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Phasor Circuit analysis

  1. Dec 10, 2015 #1

    MacLaddy

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    Gold Member

    1. The problem statement, all variables and given/known data

    Hello folks. I had a recent midterm where I got a problem wrong, but I'm not sure I understand why. I am hoping someone here can shed some light. Please see the attached circuit diagram.


    2. Relevant equations

    Phasor notation stuff... I am correct on all except one sign. Please see below.


    3. The attempt at a solution

    Using superposition, and considering the sources as Case 1, and Case 2, as noted in the image.

    Case 1:

    [itex]12cos(2000t)=12\angle{0^\circ}[/itex]
    [itex]cap= -40j[/itex]
    [itex]Ind= 10j[/itex]

    Using KVL

    [itex]-12\angle{0^\circ}+(-40j)I_1+30I_1+10jI_1=0[/itex]
    [itex]I_1=0.28\angle{45^\circ}[/itex]
    [itex]I_1=0.28cos(2000t+45^\circ)[/itex]

    Case 2:

    [itex]18cos(4000t)=18\angle{0^\circ}[/itex]
    [itex]cap= -20j[/itex]
    [itex]Ind= 20j[/itex]

    Using KVL

    [itex]-18\angle{0^\circ}+(-20j)I_2+20jI_2+30I_2=0[/itex]
    [itex]I_1=0.6\angle{0^\circ}[/itex]
    [itex]I_1=0.6cos(4000t)[/itex]

    Final answer

    [itex]i(t)=0.28cos(2000t+45^\circ)+0.6cos(4000t)[/itex]

    However, my instructor says that the answer above should have a negative sign instead of a positive. Like this...

    [itex]i(t)=0.28cos(2000t+45^\circ)-0.6cos(4000t)[/itex]

    He explained that he is following KVL by using the sign directions that are drawn on the diagram. This seems wrong to me, as that would make every element in the second equation positive.

    [itex]18\angle{0^\circ}+(-20j)I_2+20jI_2+30I_2=0[/itex]

    But shouldn't the power source always have a different sign than the other elements? If I apply KVL to the above it would indicate that either everything is applying voltage to the element, or everything is removing it.

    Any hints on where my thinking is incorrect?

    Thanks,
    MacLaddy
     

    Attached Files:

  2. jcsd
  3. Dec 10, 2015 #2

    gneill

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    Staff: Mentor

    When you are summing potential drops around a loop, then a potential rise due to a voltage source is written as a negative value, and a potential drop as a positive value.

    In your case for the second source you are going clockwise around the loop and when you go through the 18 V source you do it from + to -, so it's a potential drop. It gets treated like any other drop, which is to say, it is summed as a positive quantity.
     
  4. Dec 11, 2015 #3

    MacLaddy

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    Gold Member

    Thanks gneill. I suppose I'm convinced, especially since you echo my professors sentiments. It just seems unlikely that a polarity direction could be assigned to a current that is alternating. Perhaps this is more of an academic problem than a real life one? At least now I know for sure when the final comes rolling around.

    Mac
     
  5. Dec 11, 2015 #4

    gneill

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    Staff: Mentor

    Consider an AC source with a function of time given by ##v1(t) = A sin(\omega t)##. At time t = 0 the value is increasing in the positive direction. Another source ##v2(t) = -A sin(\omega t)## would be decreasing. So the sources are in fact 180° out of phase. The polarity of the symbol used in the schematic tells you the relative orientation of the sources' basic phase angles.
     
    Last edited: Dec 11, 2015
  6. Dec 11, 2015 #5

    MacLaddy

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    Gold Member

    Thank you, that I do understand. But aren't the sources above out of phase--have different frequencies? Truly, it doesn't matter too much. I'm a mechanical guy taking an electrical class. I believe I understand well enough now for my purposes.

    Mac
     
  7. Dec 11, 2015 #6

    gneill

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    Staff: Mentor

    Their relative phases will change over time due to having different fundamental frequencies, yes. It's not always the case though, and relative phase matters a great deal in many cases.
     
  8. Dec 11, 2015 #7

    NascentOxygen

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    Staff: Mentor

    A.sin(ωt) and -A.sin(ωt)
    have the same peak amplitude, A
    have opposite phase, meaning they differ by 180°

    The pair could be written, equivalently, as A.sin(ωt) and A.sin(ωt - 180°)
    because the minus sign for amplitude is, basically, a shorthand way of showing a 180° phase shift
     
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