# Homework Help: Using complex impedances

1. May 7, 2014

### physiks

I have reached a stage in a problem where I have a complex current through an inductor in a parallel LCR circuit with a current source (the circuit is a parallel LC circuit with a resistance R in series as well as a driving voltage V0sinωt, converted into the above via a Norton equivalent).

I have obtained IL=V0sinωt/R[1+i(ωL/R)-(ω2LC)] by first finding the common voltage across each component.

The problem then says that the circuit is excited at the resonant frequency ω0=1/√LC by a voltage cosω0t. I need to calculate IL(t). This reduces my expression to IL=cosω0t/iω0L.

Now I need to get this to be real. I just wrote cosω0t=e0t and i=eiπ/2 giving IL=cos(ω0t-π/2)/ω0L after taking the real part.

Now I'm not sure if this is correct. Besides that, if it is, I don't quite understand why I would be allowed to do that. Why would I just take the real part at the end. It just doesn't seem mathematiclly rigorous and so if somebody could explain the maths behind the approach I would feel more comfortable. Thanks.

2. May 7, 2014

### ehild

If you use complex impedances, use also complex currents and voltages, in the form I=I0 eiωt and U=U0eiωt, where I0
and U0 are complex amplitudes, including phases. The impedance is the ratio of the complex voltage and complex current. If you use the sine or cosine form, the ratio U/I depends on time.
If the real driving voltage is Vosin(ωt), which is the imaginary part of V0eiωt, write it in the form V0eiωt solve the problem, and then take the imaginary part of the solution.

ehild

3. May 7, 2014

### physiks

Thanks! Just to confirm, the above method is correct then right?

Last edited: May 7, 2014
4. May 7, 2014

### ehild

If two complex numbers are equal both the real parts and the imaginary parts are also equal.

ehild

5. May 8, 2014

### physiks

I can't see how that helps?

6. May 8, 2014

### ehild

I did not understand your question
I thought you ask if the method, solving a complex equation, and then taking the real of imaginary part of the solution works.
So what did you mean?

ehild

7. May 8, 2014

### physiks

Oh I see! I was talking about the specific example problem in my original post, i.e is that correct (I believe it agrees with your method).

8. May 8, 2014

### ehild

No, that solution is not good as you wrote it. Do not use sin(ωt). It should be V=Vo eiωt
What was the circuit at all? Was the resistor in series with the inductor or in series with the parallel LC circuit? If so, your formula for IL is almost correct, but use V instead of Vosin(ωt) and you miss a pair of parentheses. V=Voeiωt, and

$$I_L=\frac{V}{R(\frac{iωL}{R}+1-ω^2LC)}$$.
It is correct, that 1-ω^2LC = 0 at resonance. Find the phase, and add to iωt. You get IL as Ioei(ωt+θ).

You final result is correct. But do not write cos or sin. Keep V=Voeiωt up to the final step, then take the real part.
ehild

Last edited: May 8, 2014