RLC (Impedance) Series/Parallel Question

Discussion in 'Homework Help' started by johncena, Dec 6, 2010.

  1. johncena

    Thread Starter Member

    Aug 15, 2010
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    0
    Here's the prob.

    [​IMG]



    My progress so far is:


    i converted V(t) to: v(t) = 120cos(377t -30°) just minues 90° to make it cos

    then frequency is
    F = 377/ 2∏
    F = 60
    XL = 2.pi(60)(40mH)
    XL = j15.08 or (0+j15.08)

    XC = - [1/(2∏(60)(50uF))]
    XC = -j53.05 or 0-j53.05

    then add XL and R
    XLR=20 + j15.08

    so this is the part i am confused...
    We need to parallel XLR and XC so the equation is this

    ZT = (XLR)(XC) / (XLR + XC)


    is this correct? or i need to convert it to polar form? but i get infinity at :
    Φ = tan^-1 (53.05 / 0 )

    How to solve this?
    i can solve i(t) by the formula i(t) = V/ZT
    but i can't solve ZT.
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Eiter method, polar or cartesia representation must yield the same result. I would say, use cartesian form to find the sum on the denominator and polar to find the product on the nominator. Then polar again to find the ratio.

    The conversions can be done very easilly with an adequate pocket calculator.
     
  3. johncena

    Thread Starter Member

    Aug 15, 2010
    32
    0
    is this correct?
    XC = - [1/(2∏(60)(50uF))]
    XC = -j53.05 or 0-j53.05

    is it negative? because the is formula positive to some books/websites.
    and.. if it is -j53.05, the polar form will be (-53.05<-90°)???
    (Note: < means right angle)
     
  4. johncena

    Thread Starter Member

    Aug 15, 2010
    32
    0
    Is Xc is negative based on the formula? because the formula is positive to some books/websites.

    Our given formula was Xc = -[ 1/(2∏fC)]


    And..

    If i converted -j53.05 to polar form.. is this correct? -> -53.05<-90°, or not?
    (note: < means right angle)
     
  5. johncena

    Thread Starter Member

    Aug 15, 2010
    32
    0
    Modified the Image: It was Dc not AC >.<

    [​IMG]


    Is Xc is negative based on the formula? because the formula is positive to some books/websites.

    Our given formula was Xc = -[ 1/(2∏fC)]


    And..

    If i converted -j53.05 to polar form.. is this correct? -> -53.05<-90°, or not?
    (note: < means right angle)
     
    Last edited: Dec 6, 2010
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Actually it is X_c=\frac{1}{j \omega C}=-j\frac{1}{\omega C}=\frac{1}{\omega C}\angle -90
     
    johncena likes this.
  7. johncena

    Thread Starter Member

    Aug 15, 2010
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  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I noticed two posts in the background.
    I made them visible for you.

    Bertus
     
  9. Georacer

    Moderator

    Nov 25, 2009
    5,142
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    -j53.05 is correct, but it is equal to 53,05\angle-90, not -53,05\angle-90.
     
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  10. johncena

    Thread Starter Member

    Aug 15, 2010
    32
    0
    ah now i see... thanks

    PS: If by any chance i have a value of XC whiich is postive... ex. +j40.. the polar form will always with -90°? like this.. --> 40<-90°
     
  11. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    V(t) is a time-varying system, so this is still steady-state AC. The Symbol for the voltage source was fine before.
     
  12. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Capacitors are always leading, so their angle is always negative. But I 'll generalize the topic for any imaginary number.

    Let's suppose we have an imaginary number X=A+jB, A and B are either positive or negative. This is a cartesian representation in a rectangular grid.
    If we wanted to turn this number into polar coordinates we would write X=M\angle \theta where M=\sqrt{A^2+B^2} and \theta=\arctan\left( \frac{B}{A} \right),\ -90^o\leq \theta \leq 90^o.
    It is also equivalent to write X=-M\angle -\theta but the convention is to have the M positive.
     
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  13. johncena

    Thread Starter Member

    Aug 15, 2010
    32
    0
    thanks to all!

    my problem is solved, my answer is correct :D
     
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