# 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 $V_{o}$.

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?

$I_{L}= \frac{V_{A}}{R+2j \omega L}$
$V_{L}=\frac{j \omega L V_{A}}{R+2j \omega L}$

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

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

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.