How Does Resonance Affect Current in a Parallel LCR Circuit?

[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 I_L=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 I_L(t). This reduces my expression to I_L=cosω₀t/iω₀L.

Now I need to get this to be real. I just wrote cosω₀t=e^iω₀t and i=e^iπ/2 giving I_L=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.

 

[ehild]

If you use complex impedances, use also complex currents and voltages, in the form I=I₀ e^iωt and U=U₀e^iωt, 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 V₀e^iωt, write it in the form V₀e^iωt solve the problem, and then take the imaginary part of the solution.

ehild

 

[physiks]

If you use complex impedances, use also complex currents and voltages, in the form I=I₀ e^iωt and U=U₀e^iωt, 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 V₀e^iωt, write it in the form V₀e^iωt solve the problem, and then take the imaginary part of the solution.

ehild

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

 

[ehild]

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

ehild

 

[physiks]

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

ehild

I can't see how that helps?

 

[ehild]

I did not understand your question

the above method is correct then right?

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

 

[physiks]

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

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).

 

[ehild]

No, that solution is not good as you wrote it. Do not use sin(ωt). It should be V=Vo e^iωt
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 I_L is almost correct, but use V instead of Vosin(ωt) and you miss a pair of parentheses. V=Voe^iωt, 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 I_L as Ioe^i(ωt+θ).

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