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feezul_93

  • 2 years ago

Complex number in circuit

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  1. feezul_93
    • 2 years ago
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    How to find the thevenin for c-d terminal ?

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

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

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

  5. ghazi
    • 2 years ago
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    hold on let me solve

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

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

  8. feezul_93
    • 2 years ago
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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 ?

  9. ghazi
    • 2 years ago
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    i think you should be using superposition principle

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

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

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

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

  14. feezul_93
    • 2 years ago
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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

  15. ghazi
    • 2 years ago
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    well, Vth, you can use voltage division rule

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

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

  18. feezul_93
    • 2 years ago
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    |dw:1356753632924:dw|

  19. feezul_93
    • 2 years ago
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    oh, some text missing ><

  20. ghazi
    • 2 years ago
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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

  21. feezul_93
    • 2 years ago
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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 @.@

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

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

  24. feezul_93
    • 2 years ago
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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 ?

  25. ghazi
    • 2 years ago
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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

  26. ghazi
    • 2 years ago
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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

  27. feezul_93
    • 2 years ago
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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 :)

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

  29. ghazi
    • 2 years ago
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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 :)

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

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

  32. feezul_93
    • 2 years ago
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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 -_-"

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

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

  35. ghazi
    • 2 years ago
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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

  36. ghazi
    • 2 years ago
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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

  37. ghazi
    • 2 years ago
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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

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