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Power Factor of an L-C-R curcuit

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A series L-C-R circuit Comprises of a 250mH coil, 0.01 MFD capacitor and a 5W resistor. A dependent voltage defined by e = 2.5cos(80000t) is applied across it. Determine the power factor.


    2. Relevant equations
    Power Factor = cos(x) = R/Z
    Z = sqrt^[R^2 + (XL - XC)^2]

    3. The attempt at a solution
    Z = sqrt^[R^2 + (wL - 1/(wC))^2] , where w = angular frequency
    w = 1/sqrt^[LC] = 1/sqrt^[0.25H * 0.1MFD]

    Power Factor = 5W/Z

    This was where I came to. I'm having trouble most with the units. I do not know how to utilize a 5W resistor and I don't know what a MFD Capacitor is. At first I thought it could be a Mega-Ferrad, but I don't know how MFD converts to Capacitance. I believe my equation manipulation is correct if I can get those straightened out though.
     
  2. jcsd
  3. Apr 26, 2015 #2

    gneill

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    Staff: Mentor

    Hi whitejac, Welcome to Physics Forums.

    MFD means micro-Farad (or μF). It's a fairly old term from the early days of radio electronics. A related term is MMFD, or micro-micro-Farad, equivalent to the modern pico-Farad (pF).

    In electronics W is usually reserved for units of Watts. Resistance is given in Ohms (Ω).

    Does that help?
     
  4. Apr 26, 2015 #3
    Thank you! The MFD is certainly a bit of trivia I'm glad to know now. I don't understand then how to solve this problem... My book actually says 5W resistor. It might be a printing error.

    I'll consider it a 5ohm resistor and see if I recieve the answer in the back: cos(x) = 1/25 x 10^-5.

    One thing I don't understand is why the voltage is given, however. If I consider it a misprint then it's a useless extra.
     
  5. Apr 26, 2015 #4

    gneill

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    Staff: Mentor

    Yes, probably an issue with fonts. If you change a W in "standard" font to a math font, it converts to Ω.
    The voltage is given so that you will know what frequency (ω) the circuit is being driven at. Note that it is not the resonant frequency!
     
  6. Apr 26, 2015 #5
    My set up is wrong then.

    If I assume what I said in the OP was correct, then:
    Power Factor = R/Z
    Z = sqrt^[R^2 + (ωL - 1/ωC)^2]

    Using the given data and a conversion in my book: ω = 1/sqrt[LC] = 0.89.

    Subbing into Z, we're given sqrt^[25 + ((0.89*0.25) - 1/(0.89 * 0.1x10^-7))^2]
    This gives an answer close to 4.99 which would imply that the Power Factor is almost 1, and that does not come anywhere near my posted solution. Am I using a wrong formula for finding ω and/or the Power Factor?
     
  7. Apr 26, 2015 #6

    gneill

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    Staff: Mentor

    That formula gives you the natural resonant frequency of the RLC circuit. It is NOT the frequency at which the circuit is being driven by the voltage source! Look in the given data to find the frequency at which it is being driven.
     
  8. Apr 26, 2015 #7
    I'm sorry. I can't see how you acquire the frequency from the voltage without knowing time prior. The problem question is all of the information I'm allowed to have. If I use v = e = 2.5cos(80000t) then I have 2 unknowns. If I use another formula, say....
    v = Vcos(ωt +∅) then I have a whole bunch of unknowns. The only sure thing I know how to find is ω at resonance. I don't know how to find it when it's not.
     
  9. Apr 26, 2015 #8

    gneill

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    Staff: Mentor

    In the expression v = Vcos(ωt +∅), what do the terms ω, t, and ∅ represent?
     
  10. Apr 26, 2015 #9
    ω = frequency, which is integral in finding Z.
    t = time
    ∅ = the angle between the voltage phasor and the current phasor (phase angle)
     
  11. Apr 26, 2015 #10

    gneill

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    Staff: Mentor

    Okay, so you were given the driving voltage as: e = 2.5cos(80000t).

    Can you pick out the (angular) frequency?
     
  12. Apr 26, 2015 #11
    Oh. Wow. Yes I can.
    ω = 80000 which will give me a what i need to find Z which will allow for me to find P.
    I can't believe it was staring me in the face that blatantly.
     
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