Superposition thinker!

Discussion in 'Homework Help' started by davidrichardson04, Feb 22, 2008.

  1. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
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    Dear all,

    Please could I have assistance with the question herein

    I have read the topics on the site but find it hard to relate this to the question posed

    any help would be greatly received as I am stuck on this full stop.

    regards

    PS specifcally the relation of polar/j numbers with the voltage phase

    ta
     
  2. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    A + jB= sqrt( A^2 + B^2)angle(tan inverse(B/A) ).
    For superposition just solve as done for DC but with complex numbers.
     
  3. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
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    many thanks

    how do i deal with the j numbers being they only have the imaginary parts to them (see pic)
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    0 + jX (if only jX is given).
    Here magnitude is j(since sqrt( 0^2 + j^2) = j ) & angle is 90 degrees.,if it was -jX then angle wud be -90.
    The procedure for finding angle is again.tan inverse of (Imaginary/real).
    Then carry out all calculations as done while solving a DC circuit. It is advisable that you learn more about how multiplications are carried out in rectangular form or in polar form..
    angles don't multiply or divide.while multiplication or division is carried out.
    see If this (http://books.google.com/books?id=mO...7ue&sig=HMvCxCMfh51w98T1MqtEITQ6efs#PPA431,M1) helps you with that.
     
  5. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
    9
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    i will read on

    for this problem i got about 4A through coil

    Is that correct do you think?
    I can illustrate the working out but can you go through question fully, as an example

    many thanks - its been a while since I studied this
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    there is nothing much to it.

    first kill any voltage source by shorting it and find equivalent impedances of the impedances in parallel, which will be either 2<0*6<90/(2<0 + 6<90)..or..8<-90*6<90/(8<-90+6<90)..depending on the source killed.Then find total current by the only source which wasn't killed.
    For this find total impedance of circuit...lets say the source with zero phase angle was killed. Then the net impedance is 2<0*6<90/(2<0 + 6<90) + 8<-90
    and total current is given by voltage/impedance.
    i;e 40<30/{2<0*6<90/(2<0 + 6<90) + 8<-90}.

    Again, let me remind you it is required to know how mathematical operations are performed in polar or rectangular form...you can simply use rectangular form and do the calculation as in case of complex numbers...this might help you..http://forum.allaboutcircuits.com/showthread.php?t=3168...
    and read this too.. http://en.wikipedia.org/wiki/Electrical_impedance.

    Then use current divider rule to find the current in each branches in the original circuit caused by this source alone....http://www.wisc-online.com/objects/index.asp?objID=DCE3502
    similarly for other source.

    Add voltages across elements( remember all operations are done considering phase angles...we can also consider all voltages as vectors and add as vectors) and we have the equivalent voltage across that element due to the two sources & since vectors are being added we don't need to concern ourselves with sign or polarity ..simple phasor addition will take care of it.
     
  7. mrally

    New Member

    Mar 10, 2008
    1
    0
    yeah recca02 is on with this one

    recca please can you lay this whole question out for me too?!

    I've been reading this but cant quite nail it

    I've shorted the circuit to eliminate the voltage and summed up resistances.

    Do you have an answer

    please will you lay this one out in full from start to finish

    It would really help me with a similar one

    If not can anyone else - daverich04? Please im deseperate to have this one nailed and cant look my kid in the eye at mo!!

    many thanks alll
     
  8. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
    9
    0
    yea im still working on it but yea could do with a full run through of one to answer like you

    btw anyone who thinks they can work this one out in full is on for one!!! PS. also there is a capacitor in line with the 0angle voltage phase at 40 on the right bridge

    go for it anyone (Im waiting stuck like you mate)
     
  9. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Here is another thread with a similar requirement to your. The author has not solved it completely but he has made a valiant effort.

    See if it helps you get a bit further on your problem.

    hgmjr
     
  10. recca02

    Senior Member

    Apr 2, 2007
    1,211
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  11. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Greetings DavidRichardson04,

    Did you ever solve the problem using superposition?

    Rather than solve the problem for you using superposition (your homework assignment), I have provided a solution based on Millman's theorem to supply you an answer to the problem as a point of reference.

    hgmjr
     
  12. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
    9
    0
    can anyone produce a complete answer to this. It is not a homework question but I have a similar one. I cant find a txt book with one in.

    please just the whole question worked through (perhaps in a jpeg) I will post mineas soon as i find out how to

    the question is above
     
  13. davidrichardson04

    Thread Starter New Member

    Feb 21, 2008
    9
    0
    [​IMG]
    [​IMG]

    Here is the original question and how far ive got

    can anyone correct me and help finish?

    thanks
     
  14. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    The above expression is for the equivalent impedance of the circuit when 40<30V is considered alone and the other source is shorted.

    Now for this source find total current sent into the circuit by V/Z.

    The using either current divider rule or by calculating voltage across the coil (V- IR) find the current in coil.

    Similarly for the other source.
     
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