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
    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}.


    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.


    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
    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 />

    Next you use a current divider between jwL and R:

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