# Using Complex Impedances in these RLC Circuit Calculations

What happened to the denominator in the expressions for ## a ## and ## b ##? I think you have forgotten those?

[EDIT]: I realised that you included them in the working, but not the initial expression. Please ignore the above statement.

That 11.5k V sounds abit to high
That is true and does definitely stand out to me. However, this is what can happen in resonant circuits with very high Q-factors... I just did a quick calculation, and the number I got was slightly larger than your answer (but on the same order of magnitude). Maybe another member might notice an error that I have overlooked

Also, in the expression for ## b##, I think the central term in the numerator should have the ## C ## as ## C^2 ##. However, I am not sure whether that makes much difference given the order of magnitudes. I will go back to check the original expression to ensure that your formula is exactly the same as what I got

Also, in the expression for ## b##, I think the central term in the numerator should have the ## C ## as ## C^2 ##. However, I am not sure whether that makes much difference given the order of magnitudes. I will go back to check the original expression to ensure that your formula is exactly the same as what I got
Master thx a lot for the help man. You really are a lifesaver

DaveE
Gold Member
It is also useful to think about dimensions when working through these types of problems. Often seeing dimensions that match can be indicative of correct working (e.g. seeing a ## \omega ^2 LC ## or ## \omega ^ 2 RC ##)
This is an incredibly useful tool when working with this sort of complex algebra. OK, actually, useful in most of physics, really. This allows you to spot errors in equations without having to go through the whole derivation. Maybe you won't know why it's wrong, but you will know it's not correct.

So, to elaborate on @Master1022s method:
R, 1/(ωC), ωL all have units of ohms. If I see a term in your equation like (R + L), I know without much analysis that it's wrong; that should be something like (R + ωL), or (R/ω + L). You just won't see solutions in physics where you have (ohms + volts), (force + energy), (flow + pressure), etc.; these rarely have any useful physical meaning. In the (R + L) example the units are (ohms + ohms*seconds).

BTW, you can't use this to verify correctness completely. It won't spot some errors, like the difference between (R + ωL) and (R + 1/(ωC)), or R vs. 2R.

If you want to know more look up "dimensional analysis".

etotheipi and Master1022
This is an incredibly useful tool when working with this sort of complex algebra. OK, actually, useful in most of physics, really. This allows you to spot errors in equations without having to go through the whole derivation. Maybe you won't know why it's wrong, but you will know it's not correct.

So, to elaborate on @Master1022s method:
R, 1/(ωC), ωL all have units of ohms. If I see a term in your equation like (R + L), I know without much analysis that it's wrong; that should be something like (R + ωL), or (R/ω + L). You just won't see solutions in physics where you have (ohms + volts), (force + energy), (flow + pressure), etc.; these rarely have any useful physical meaning. In the (R + L) example the units are (ohms + ohms*seconds).

BTW, you can't use this to verify correctness completely. It won't spot some errors, like the difference between (R + ωL) and (R + 1/(ωC)), or R vs. 2R.

If you want to know more look up "dimensional analysis".
Thx man. I’ll look up dimensional analysis

DaveE
Gold Member
Thx man. I’ll look up dimensional analysis
It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.

Just a quick update, I have checked my expression:
$$Z = \left( \frac{\omega^4 R L^2 C^2}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) + j \left( \frac{\omega L (1 - \omega^2 LC + (\omega R C)^2)}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right)$$
and think it is equal to yours (thus we are either both correct... or both wrong !)

Quick question about the problem statement: are the units of ## \omega ## a typo when it says Hz? Usually we expect ## \omega ## to be in rad/s (a conversion factor of ## 2 \pi ## is needed).

Am doing the calculations in a Python notebook, so will post pictures here to show how my calculations are resulting

Just a quick update, I have checked my expression:
$$Z = \left( \frac{\omega^4 R L^2 C^2}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) + j \left( \frac{\omega L (1 - \omega^2 LC + (\omega R C)^2)}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right)$$
and think it is equal to yours (thus we are either both correct... or both wrong !)

Quick question about the problem statement: are the units of ## \omega ## a typo when it says Hz? Usually we expect ## \omega ## to be in rad/s (a conversion factor of ## 2 \pi ## is needed).

Am doing the calculations in a Python notebook, so will post pictures here to show how my calculations are resulting
Yeah the w = 2(pi)f but since the question tells us that w is 200

Yeah the w = 2(pi)f but since the question tells us that w is 200
Ah okay, that explains the discrepancy I had earlier. After setting ## \omega = 200 ##, I did get the same value of the gain and the final voltage as you (give or take some rounding errors) - the computer output ## 11551.24... ## (V), so this does agree with your result.

Hope that is of some help.

Ah okay, that explains the discrepancy I had earlier. After setting ## \omega = 200 ##, I did get the same value of the gain and the final voltage as you (give or take some rounding errors) - the computer output ## 11551.24... ## (V), so this does agree with your result.

Hope that is of some help.
Master1022 I got nothing but respect for you man. Really makes me happy to see people like you helping out people like me

DaveE
Gold Member
Quick question about the problem statement: are the units of ω a typo when it says Hz? Usually we expect ω to be in rad/s (a conversion factor of 2π is needed).
It's very common for people to be sloppy about that. I guess they assume you'll know about the 2π correction.

Master1022
Master1022 I got nothing but respect for you man. Really makes me happy to see people like you helping out people like me
Happy to help, hopefully that answer is correct. One other sanity check is to see whether you can find a formula for the damping ratio of an RLC circuit with this setup and see whether it is very small (which is the only way such a large value of the gain is possible). I did have a quick glance on wikipedia, but I cannot see any similar variation. Regardless of the numerical values of the question, you did get the overall method correct (finding the transfer function and then calculating the gain).

I am going to ask @DaveE as I think he is more knowledgeable: can this topology of the RLC circuit be re-arranged into another form (eg. from the list on wiki here) such we can use the same formula for the damping ratio ## \zeta ##?

DaveE
Gold Member
The canonical form for this particular LCR impedance is:

Z(s) = sL⋅(1+sRC)/(1+sRC+s2LC), where s=jω (sorry, I have habits that are hard to break!)

Then if you define some common features: ω=1/sqrt(LC), Zo=sqrt(L/C), Q=1/(2ζ)=Zo/R

The quadratic term looks like 1+(1/Q)(s/ω)+(s/ω)2.

This looks very different from the homework problem which wants a solution as a+ib. Hence a lot of algebra ensues.

The "factored pole-zero" or "ratio of factored polynominals" form above is how we really do this in EE/Controls world. I gives much better insight into the dynamics of this circuit. It's also easier to derive, IMO. Also notice that the polynomials are dimensionless, except the leading sL term, which is ohms, of course.

edit: all of my ω's above (except s=jω) should be ωo, it's not that ω, it's a constant.

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Thanks for the reply @DaveE !

The canonical form for this particular LCR impedance is:

Z(s) = sL⋅(1+sRC)/(1+sRC+s2LC), where s=jω (sorry, I have habits that are hard to break!)
Agreed! (I also prefer working in the ##s##-domain)

Then if you define some common features: ω=1/sqrt(LC), Zo=sqrt(L/C), Q=1/(2ζ)=Zo/R

The quadratic term looks like 1+(1/Q)(s/ω)+(s/ω)2.
Yes, that is correct. This leads to
$$\zeta = \frac{R}{2} \sqrt \frac{C}{L} = \frac{10^4}{2} \sqrt{\frac{5 \cdot 10^{-6}}{20}} = 2.5$$
which is odd because that doesn't suggest gain on the order of ## 10^3 ##...

Hmm, perhaps I ought to have another look over my algebra tomorrow to see if I come across any errors.

Thanks once again.

DaveE