Series RLC circuit

Discussion in 'Homework Help' started by jetpac, May 3, 2010.

  1. jetpac

    Thread Starter New Member

    Feb 5, 2010
    7
    1
    Hello all, I'm having some difficulty doing this problem. I feel like I'm nearly there so if anyone has a minute to help me out with the last few steps I'd really appreciate it.

    I know the formatting's not great - if requested I will have another go at formatting it a bit better..


    =========================

    Question:

    In a series RLC circuit the value of the inductance is 0.02H
    Find the values of the resistance and capacitance where the voltage and current are given by:

    v(t) = 353.5 cos(3000t -10) V
    i(t) = 12.5 cos(3000t -55) A


    =========================


    Cos(x) = Sin(x + 90)

    => v(t) = 353.5 sin(3000t, -10 + 90)
    = 353.5 sin(3000t, +80)

    & i(t) = 12.5 sin(3000t, -55 + 90)
    = 12.5 sin(3000t, +35)

    =>
    ω = 3000
    => Xl = 3000 * .02 = 60 Ω
    => Zl = 60 ∠ 90

    Vpeak = 353.5
    => V = 353.5 * .707 = 249.9 ∠ 80
    = 249.9 * cos(80) + j 249.9 * sin(80)
    = 43.4 + j246.1

    Ipeak = 12.5
    => I = 12.5 * .707 = 8.8 ∠ 35
    = 12.5 * cos(35) + j 12.5 * sin(35)
    = 7.2 + j 5.0


    Ir = Il = Ic = Itotal = (7.2 + j 5.0) OR (8.8 ∠ 35)

    Ztotal = Etotal / Itotal
    = (249.9 ∠ 80) / (8.8 ∠ 35)
    = (249.9/8.8) ∠ (80 - 35)
    = 28.4 ∠ 45
    = 28.4 * cos(45) + j 28.4 * sin(45)
    = 20.1 + j 20.1

    El = Il * Zl
    = (8.8 ∠ 35) * (60 ∠ 90)
    = (8.8 * 60) ∠ (35 + 90)
    = 528 ∠ 125
    = -302 + j 432.5

    So I now have values for El, Etotal, Ir, Il, Ic, Itotal, Zl, Ztotal - basically everything except Er, Ec, Zr and Zc. I'm just not sure how to go about the last portion of splitting the impedance between the inductor, capacitor and resistor.

    Also - would anyone like to hazard a guess as to what percentage of the marks getting this far would get me in an exam situation?
     
    Last edited: May 3, 2010
  2. kingdano

    Member

    Apr 14, 2010
    377
    19
    heres a hint -

    if i read this right you were GIVEN the inductance value - and you know that inductance and capactiance account for all of the imaginary component of the impedance.

    therefore resistance is the "real" part of the impedance.

    ...does that help you at all?
     
  3. jetpac

    Thread Starter New Member

    Feb 5, 2010
    7
    1
    Thanks for the response!!

    Would this look something like the right answer then?

    Ztotal = 20.1 + j 20.1 (from above)
    Xl = 60Ω (from above)
    =>
    R = 20.1Ω
    Xc = 60 - 20.1 = 39.9Ω


    Also - could you just verify that I'm handling the cosine functions correctly above (ie. adding 90 to the phase angle and dealing with as a sine)?

    Thanks!
     
  4. kingdano

    Member

    Apr 14, 2010
    377
    19
    to be honest i havent done that stuff in about 6 or 7 years - so i wont be much help checking your work - i only remember the basic ideas and theories.

    heres a bump though.
     
  5. jetpac

    Thread Starter New Member

    Feb 5, 2010
    7
    1
    OK thanks anyway, useful tip - solved the problem I think!
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You haven't quite finished - you've found Xc but not C.
     
  7. jetpac

    Thread Starter New Member

    Feb 5, 2010
    7
    1
    Ah good point.

    Xc = 1/ωC

    => 39.9 = 1/3000 * C
    => C = 1/3000 * 39.9
    => C = 8.35421888 μF

    Cheers!
     
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