AC Circuit Theory Help? - Page 1
Forum section: electronics / beginners

author: Syko_
author_rank: Contributor
timestamp: September 22, 2013, 10:45:05 am
content: Hey all,First off, apologies if this isn't in the right section...I'm currently taking an engineering degree at an Aussie university, and one of the units is a basic introduction to electronic systems. We have an assignment due on Thursday, and I'm not certain about one of the questions, I was hoping some of you might be able to point me in the right direction?It's an abstract problem, to do with the resonant frequency of an AC parallel circuit... Here it is:Thanks for your help

author: jahonen
author_rank: Super Contributor
timestamp: September 22, 2013, 10:53:58 am
content: You'll need to solve the frequency where that LRC circuit looks purely resistive from the source point of view.Regards,Janne

author: Rerouter
author_rank: Super Contributor
timestamp: September 22, 2013, 11:31:23 am
content: A circuit is at resonance when Xc and Xl are equal, (there reactance) the inductors reactance is 2*pi*F*L, and the capacitors reactance is 1/(2*pi*F*C)now the normal equation i am used to seeing is only 1/(2*pi*Root(L*C)) and have yet to come across the secondary part,

author: Syko_
author_rank: Contributor
timestamp: September 22, 2013, 11:43:46 am
content: Thanks both,Any idea what the second part of the equation is for?

author: Wytnucls
author_rank: Super Contributor
timestamp: September 22, 2013, 12:40:31 pm
content: The inclusion of resistance in the circuit, either in series or parallel affects the resonant frequency. It is called antiresonance. In this particular case, the frequency is shifted to the lower side. The second term of the equation characterizes that antiresonance, due to the resistance.

author: Syko_
author_rank: Contributor
timestamp: September 22, 2013, 12:46:34 pm
content: Ah, Thank you very much!So I guess now I just have to find the first principles equations and try to derive this put of them... Might be a bit of a task haha.

author: JackOfVA
author_rank: Unknown
timestamp: September 22, 2013, 03:43:50 pm
content: Part of your problem is that there are three definitions of "parallel resonance."1) Frequency at which the impedance of the parallel circuit is maximum2) Frequency at which the phase difference between voltage and current is 03) Frequency at which wL=1/wC - i.e., the frequency at which the L and C would be series resonant.If R is small compared with wL at resonance, then all three parallel resonance frequencies will be close.

author: Syko_
author_rank: Contributor
timestamp: September 22, 2013, 04:05:22 pm
content: I think what's being asked is "from first principles, show how you would derive the following equation for resonant frequency" so, like was mentioned earlier, the first part of the equation pertains to the (was it reactivity?) of the components and the second, the resistance... There are no specified values for the components, he doesn't want me to solve the equation to find the actual resonance value, rather he just wants me to show how I would derive that particular equation (the one shown) from first principles. My apologies if I'm misunderstanding something here and making a dick of myself...

author: sub
author_rank: Regular Contributor
timestamp: September 22, 2013, 09:28:33 pm
content: Because this is a parallel circuit, you need the imaginary part of the admittances to be equal rather than the reactances.So, compute the impedances Z_i of each branch and set I(1/Z_C) = I(1/Z_RL).

author: IanB
author_rank: Super Contributor
timestamp: September 22, 2013, 09:52:40 pm
content: Quote from: Syko_ on September 22, 2013, 10:45:05 amI think it goes as follows (hopefully an expert will correct me if I'm wrong):Imagine for a moment that only the inductor is present (ignore the resistor for the time being).You can write down the reactance of the inductor as a function of frequency using the standard formula. The reactance will be zero at DC and will increase towards infinity as the frequency increases.Now do the same thing for the capacitor alone in the circuit. For the capacitor the reactance will be infinite at DC and will decrease towards zero as the frequency increases.Next, put the inductor and capacitor in parallel. You will now have a combined impedance from the parallel components that is zero at DC and zero at infinite frequency. At some frequency in the middle the combined reactance will rise to a maximum. This is where resonance occurs.If you differentiate the parallel reactance formula with frequency and set the slope to zero you will find the resonant frequency.Adding the resistor changes things slightly, but not much. Now you have to deal with impedance instead of pure reactance, but following the same principles. Write down an expression for the combined impedance of the system and differentiate it to find the frequency where it is at a maximum.

author: JackOfVA
author_rank: Unknown
timestamp: September 22, 2013, 10:04:56 pm
content: Here's the simple way to do it.Assume that the definition of parallel resonance desired is the 0 phase angle one. (That seems to be the one in formula provided.)C and the R+L arms have the same voltage across them. Calculate the admittance of the R L C network. At resonance, the imaginary part must be 0. This can be easily solved for.The math will go a lot easier if you convert to admittance (Y+jB) instead of trying to keep everything in series (R+jX) form.First convert R + jwL to admittance. Then 1/jwc to admittance. Add the two together. This will give you the admittance of the R L C network. You can then convert it back to impedance and set the imaginary part to zero, or you can keep it in admittance and do the same.It will be an excellent exercise in converting from impedance to admittance in the complex case and the reverse.I suppose you could also solve this by mesh current analysis if you prefer but that seems unnecessary.

author: IanB
author_rank: Super Contributor
timestamp: September 22, 2013, 10:30:44 pm
content: Quote from: IanB on September 22, 2013, 09:52:40 pmHaving just attempted to do this I found it is not that easy. You have to deal with the complex impedances and can't just consider it to be a scalar problem.The hints from others here seem to be a way of simplifying the algebra and making it more tractable.

author: amiq
author_rank: Regular Contributor
timestamp: September 23, 2013, 12:41:43 am
content: Just replace the series RL branch with a parallel RL equivalent (hint - the Q will remain the same) and the answer drops right out.

author: Zero999
author_rank: Super Contributor
timestamp: September 23, 2013, 04:44:14 am
content: Yes, the circuit is resonant when the impedance of the RL branch is equal to that of the capacitor.This only applies when the circuit is under-damped, otherwise R will dominate.

author: Wytnucls
author_rank: Super Contributor
timestamp: September 24, 2013, 08:15:39 am
content: The admittance is jwC+1/(R+jwL) with w=2Pi*fSolve for real part(conductance)+imaginary part(susceptance) and then, to obtain the resonance frequency, the imaginary part has to be equal to 0.Solving the imaginary part=0 for frequency should give you the required expression.

author: Syko_
author_rank: Contributor
timestamp: September 24, 2013, 12:18:15 pm
content: Hey all, thanks for the replies.So, solving for the first part of the expression seems simple enough, but I'm having trouble solving for the second part, where the damping comes into play... I need to be able to derive the equation from the first principles (most basic) equations, but I just can't seem to figure it out. Any pointers, or an explanation would be great thanks

author: Wytnucls
author_rank: Super Contributor
timestamp: September 24, 2013, 12:44:46 pm
content: Here you go:This is the imaginary part of the admittance, susceptance (B) of the circuit, which is equal to zero for resonance.Replace w in the last equation by 2Pi*f, to work out the resonant frequency:

author: Syko_
author_rank: Contributor
timestamp: September 24, 2013, 01:23:24 pm
content: Aha. Thank you very muchI might just pass this unit yet...

author: Wytnucls
author_rank: Super Contributor
timestamp: September 24, 2013, 01:47:46 pm
content: Yes, the trick is to use admittance for calculations with parallel RLC circuits. The math is easier. A good knowledge of complex numbers operations is also a must.

author: arkanix38
author_rank: Contributor
timestamp: October 06, 2013, 11:47:02 pm
content: Easiest way to derive from first principles is to sum the voltages and differentiate with respect to current. We use this in our physics labs to create an electrical analog to a mechanical oscillator. Plug in all your values and rearrange a bit and you should be able to get out the required equation.