Derive a general expression for the phasor response I_L and V_o

Discussion in 'Homework Help' started by cursed, Feb 24, 2011.

  1. cursed

    Thread Starter New Member

    Oct 3, 2010
    2
    0
    The circuit in the figure below is operating in the sinusoidal steady state with V_{s}(t)=V_{A} cos(\omega t). Derive a general expression for the phasor response I_{L} and the voltage V_{o}.

    [​IMG]

    Relevant equations:

    I= \frac{V_{A}}{Z_{EQ}}, where Z is the impedance and V is the volage.

    My attempt:

    I converted everything to the phasor ("frequency") domain. So the value of the inductor is now Z=j \omega L. The resistors do not change.

    Now,


    Z_{EQ} = R + (j \omega L || R), where || represents "parallel." (i.e. j \omega L in parallel with R.)

    Z_{EQ} = R + \frac{j \omega LR}{j \omega L + R}

    I stopped there because I realized I would be nowhere near close to the correct answer. Am I even on the right track with this logic first of all?

    Actual answers:
    I_{L}= \frac{V_{A}}{R+2j \omega L}
    V_{L}=\frac{j \omega L V_{A}}{R+2j \omega L}
     
  2. Robert.Adams

    Active Member

    Feb 16, 2010
    112
    5
    You were headed in the right direction with that idea.

    Converting to the frequency domain, you get:
    Z=R+(jwL||R)=R+jwLR/(jwL+R)=\frac{R(jwL+R)+jwLR}{jwL+R}<br />
Itotal=\frac{Va}{Z}=\frac{Va(jwL+R)}{R^{2}+2jwLR}

    Next you use a current divider between jwL and R:
    I=Itotal*\frac{R}{jwL+R}=\frac{Va(jwL+R)}{R^{2}+2jwLR}*\frac{R}{jwL+R}=\frac{Va}{R+2jwL}

    Try the voltage yourself and don't give up prematurely because you think you might be wrong.
     
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