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Determining total impedance, Voltage V2 and current IL of circuit

  1. Apr 28, 2014 #1
    I am asked to find the total impendance ZT, voltage V2 and IL for the circuit attached







    I am not sure if i am correct, but this is how i approached this problem:

    So i first found the total impendance in the first parallel branch and then the second
    [itex]Z=\frac{1}{\frac{1}{4<-90^o}+\frac{1}{6<90^o}}=2.4<0^o[/itex]

    Then i found impendance for 2nd parallel branch:
    [itex]Z=\frac{1}{\frac{1}{6.8<0^o}+\frac{1}{6.8<0^o}+\frac{1}{8<90^o}}[/itex]=8.69<66.97^o

    So, [itex]ZT= 8.69<66.97^o+2<0^o+2.4<0^o = 13.09<66.97^o[/itex]

    Finding V2: Knowing [itex]I=4mA<0^o, Z_T=13.09<66.97^o[/itex]

    [itex]E=4<0^o*13.09<66.97^o=52.36V<66.97^o[/itex]

    Therefore [itex]V_2=52.36V<66.97^o[/itex]

    Finding [itex]I_L[/itex]
    [itex]I_L = \frac{V_L}{Z_L}=\frac{52.36<66.97^o}{8<90^o}=5.76<23.03^o mA[/itex]
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2014 #2

    gneill

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    Staff: Mentor

    Redo your calculations for the parallel branch impedances. The formulas look good but your results do not look right. For example, for the first parallel group you have two purely reactive components (inductor and capacitor) in parallel. The resulting combined impedance should be purely imaginary, yet you've come up with purely real result (angle 0°). So something's going awry with your calculations.
     
  4. Apr 29, 2014 #3
    When you add the two reactive components together, the angle becomes 90, since the inductor is 6kohm<90 and capacitor is 4kohm<-90?????
     
  5. Apr 29, 2014 #4

    gneill

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    Are you asking, concluding, or guessing? Why not show your calculation steps?

    For the first parallel combination both components are reactive (so their impedances have only imaginary components). Carrying out the arithmetic should be fairly straightforward.
     
  6. Apr 29, 2014 #5
    So with what your saying would the total impedance for the first branch be Z=(6-4)j, how do we obtain an angle from that when they are both on the same axis? Isn't the angle between them 0?
     
  7. Apr 29, 2014 #6

    gneill

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    You had the right formula for parallel impedance in your first post. Just plug in the rectangular form of the impedances for the two components and carry through the math to arrive at a single value.

    Note that the angle is specified with respect to the positive real axis. If the result lies along the imaginary axis then the angle must be either +90° or -90°.
     
  8. Apr 29, 2014 #7
    So [itex]Z=\frac{1}{\frac{1}{6kohm<90^o}+\frac{1}{4kohm<-90^o}}?[/itex]
     
  9. Apr 29, 2014 #8

    gneill

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    Sure. Do the math. You might find that using rectangular form for the complex values will be convenient.
     
  10. Apr 29, 2014 #9
    [itex]z=\frac{12}{(2+3)j}
    =12*(\frac{1}{(2+3)j})[/itex]
     
  11. Apr 29, 2014 #10

    gneill

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    Can you explain what the above is all about? The numbers don't seem to correspond to any given in the problem statement. What are you calculating?
     
  12. Apr 29, 2014 #11
    Disregard previous post:

    Correct me if i am wrong.

    [itex]6kohm<90^o=(6sin(90^o)+6cos(90^o))j
    =6j[/itex]
    [itex]4kohm<-90^o=(4sin(-90^o)+4cos(-90^o))j
    =-4j[/itex]

    Therefore:

    [itex]Z= \frac{1}{\frac{1}{6j}-\frac{1}{4j}}
    =-12j[/itex]
     
    Last edited: Apr 29, 2014
  13. Apr 30, 2014 #12

    gneill

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    That's better! Note that the units of the result are kΩ.

    Can you do the same for the second parallel grouping?
     
  14. Apr 30, 2014 #13
    Thanks!

    I also have this other problem which i cannot see to get right and i have attempted it 100 times :/

    I have to determine the Thevenin's equivalent circuit for the circuit attached.
    I am stuck on the first and essential part, finding the thevenin's resistance.

    Answer should be 10ohms, my attempt:

    [itex]R_2||4 = \frac{5*16}{5+16} = 3.81ohms[/itex]

    [itex]R_1,3,5 = 20 + 12 + 2 = 34ohms[/itex]

    [itex]R_th= \frac{34*3.81}{3.81+34} = 3.42ohms[/itex]
     

    Attached Files:

  15. Apr 30, 2014 #14

    gneill

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    You should start a new thread for a new problem. Please do so.

    One hint: R2 and R4 are not in parallel.
     
  16. Apr 30, 2014 #15
    I did :)
    They look like they are in parallel though :/
     
  17. Apr 30, 2014 #16
    For this circuit, does V2=E?
     
  18. Apr 30, 2014 #17

    gneill

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    In general I won't confirm or refute unsubstantiated guesses. What logic lead you to your conclusion?
     
  19. Apr 30, 2014 #18
    The parallel the branch that V2 is on seems like it is parallel to the origin source, otherwise called the current source in this case.
     
  20. Apr 30, 2014 #19

    gneill

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    Sketch the "reduced" circuit, replacing the simplified parallel combinations with boxes and label them with their reduced impedances. How does the circuit look? Where does V2 appear?
     
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