Derive a general expression for the phasor response I_L and V_o

Thread Starter

cursed

Joined Oct 3, 2010
2
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?

Actual answers:
\(I_{L}= \frac{V_{A}}{R+2j \omega L}\)
\(V_{L}=\frac{j \omega L V_{A}}{R+2j \omega L}\)
 

Robert.Adams

Joined Feb 16, 2010
112
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.
 
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