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Finding unknown Phasor in a series-parallel circuit

  1. Apr 30, 2017 #1
    • Thread moved from the technical forums, so no Homework Template is shown
    So this is a "bonus" question from a past final exam for my Electrical Circuits course. I've spent a good few hours trying to solve it but haven't really had much luck. Any method can be used I figured if I found the voltage of the known XL, it would be equivalent to the voltage of the unknown ZL (using thevenin) but then I'm missing the current, in which if I used the general form, I don't have the resistance for ZL. which leaves finding the Zt from the power and then finding the ZL by back tracking it but the unknown resistance is parallel with the XL so I'm finding it impossible to do it algebraically.
    Also tried to use the power (i'm guessing apparent power since it's mentioned in VA) to find the I conjucate, and I, and Zt for the circuit but still, no luck.

    Attached is the question I have in mind. I still have 10~ days left for the final but I rather know how to solve it as soon as possible and get it out of the way.
    Any help would be appreciated!
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2017 #2

    Charles Link

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    Suggestion is to characterize the inductor and 15 ohm resistor as a parallel combination: ## 1/Z=1/(15 \Omega)+1/(j \omega L) ## . The problem is ultimately to find ## \omega L ##, but work through the algebra. It should be straightforward. Meanwhile ## X_c=25 \, \Omega ## ==>> ## Z_c=-j/(\omega C) ##, so that ## Z_c=-j \, 25 \, \Omega ## is a more accurate description.
     
  4. Apr 30, 2017 #3

    tech99

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    It looks as if 15 Ohms refers to an inductor and ZL means an unknown load impedance.
     
  5. Apr 30, 2017 #4

    Charles Link

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    Thank you. Include a "j" on the ## 15 \Omega ## above.
     
  6. May 1, 2017 #5
    Solving it with ohms means doing it in polar form in which inductors are Resistance/angle 90 and capacitors are resistance/angle -90 where omega isn't needed as resistances are given for such.
     
  7. May 1, 2017 #6

    tech99

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    Apologies if I am in error with this vast working! There must be an easier way. Where there are combinations of series and parallel, you need to convert the X and R values between series and parallel equivalents as you work along the network. It is very hard to keep a clear head.
    1) Use a phasor diagram to start with. It looks as if the generator sees an impedance in the First Quadrant, which is inductive. The generator gives 30 volts at 1 Amp at an angle of 68 degrees, inductive, lagging.
    The total circuit has a shunt X drawing 1 x sin 68 = 0.93A. However, the 25 Ohm capacitor draws 1.2A, so the rest of the circuit draws 1.2 - 0.93 = 0.27A inductive, which is 30/0.27 = 111 Ohms inductive.
    The total circuit has an equiv shunt R drawing 1 x cos 68 = 0.37A, a shunt R of 30/0.37 = 81 Ohms. Allowing for the two resistors, the inductor and ZL must together present an equiv series (AC) resistance of 81 - 10 - 20 = 51 Ohms.
    2) To recap, looking into the circuit to the right of the resistors, we see a resistance of 51 Ohms in series with an inductance of 111 Ohms. Now we must find the equivalent shunt values.
    (i) Find Q = 111/51 = 2.18
    (ii) Rp = (Q^2 + 1 ) x Rs = (4.73 +1) x 51 = 292 Ohms (this is the shunt equiv resistance of the load)
    (iii) Xp = Rs / Q = 134 Ohms inductive (this is the reactance of the load in parallel with the 15 Ohm inductor).
    3) Find shunt reactance of ZL alone (not with 15 Ohm inductor).
    The reactance of ZL in parallel with the 15 Ohm inductor gives an inductive reactance of 134 Ohms. So the reactance of ZL is capacitive.
    1/134 = 1/15 - 1/Xload
    Xp load = 16.9 Ohms capacitive
    4) The impedance of the load is therefore 292 Ohms in parallel with a capacitive reactance of 16.9 Ohms.
    Convert to series form.
    (i) Find Q = 292/16.9 = 17.3
    (ii) Rs = Rp /(Q^2 + 1) = 292 / (17.3^2 +1) = 0.97 Ohm
    (iii) Xs = Rs x Q = 0.97 x 17.3 = 16.8 Ohms
    Answer: ZL = 0.97 - j16.8
     
  8. May 1, 2017 #7

    jim hardy

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    Old Fashioned Approach , see if i missed something:

    others have probably solved it already i just had to look....


    Question6.jpg
    Ohm's Law:
    I1 = 30∠0 volts / 25∠-90 ohms = 1.25 ∠+90 amps

    I2 + I3 = (1∠68 - 1.25∠+90) amps , change into rectangular and calculate

    KVL
    volts across XL and ZL = (30∠0volts - (I2 + I3)amps X 30∠0 ohms) volts
    so I2 is (that voltage / 15∠+90 ohms) amps

    and I3 is readily found by KCL

    back to Ohm
    ZL = ( (30∠0volts - (I2 + I3) amps X 30∠0 ohms ) volts / I3 amps ) ohms
    ..............................................................^^^^ that's R1 + R2..........

    lots of polar ↔ rectangular but that'd be fun.


    Check my thinking ?

    old jim
     
  9. May 2, 2017 #8
    Not sure I follow with either responses, as the VA (S phasor) gives the current conjugate, in which the angle for the actual current is the opposite? I tried finding ZL using thevenin (Zth) circuit and then using the S (VA) to get the total impedance from the current (S= I^2non phasor * Z phasor) and then reduced the Zt-Zth in the thevinin circuit (series) to get ZL.

    where Is = 1∠-68
    Ztotal = 30∠136 (30∠68/1∠-68)
    Zth = 26.52∠-45 from ((10∠0+15∠90+20∠0)//25∠-90)

    therefore, series circuit from thevenin
    ZL= 30∠136-26.52∠-45 = 56.517∠135.5 ohms?


    you can use your calculator to subtract and add complex numbers (impedances) in polar form as they are, with the ∠. Which is what we were taught in my university. Makes it much easier and faster to solve these problems.
     
  10. May 2, 2017 #9
    Another calculation way is to consider parallel impedances.

    If Ztot=Vs^2/S* total impedance seen from Vs[source]

    [Since S=Vs*I*[I conjugate] then I=S*/Vs and Ztot=Vs/I=Vs^2/S*[conjugate S].

    Ztot=30^2/(30*(cos(68)-sin(68)i))=11.238+27.81i

    On the other hand Ztot=Zeq*Xc/(Zeq+Xc) where Zeq=ZL*XL/(ZL+XL)+R1+R2

    From Ztot equation Zeq=Ztot*Xc/(Xc-Ztot)

    Zeq=(11.238+27.81i)*(-25i)/(-25i-11.238+27.81i)=2.409+13.68i

    From Zeq equation [Zeq=ZL*XL/(ZL+XL)+R1+R2] :

    ZL=[(R1+R2)*XL-Zeq*XL]/[(Zeq-XL-(R1+R2)]

    ZL=[(10+20)*15i-(2.409+13.68i)*15i]/[(2.409+13.68i-15i-(10+20)]

    ZL=-8.137-14.61i
     
  11. May 2, 2017 #10

    jim hardy

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    Why do you say it gives conjugate ?
    VA = V X A , so A = VA / V , VA has angle +68 and and V has angle zero.
     
  12. May 2, 2017 #11
    I think a good explanation it is in
    IEEE Std 1459-2010 IEEE Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions CH. 3.1.1.6 Complex power (VA):
    The complex power is a complex quantity in which the active power is the real part and the reactive power is the imaginary part S=P+jQ=VI* where:
    V =V< 0 is the voltage phasor; I=I<-θ is the current phasor ; I*=I<θ is the complex conjugate of the current phasor.
    That means since usually the current is inductive [lags the voltage θ<0] the reactive power has to be positive.
     
  13. May 2, 2017 #12

    tech99

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    [


    ZL=-8.137-14.61i[/QUOTE]
    How can the resistance be negative?
     
  14. May 2, 2017 #13

    jim hardy

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    Hmmm so it's a definition?

    I learned to work phasor notation in 1962 , in high school electronic circuits class not power. To us boys an angle was positive unless it had a negative sign. So i've probably worked power problems wrong ever since . Makes sense though to define lagging reactive power as positive.

    Thanks @Babadag , if i learn something every day i should some day become learned.

    old jim
     
    Last edited: May 3, 2017
  15. May 3, 2017 #14

    jim hardy

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    This picture made sense to me before Babadaq made me aware of the IEEE definition that swaps sign of θ

    Question6.jpg
    With that big capacitor hanging right across the supply an angle of +68 (leading) made intuitive sense

    Instead, my current I1 +I2 +I3 should be 1∠-68 , lagging

    Now, 1∠-68 in rectangular is 0.37 - j0.93

    and current I1 through the 25 Ω capacitor should be V/Xc= 30/-j25 = +j1.25
    Subtracting I1 of +j1.25 from sum (I1 + I2 + I3) yields (0.37 - j0.93) -j1.25 = 0.37 - j 2.18 = 2.21 ∠-80.37

    meaning 2.21 amps at angle -80.37 must flow through the 30 ohm sum of R1 and R2.
    That would produce 66.3 volts at angle -80.37, 22.1 of them across R1 and the other 44.2 across R2..

    So let's assign some polarity signs and write KVL:

    Question6Rev1.jpg

    I'll start in lower left corner and walk clockwise :

    -Vs + (I2 + I3) X R1 +(Volts across paralleled Zl and j15Ω) + (I2 + I3) X R2 = 0


    (Volts across paralleled ZL and j15Ω) = Vs - (I2 + I3) X (R1 + R2)

    (Volts across paralleled ZL and j15) = 30∠0 - (66.3∠-80.37) = 30 - ( 11.09 -j65.37) = 18.91 +j65.37 = 68.05∠+73.9

    Now we can calculate I2. It's (Volts across paralleled ZLand j15Ω) / j15Ω = 68.05∠+73.9 / 15∠+90 = 4.537 ∠-16.1 = 4.36 -j1/26

    hmmmm I2 = 4.36 -j1.26

    Now we can calculate I3
    I3 = Itotal - I1 - I2
    I3 = (0.37 -j0.93) - j1.25 - (4.36 -j1.26) = -3.99 -j0.92 = 4.09∠-167

    so ZL must be Volts across itself divided by current through itself I3
    which is (68.05∠+73.9) / ( 4.09∠-167) = 16.64∠240.9 = -8.09 -j14.54

    That's how i'd have worked this with a sliderule fifty five years ago.
    What ? OMG how can that be? Seems like yesterday .


    How can the resistance be negative?[/QUOTE]

    My real component differs from Babadaq's by roughly four parts in 811, ~½%
    and my imaginary by roughly seven parts in 1457, about ten times less oops make that also ~½% (i said i was pooped)
    @Babadag 's skill with math i revere
    so I am very pleased with the small difference that resulted from our diverse methods, especially considering how much rounding i did in polar ↔ rectangular conversions.
    ( i cheated and used an online calculator,
    http://www.intmath.com/complex-numbers/convert-polar-rectangular-interactive.php
    so my errors might be rounding or mis-transcription)

    If anybody sees a mistake in my arithmetic PLEASE flag it.

    Now -
    That the solution to the problem as given requires an active load(negative resistance) makes me question - was that author's intent?
    It's late and i'm pooped and i'm not going to work it again right now with assumption author didn't follow IEEE definition for VA and total current is leading not lagging.

    IF anybody has this one in a spreadsheet - a question---
    When you assume total load is leading not lagging , as it looks to me like it should be with a big capacitor separated from rest of load by resistors,
    ..........
    Do you get a passive load ie one whose real term is positive?

    old jim
     
    Last edited: May 3, 2017
  16. May 3, 2017 #15

    cnh1995

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    Yes I believe.

    Take input voltage V=1∠0° pu and Z=0.3-j0.4 pu (capacitive impedance). The current is 2∠53.13° i.e. 1.2+j1.6 pu, leading.
    Complex power=VI*=(1+j0)(1.2-j1.6)
    =1.2-j1.6 pu.
    This has positive real part and negative reactive part.
    I think in passive circuits, real power flows from source to load and gets dissipated there. So the real part must be positive.

    But the conjugate part doesn't only decide the sign of the reactive term. It decides the magnitude of the real term too.

    In the example above, if we shift the voltage phasor by 30° and take V=1∠30° pu and keep Z unchanged,
    we'll have current I= 2∠83.13° pu.
    Now, if we use S=VI instead of I*, we'll get this:
    S=(0.866+j0.5)(0.2392+j1.985) whose real part is -0.7853. So the real power changes with the change in the phase of the voltage phasor, which can't be right. Since the phase difference is constant, active and reactive power must be constant.

    And S=VI* gives this:
    S=(0.866+j0.5)(0.2392-j1.985)
    whose real part is +1.2, the exact amount of real power that the source is supplying.
     
  17. May 3, 2017 #16

    The Electrician

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    Starting from the impedance seen by Vs and working to the right, removing parallel and series components as we go, Zl is the final result:

    GetZl.png

    Perhaps the problem originator didn't use the IEEE convention for complex power as VI*; maybe they just used VI. An active load seems a little far-fetched for a problem like this.
     
    Last edited: May 3, 2017
  18. May 3, 2017 #17

    jim hardy

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  19. May 3, 2017 #18

    Yeah what you said, plus the student handbook we use says pretty much the same thing.
     
  20. May 3, 2017 #19

    jim hardy

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    So THAT's what the asterisk means .

    Thanks , guys !!!!!
    Well now - three diverse methods all agreed .within reason.

    I'd resolve with teacher whether he intended for you to use the comjugate current I* or I , since it's not mentioned in the Question 6 image .

    old jim
     
  21. May 4, 2017 #20
    oh in the content we use per the lectures, we use the conjugate so it's a definite yes. though everyone is giving a solution with a different answer so it's getting confused lol
     
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