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Homework Help: Past paper question

  1. Nov 10, 2006 #1
    Past paper question - please help, still unresolved! (11/11)

    Hey guys

    Im studyin for my physics end of year exam (only first year so its prob easy peasy for you guys) but unfortunately i have no answer key as yet so i dunno if what im doing is right. I get a plausable answer but i just wanted to be sure incase i never get the answers from lecturers (im badgering em heaps dont worry)

    Ok this might take a while to set out in latex thing but ill give it a go

    The question is about power factor of a parallel LCR ciruit but in a nutshell i need to find C such that the imaginary part of the complex impedance vanishes (ie angle is zero)

    Were given that
    [tex]L= 5 \\
    R = 4 [/tex] and were doing this at mains freq of [tex] \omega = 50Hz[/tex]

    Unfortunately i cant upload the image right now but all you need to know is that this parallel circuit has 2 arms, one just has the capacitor the other the resistor and inductor in series.

    So, working out impedances:
    [tex] Z = \frac{1}{(R+\omega Li)} - \omega Ci[/tex] Now ive always used minus i for the capacitor part because of the angle difference in the phasor, i assume i do that here.

    Next i try to get the imaginary parts off the bottom.
    [tex] \frac{1}{(R+ \omega Li)} * \frac{(R- \omega Li)}{(R-\omega Li)} = \frac{R- \omega Li}{((R^2)+(\omega L)^2)}[/tex]

    Then putting it back in and making it one big happy fraction
    [tex]\frac{(R-\omega Li - (\omega Ci)(R^2 +(\omega L)^2)}{(R^2 +(\omega L)^2)}

    = \frac{(R-\omega Li - \omega CR^2 i - \omega ^3 L^2Ci)}{(R^2 +(\omega L)^2)}
    [/tex]
    Now just taking the imaginary bit and equating to zero (getting rid of denominator for obvious reasons)
    [tex]
    (-\omega L-\omega CR^2-(\omega )^3 L^2C = 0
    [/tex] Then just shoving in those numbers and ignoring the minus sign (which has me rather worried) i get C = 7.9 x 10^-5 farad.

    Hope this makes sense and im right!

    Cheers for help (please dont flame my stupidity if i made a stupid mistake, first time with latex too)
    -G
     
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2

    dextercioby

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    The [tex] tag closes with [ / tex] (without the spaces of course)

    Daniel.
     
  4. Nov 10, 2006 #3
    yah just fixing that now cheers :D tryin to get omega to work too :S

    hmm some of the changes arent saving. anyway
     
    Last edited: Nov 10, 2006
  5. Nov 10, 2006 #4

    dextercioby

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    It's \omega and the same for the other symbols and functions.

    Daniel.
     
  6. Nov 10, 2006 #5
    yah its not saving the changes


    EDIT... ah there we go
     
  7. Nov 12, 2006 #6
    Ok i found out whats wrong, its the fact that i used -iwC instead of +iwC, which screws it all up. However i always thought that due to phasors and phase differences etc your supposed to use the minus. Anyone?
     
    Last edited: Nov 12, 2006
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