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Curry

  • one year ago

Question with circuits

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  1. Curry
    • one year ago
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  2. Curry
    • one year ago
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    @IrishBoy123

  3. Curry
    • one year ago
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    @jim_thompson5910

  4. Curry
    • one year ago
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    @amistre64

  5. Curry
    • one year ago
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    @campbell_st

  6. campbell_st
    • one year ago
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    sorry not good with circuits

  7. Curry
    • one year ago
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    @radar

  8. Curry
    • one year ago
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    @inkyvoyd

  9. perl
    • one year ago
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    @Michele_Laino

  10. sidsiddhartha
    • one year ago
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    ok u should try first calculate V_c(0-) then V(infity ) right?

  11. sidsiddhartha
    • one year ago
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    at t<0 switch was open|dw:1433646718816:dw|

  12. sidsiddhartha
    • one year ago
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    and at t<0 capacitor will act as an open circuit

  13. sidsiddhartha
    • one year ago
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    so use nodal at point A|dw:1433646864478:dw| makes sense?

  14. sidsiddhartha
    • one year ago
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    that gives Vc(0)=25/3 now at steady state the capacitor will act as an open circuit again \[V(\infty)=0\\ \Large v(t)=v(\infty)+[v(0)-v(\infty)]*e^{\frac{- t }{ R_{th}C }}\] is that making any sense?

  15. radar
    • one year ago
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    It appears that the load real value is 30 Ohms with a parallel " Z matching capacitor" which we are to determine the value of capacitance. In a previous post, I mentioned that the internal impedance (10 + j20) is equivalent to an impedance of 22.36 Ohm at 63 degrees inductive. The capacitor needs to be a value such that combined in parallel with the load results in a value of 22.36 Ohms at 63 degrees capacitive.|dw:1433646405417:dw|This results in a capacitance reactance of 82.5 Ohms ( I realize that i did the math as if both of the values were real creating an error) Now calculate the value of the capacitor|dw:1433646874269:dw|

  16. Curry
    • one year ago
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    Well when I calculate the voltage across teh capacitor, i get the right answer. But, when I calculate the voltage across the resistor, i'm not able to do it.

  17. radar
    • one year ago
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    Sorry @Curry I was on a previous problem lol.

  18. Curry
    • one year ago
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    I understand, I'm looking at your posts in relevance to my previous post. Let me read, one second.

  19. Curry
    • one year ago
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    @radar but that still doesn't give us a value of 0.8mF for the capacitor. @sidsiddhartha did you see me previous post?

  20. Curry
    • one year ago
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    @sidsiddhartha , the V(0) isn't 25/3 actually. It is 5. Because if the siwtch is open, then there is no current going through the 40ohm resistor, and therefore it is just 2 resistors acting on that circuit.

  21. radar
    • one year ago
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    You could "theviens" theorem (if you have studied it (spelling may be wrong)). The circuit would then look like this|dw:1433648004142:dw|

  22. radar
    • one year ago
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    |dw:1433648176592:dw|

  23. Curry
    • one year ago
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    well ofc i've studied thevenin's and Norton's. But um, let's see. there seems to be a small flaw.

  24. Curry
    • one year ago
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    So when siwtch is open, the initial voltage is 5v.

  25. Curry
    • one year ago
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    and then, the switch is closed, so (5v/Rt)[40 + Zc),

  26. Curry
    • one year ago
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    but we can't do that, because we can't calculate inpedance, so we just calculate voltage of capacitor. ok. so then,

  27. radar
    • one year ago
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    |dw:1433648508210:dw|

  28. Curry
    • one year ago
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    the Vc = 5 + (0 - 5)e^(-t/(48*0.1)

  29. radar
    • one year ago
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    After 240.5 seconds (5 time constants 5 X 4.8) it would build up in an expotential manner to 5 volts.

  30. Curry
    • one year ago
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    yes, i too got 5. But the second part of the equation is -5/6e^(-t/RtC)

  31. Curry
    • one year ago
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    the -5/6 is the problem i'm having. Ofc, solving the voltage across the capacitor gives me taht value. but they're asking for voltage across resistor...

  32. radar
    • one year ago
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    Yes, the voltage across the resistor would vary with t until the capacitor becomes fully charged, which would occur practically in 5 RC seconds. Calculate the current and V(t) would equal 5V - 8I(t)

  33. radar
    • one year ago
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    |dw:1433649839635:dw|

  34. Curry
    • one year ago
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    So did you use the formula I(t) = cdv/dt?

  35. radar
    • one year ago
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    No, I used the formulae:\[i _{c}(t)=(E/R)e ^{-t/\tau}\]

  36. Curry
    • one year ago
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    oh ok ok, i just checked it with the formula i posted, and it works too. so for future reference, :)

  37. Curry
    • one year ago
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    and why are we subtracting 8i?

  38. radar
    • one year ago
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    Refer to the "thevenized" circuit and note that the first two resistors (40 and 10 Ohm) become parallel in the thenize process and become 8 Ohm . See the 12th post above this one for the drawing.

  39. Curry
    • one year ago
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    sorry, i meant, why are we subtracting?

  40. radar
    • one year ago
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    The voltage drop across that resistor is subtracted from the thevinized 5 volt source that gives you the V(t).

  41. Curry
    • one year ago
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    Hmmm, i figured. But don't we have to account for the inital value for 5v also?

  42. radar
    • one year ago
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    We do, and when t=0 (switch first closed) you will see that the equation will give you a value of 5 - 40/48 = 4.167 whic is v(t) when t = 0

  43. Curry
    • one year ago
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    so initial value is 5v. and then as t fgoes to inifinty it's 5 - 5/6e^(something). And then we should be adding these values right?

  44. Curry
    • one year ago
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    Wait actually, shuoldn't the inital value be 5v, not 4.167?

  45. Curry
    • one year ago
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    in that circuit, when switch is open, the current going trhough that 40 ohm resistor is 0. which means there is no voltage trhough that resistor.

  46. Curry
    • one year ago
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    so we can just avoid that 40ohm R. And so the inital voltage would be (25/40)(10)

  47. Curry
    • one year ago
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    i mean, 25/50 * 10

  48. radar
    • one year ago
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    With the switch open v(t)= 5 volts yes......but at t=0, the uncharged capacitor is connected to the circuit by the switch closure and now v(t) is 4.167 volts and increasing with t. As t increases to 5 RC (5 time constants) and more the capacitor becomes fully charged for all practical purposes and it will be fully charged and current will cease thru the rightmost 40 ohm resistor and v(t) will be back to 5 volts.

  49. radar
    • one year ago
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    Or as you say back to (25/40) (10) which is 5 volts.

  50. radar
    • one year ago
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    |dw:1433651558519:dw|Kind of like this.

  51. radar
    • one year ago
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    I am still not totally satisfied with the previous problem regarding maximum power transfer - impedance matching. But I have to call it a day and retire for the night.

  52. Curry
    • one year ago
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    Yes, that problem is a stickler. I'll figure oout how to solve it, as i'm meeting with my friends. I'll be sure to let you know tomorrow!

  53. radar
    • one year ago
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    Yes please do, and I believe you will get it, I am 76 yrs old and gotten rusty with these things, but you stick with it and you will overcome. Good luck in your studies. good night.

  54. sidsiddhartha
    • one year ago
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    yeah sorry i made a mistake took that 10 ohm as 40 yeah v(0)=5

  55. sidsiddhartha
    • one year ago
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    i guess the problem is solved then, my internet connection got stuck sorry :(

  56. Curry
    • one year ago
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    So, if that capacitor was replaced with a inductor, would I treat the intial as if there's an open circuit there?

  57. radar
    • one year ago
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    Yes, an inductor would behave as initially as an open at switch closure, however the current would increase to a value as if the opposition was only the DC resistance of the inductor (wire resistance). In other words inductor oppose any change in current. The change from 0 current (switch open) to the initial attempt of current condition (switch closed) is maximum as that is the greatest change (di/dt). Hard to explain......as they say "it gets complicated" lol.

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