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Homework Help: Rlc problem: calculations of components and phasor drawing

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Find impedance and Io.
    draw phasor diagram for rlc circuit
    V = 120cos(2π525t) where R = 10Ω, C = 1μF, L = 100mH, Vo = 120V


    2. Relevant equations

    Z = √(R2 + (ωL-1/(ωC))^2) = √(10^2 + (2π * 525 Hz * 100 mH - 1/(2π * 525 Hz * 1 μF ))^2) = 28.525Ω

    Io = Vo/Z = 120 V / 28.525 Ω = 4.207 A

    tan ϕ = (ωL - 1/(ωC))/R = (2π525 * 100 mH - 1/(2π525 *1 μF)) / 10 Ω = -15882.508
    ϕ = -1.5707 rad = -89.996°

    VRo = IoR = 4.207 A * 10Ω = 42.07 V
    VCo = Io/(ωC) = 4.207 A / (2π525 * 1 μF) = 1275.361611 V
    VLo = IoωL = 4.207 A * 2π525 * 100 mH = 1387.751431 V

    Vo = √(VR02 + (VLo- VCo)^2 ) = 120.0056523 V


    3. The attempt at a solution

    I get a negative phase angle which should mean that capacitance is greater than inductance but this is not the case based on values above.
    when I draw out a rough phasor diagram I have voltage ahead of current.
    not sure where i went wrong.
    note: diagram attached is rough and not to perfect scale
     

    Attached Files:

    • rlc.jpg
      rlc.jpg
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  2. jcsd
  3. Dec 8, 2011 #2
    I calculated Xl and Xc separately and got Xl = 329Ω and Xc = 303Ω
    This gave me an impedance of 28Ω and I = 4.3A
    My (Xl-Xc) = 26Ω which gives Tan∅ = 2.6 (leading)
    My phasor diagram would be 329Ω on the +y axis, 10Ω on the + x axis and 303Ω on the -y axis (I dont know how to get drawings on here yet !!
    My values are pretty much the same as yours !!!
    I prefer to work out individual quantities rather than lump every thing together in one equation.
    I cannot see where our answers differ !
    Hope this helps
    Just ocurred to me.... did you change mH into H and μF into F in your phase angle calculation?
     
  4. Dec 8, 2011 #3
    thanks
    i did convert mH and μF in my calculations to H and F....
    I finally realized that I had left out the 525 in calculating XC. now i get tanϕ = 2.671 , which is voltage leading....which soothes my brain

    working out individual quantities probably would have saved me the headache. next time i won't plug in the whole thing in excel.
     
  5. Dec 8, 2011 #4
    Well done.... no lack of understanding, that is the main thing.
     
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