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Homework Help: AC RCL circuit

  1. Apr 9, 2014 #1
    Hello there, I'm not sure if my solution is correct for [itex]\hat{I}_{C}[/itex]

    1. The problem statement, all variables and given/known data

    In the given circuit, calculate the current in each circuit element given that V = V[itex]_{o}[/itex]sin(ωt)
    R, L and C are given.

    2. Relevant equations
    Z = R + jX (j[itex]^{2}[/itex] = -1)
    X[itex]_{L}[/itex] = jωL
    X[itex]_{C}[/itex] = [itex]\frac{-j}{ωC}[/itex]
    [itex]\hat{V}[/itex] = V[itex]_{o}[/itex]e[itex]^{jωt}[/itex]
    [itex]\hat{I}[/itex] = I[itex]_{o}[/itex]e[itex]^{j(ωt-ø)}[/itex]
    I[itex]_{o}[/itex] = [itex]\frac{V_{}}{|Z|}[/itex]
    tan(ø) = [itex]\frac{\Im(Z)}{\Re(Z)}[/itex]

    3. The attempt at a solution
    First for the impedance, [itex]1/Z_{XL} = 1/X_{L} + 1/X_{C}[/itex]
    [itex] Z_{XL} = j\frac{ωL}{1- ω^{2}CL} [/itex]
    [itex] Z = R + Z_{XL} = R + j\frac{ωL}{1- ω^{2}CL} [/itex]
    The current phasor in the resistor R: [itex]\hat{I_{R}}[/itex]= [itex]\hat{I}[/itex]= [itex]\hat{I_{C}} + \hat{I_{L}}[/itex]
    [itex]\hat{V_{C}} = \hat{V} - \hat{V_{R}} [/itex]
    [itex]\hat{I_{C}} = (\hat{V} - \hat{V_{R}} )/X_{C}[/itex]
    [itex]\hat{I_{C}} = (V_{o}e^{jωt} - I_{o}e^{j(ωt-ø)})(jωc) = V_{o}ωCe^{jwt}(1-\frac{R}{|Z|}e^{-jø})(j) [/itex]
    [itex]\hat{I_{C}} = V_{o}ωCe^{j(ωt+\pi/2)}(1-\frac{R}{|Z|}e^{-jø})[/itex] ø and |Z| are known.
    And[itex]I_{L}[/itex] can be found the same way.
    I'm not entirely sure my solution is correct. Also, since [itex]I_{R}[/itex] is always in phase with V, does that mean that ø is 0? If so, then tan(ø) is also 0 but that would mean that either ω or L are 0...

    Note: sorry if it seems slobby and for the skipped steps, my exam is in less than 3 hours and I'm really nervous and running out of time.
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2


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    Science Advisor
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    2017 Award

    Good luck with your exam! Don't worry too much about this last exercise. The assumption Ir is in phase with V seems incorrect to me.
  4. Apr 9, 2014 #3


    User Avatar
    Gold Member

    well i'm still working through it, but one thing I want to point out is your source voltage is wrong. e^iwc =cos(wc) +isin(wc)
    if you want to use eulers id. you want (e^iwc - e^-iwc)/2 , but that seems unnecessary, i'd just keep it v_s or something until you have your final answer, then sub in V(not)sin(wt). also it looks like complex analysis only? no time dependent ? just wondering.
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