1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    IMG_2958.JPG
    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

    NascentOxygen

    User Avatar

    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

    cnh1995

    User Avatar
    Homework Helper

    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

    gneill

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted