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Homework Help: Help understanding phasor solution

  1. May 10, 2017 #1
    1. The problem statement, all variables and given/known data
    From the given circuit and provided information draw a phasor and calculate the inductance.
    2. Relevant equations
    3. The attempt at a solution
    IMG_2959.JPG IMG_2960.JPG

    The paper was given to me at class as a solution to our homework and i cant understand how they drew the phasor and drew the conclusion out of it.
    I see that they took ##U_{12}## as their phasor axis since its the same for both sides but why do they draw the ##I_c## the way they did? How did they know that it goes up and how did they get the ##I_c/2## part?
    The continuation of the left side of the third equation i do get but not the right side. Why is the right side like that?
  2. jcsd
  3. May 11, 2017 #2


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    Staff: Mentor

    If the inductor were ideal (with R=0) then IC would lead I by 180°. If the inductance were zero, making it a simple parallel RC arrangement, then IC would lead I by 90°. So with both R and L then IC will lead I by some angle between those 2 extremes. That constrains the shape of triangle that you draw.

    You also need to bring into the geometry the condition stated in your data of 2 currents having equal magnitude, viz., I = Ig
  4. May 12, 2017 #3
    Do you mean 90° ?
  5. May 12, 2017 #4


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    Homework Helper
    Gold Member

    No. I is the inductor current. Ic would lead U12 by 90° and if the inductor were ideal, inductor current I would lag behind U12 by 90°. This means Ic would lead I by 180°.
  6. May 13, 2017 #5


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    Staff: Mentor

    I believe that there is a flaw in the question as posed. Perhaps the angular frequency given is to high? If it is truly ##10^7 rad/sec## then the capacitor reactance is just 10 Ω. The real resistance of 100 Ω in the other branch will make it impossible for it to conduct a current with the same magnitude as the source current, even taking into account resonance effects. If I'm not mistaken the maximum magnitude for the current ##I## would be just 1 mA.
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