: u can guess the answer just by watching it ? awesome ! D:

    • one year ago
  29. ghazi
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    you are welcome :D and yes the answer is around what i said, but i have to check it again , only reason is i hate this complex calculation lol i'll have to use calculator , dont worry i'll do it :D :)

    • one year ago
  30. feezul_93
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    I don't get it, why we do we have to let V1 - V2 ...

    • one year ago
  31. ghazi
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    we are finding V1-V2 because just take off the capacitor and you are supposed to find the terminal voltage

    • one year ago
  32. feezul_93
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    soo, deducting v2 by v1 is equal too ... V at terminal ? D: I don't see relation between it, can't even imagine it -_-"

    • one year ago
  33. ghazi
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    |dw:1356762940423:dw| look here these are V1 and V2

    • one year ago
  34. feezul_93
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    yeap2, but then ... v1 isn't connected with v2 D:

    • one year ago
  35. ghazi
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    when we measure potential difference we measure the potential difference at two terminals, that is why we have eliminated capacitor it is not needed to be connected to measure the potential drop across those terminals

    • one year ago
  36. ghazi
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    if like you will join that two points by wire there will be zero potential difference which means both the points at same potential or short circuit

    • one year ago
  37. ghazi
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    just think of using multimeter for the measurement of potential difference across a resistor or any element , we place two terminals of multimeter at the two ends

    • one year ago
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