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!

Basic RL circuit -- Power example

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A simple RL circuit in series connected to a source with V(t) = 30Cos(3t+75°)V and is delivering 14.12 W of power. I'm also given the power factor which is pf=0.857 and is lagging.

    Find the values of R and L

    2. Relevant equations

    P=Vm*Im/2 * pf for power
    theta = Cos^-1(pf)



    3. The attempt at a solution

    I've tried the problem and I already have the solution which I will link to this post. But I do not understand some part of the problem.
    So what I've basically done is I found the angle of the power factor by using the equation above.
    Theta = Cos^-1(0.857) = 31.01° ( positive because the powerfactor is lagging)

    Then by Ohms law I find the Current which is I = V/Z and Z = R + jwl
    so I = V/(R + jwl). We know that the Voltage is sinusoidal and therefor Vm = 30 V .
    Then we have I = 30/(R +jwl)

    I can also find maximum current by using the eqation stated above. Which is
    P=Vm * Im / 2 * pf ----> algebra ---> Im = 2*P/(Vm*pf) = 1.098 A

    This is how far I could get and I don't know how to get anything else. But in the solution I have it says that
    | I | = 30/(R^2 + w^2 * L^2) .... I have absolutely no idea how this is true? But what I find the most disturbing is that I thought that the impedance of the Inductor would be "Jwl" not only "wL", Where is the "J" in that eqation?


    same thing with given solution that (theta)v - (theta)i = avg{R + jwl} = atan(wl/R} .... <-- what is this. Where does this come from and where is the "J"

    In the end my solution tries to find two other variables called X and Y. I assume that X is the reactance, and once again I do not know what Y stands for.
    The solutions says R^2 + w^2 * L^2 = 30/| I | = 27.3123 = X <--- again where is that J and why is it not there. I need the explanation.
    To find Y variable you use this equation R = wl/X and from that w^2 * L^2 / Y ^2 + w^2 L^2 = X
    I do not understand what was happening there.

    So to sum it all up. I am confused why there is no 'J' in some of the equations and why they are squared in others.
    It would be greatly appreciated if I could get some explanation on how this power transfer works.

    I will link to pictures with this post. One with my attempt and one of the solution.
    Lausn+12_Page_1.png 20141218_215432.jpg
     
  2. jcsd
  3. Dec 18, 2014 #2
    It is not true. Correct formula is:
    gif.latex?\mid%20I\mid%20%3D\frac{30}{\sqrt{R^{2}&plus;\omega%20^{2}L^{2}}}.gif
    Notice also this is a peak current since peak voltage is 30 V.
     
  4. Dec 18, 2014 #3
    Alright thanks for that, but isn't the impedance for inductor Zl = Jwl?
     
  5. Dec 18, 2014 #4
    It is. Total impedance is Z=R+jωL, and it's modulus IZI2=R22L2
     
  6. Dec 18, 2014 #5
    Thank you very much. Problem solved. I just couldn't figure out what the solution was telling me but once you said they were wrong It all came together.

    Thanks again for the help :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Basic RL circuit -- Power example
  1. RL circuit (Replies: 1)

  2. RL Circuit (Replies: 1)

  3. RL Circuit (Replies: 2)

  4. RL Circuit (Replies: 3)

  5. RL circuit (Replies: 7)

Loading...