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Curry

  • one year ago

Question regarding circuits.

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  1. Curry
    • one year ago
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  2. Curry
    • one year ago
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    the answer is 4/3 + 27.7 cos(20t - 56deg). I was able to get the second part of the answer using super position. However, i don't know where the 4/3 is coming from.

  3. freckles
    • one year ago
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    what differential equation did you solve? and how do you tell the initial condition? if you don't know I'm still browsing the internet

  4. Curry
    • one year ago
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    well, using super position, I first turned off the 2v voltage source. and then solved for v(t). then i turned off the current source and then solved for v(t) again. Then added these 2 v(t). That gives the second part of the answer.

  5. Curry
    • one year ago
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    and I didn't use any differential equation, i used phasors.

  6. freckles
    • one year ago
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    oh looking up that term :p

  7. freckles
    • one year ago
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    I'm not sure if @ganeshie8 knows about circuits but I will call him here just in case

  8. Curry
    • one year ago
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    |dw:1433569338362:dw|

  9. Curry
    • one year ago
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    so solving that for current gives 2v([(10+10j)+(20||-5j)] x (20||-5j) = V1.

  10. Curry
    • one year ago
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    |dw:1433569640632:dw|

  11. Curry
    • one year ago
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    Solving that gives V2. V1 + V2 gives the 27.7cos(20t - 56). But i don't get where the 4/3 comes frmo.

  12. Curry
    • one year ago
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    @dan815 @satellite73 @ganeshie8 @perl

  13. freckles
    • one year ago
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    http://www.labsanywhere.net/circuit/lectures/lect10/lecture10.php this seems to be similar to what you are doing

  14. freckles
    • one year ago
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    except they solved for i

  15. Curry
    • one year ago
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    well that circuit is different because it's both AC. My problem is AC and DC. :(

  16. Curry
    • one year ago
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    well that circuit is different because it's both AC. My problem is AC and DC. :(

  17. Curry
    • one year ago
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    @perl

  18. Curry
    • one year ago
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    @robtobey @UnkleRhaukus

  19. Curry
    • one year ago
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    @Adi3

  20. anonymous
    • one year ago
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    CURRY plz come back and help me

  21. IrishBoy123
    • one year ago
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    2 * 20/30 = 4/3 ie not sure what you are doing but the 2V source surely only goes through the resistors total impedance to AC is \(5.55 e^{-56^0j}\) giving the right hand part of the desired solution.

  22. Curry
    • one year ago
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    Why doesn't the 2v go through the inductor and capacitor? I've done a lot of problems with DC circuits and Capacitors and inductors.

  23. Curry
    • one year ago
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    And calculating the second part of the equation requires both the AC and DC value. When doing it through super position ofc.

  24. Curry
    • one year ago
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    btw, what major are you Irish boy?

  25. IrishBoy123
    • one year ago
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    direct current goes through a capacitor? in a dc-C circuit you will get that initial flow as the cap charges up but that's it. it "will" go through the inductor but what is the inductor's resistance? so the dc to superimpose is flowing through a 10\( \Omega\) and 20\( \Omega\) resistor and you're measuring the drop on the latter. the AC is a RL/R/C all in parallel. btw, if he's around, this is "the" chap to tag on this type of stuff: @radar

  26. Curry
    • one year ago
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    So, when i have a DC and an AC, what should my appraoch be? I usually start with super position. I calculate the voltage drop from the DC and then the AC. and then i add them up. So for this problem, i did that and i got the correct answer for the latter of the equation. but the initial part i couldn't. So generally, is it just, ignore the AC, and short inductors. and open capacitors?

  27. Curry
    • one year ago
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    @IrishBoy123

  28. Curry
    • one year ago
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    @IrishBoy123 Hey can you clarify whenever you can?

  29. Curry
    • one year ago
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    wait nvm. I got it. Thanks so much. you saved my life.

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