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Homework Help: Power Factor of RL circuit

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    See attachment.

    2. Relevant equations
    V(t) = Vmcos([itex]\omega[/itex]t + [itex]\theta[/itex])
    P(t) = VmImcos([itex]\omega[/itex]t + [itex]\phi[/itex])cos[itex]\omega[/itex]t
    PR(t) = [itex]\frac{V^{2}(t)}{R}[/itex]
    PF = [itex]\frac{P}{VeffIeff}[/itex]
    S = VeffIeff
    Pavg = Ieff2(Re(Zeq)

    3. The attempt at a solution
    Ieff = 5 A
    f = 50 Hz or [itex]\omega[/itex] [itex]\approx[/itex] 314 rad/s
    Zeq = [itex]\frac{80(j60)}{80+j60}[/itex] = 28.8 + j38.4 Ω(48[itex]\angle[/itex]53° Ω)
    V = (5[itex]\angle[/itex]0)(48[itex]\angle[/itex]53) = 240[itex]\angle[/itex]53.13010235° V
    Pavg = (5)[itex]^{2}[/itex](28.8) = 720 W

    PF = [itex]\frac{720}{1200}[/itex] = 0.6

    Pavg = 720 W

    PR(2 ms) = [itex]\frac{57600cos^{2}(314(0.002)+53°)}{80}[/itex] [itex]\approx[/itex] 253 W
    -tan([itex]\frac{{\omega}L}{R}[/itex]) [itex]\approx[/itex] -36.85°
    P(2 ms) = 1200cos(314(0.002)-36.85°)cos(314(0.002)) [itex]\approx[/itex] 968 W

    PL(2 ms) = P(2 ms) - PR(2 ms) = 715 W

    Have not done yet.

    I want to know if I am going in the right direction or if I'm doing it all completely wrong haha. I don't want to do part (d) until I know the other parts are headed in the right direction.

    Any help is appreciated.

    Attached Files:

    Last edited: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2


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    Staff: Mentor

    The image is not appearing for me. Can you try to post it again?
  4. Oct 1, 2012 #3
    It should show up now (attached to the first post).
  5. Oct 2, 2012 #4


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    Staff: Mentor

    (a) and (b) are correct. I haven't figured out what is being asked in (c). I read it as being asked to express instantaneous power as a function of time, specifically t - 2msec, but you seem to have read it as t = 2msec. Is precisely half of the equals sign missing?
  6. Oct 2, 2012 #5
    Yeah, it is t = 2 ms. For some reason, it cut off half of it.
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