Inductive, Capacitive, and Resistive Circuits

Thread Starter

SamR

Joined Mar 19, 2019
1,171
Working with resonant circuits and when Xl = Xc the circuit is purely resistive. What I am having a problem picturing is what happens at f1 and f2. One is capacitive f1 and one is inductive f2. Somehow I can't work my head around why. How does raising or lowering the frequency from resonance cause the phase angle to become capacitive or inductive? Raising or lowering either the value of C or L will do the same thing. Sorry I just can't picture what it does to the phase angle. Time to play with LTS I guess...
 

MaxHeadRoom

Joined Jul 18, 2013
19,055
Resonance when inductive reactance = capacitive reactance.
leaving resistance the only form of impedance.
See effect of series and parallel tuned circuit.
Max.
 

Thread Starter

SamR

Joined Mar 19, 2019
1,171
Yes, Max, I am working on series resonance. The question is at the f1 and f2 points are they capacitive or inductive and why. I know f1 is capacitive and f2 inductive but I am having trouble visualizing in my head why? What happens to the Phase Angle theta? I can calculate theta but I just can't picture it in my head.
 

Janis59

Joined Aug 21, 2017
767
RE: SamR
Nowadays the books are teaching a resonance from energy standpoint. Okay, that is not wrong, but then is too easy to loss the rational clue, thus the older way to explain the resonance as balance between two reactances was much more fruitful. So: You have some two reactances, inductive reactance X(L)=2pi*F*L and capacitative reactance X(C)=1/(2pi*f*C). So, the real resonance happens when both them are perfectly equal, thus giving an imaginary part zeroed, tells older books. What happens lowering a frequency? Capacitative reactance will grow, and inductive will lowering, thus the current will become slight capacitative (aka "negative angle") - means current maximum will happen slightly beforehand of voltage maximum. And vice versa, when frequency is slightly raised over resonance, the capacitative X(C) will lower, but inductive X(L) will dominate, thus the current will become slightly inductive (or angle will happen positive) - means voltage maximum will happen slightly faster than current maximum.
The said is referring as well the parallel and serial network, the only difference is that serial is giving a Q-fold multiplication of voltage on the elements. For example, with 12V Vcc fed generator one may easily obtain the 5 kV on the C or on the L. Contrary is on the parallel, there resonance are giving the current Q-fold multiplication, for example, some femtoamperes in the antenna may give a many miliamperes enough to run a certain baterries-less most primitive detector radio head-loudspakers.
 

MrAl

Joined Jun 17, 2014
6,500
Working with resonant circuits and when Xl = Xc the circuit is purely resistive. What I am having a problem picturing is what happens at f1 and f2. One is capacitive f1 and one is inductive f2. Somehow I can't work my head around why. How does raising or lowering the frequency from resonance cause the phase angle to become capacitive or inductive? Raising or lowering either the value of C or L will do the same thing. Sorry I just can't picture what it does to the phase angle. Time to play with LTS I guess...
Hi,

There are actually three types of resonance the one you refer to is usually called "physical resonance" because there are two more that actually are purely electrical in nature.

For the one you talk about it's not too hard to see why the two cancel leaving just the resistive part but you have to do a little circuit analysis.

We start with the impedance which is probably enough:
Z=R+j*w*L-j/(w*C))

and if you notice here the complex impedance for C is subtracted from that of L, the key word here being "subtracted". That means that when:
j*w*L=j/(w*C)

[and dividing both sides by 'j' we are left with: w*L=1/(w*C)]

the two disappear from the equation with correct w=w0 leaving just:
Z=R

and that is of course purely resistive.

It's also a little interesting to look at the current when we apply a sine or cosine voltage E to the series combination of R, L and C:
I=E/(R+j*w*L-j/(w*C))

Again we can see that when w*L=1/(w*C) we end up with:
I=E/R

and that has no phase shift. In fact calculating the phase shift we get:
ph=atan((1/(w*C)-w*L)/R)

and we see that again when the two are equal we are left with:
ph=atan(0/R)=atan(0)

which of course equals zero. So the phase shift goes to zero.

This is yet another reason why i encourage learning some form of circuit analysis (such as Nodal) before anything else because it helps to answer SO many questions about electrical circuits.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,700
Working with resonant circuits and when Xl = Xc the circuit is purely resistive. What I am having a problem picturing is what happens at f1 and f2. One is capacitive f1 and one is inductive f2. Somehow I can't work my head around why. How does raising or lowering the frequency from resonance cause the phase angle to become capacitive or inductive? Raising or lowering either the value of C or L will do the same thing. Sorry I just can't picture what it does to the phase angle. Time to play with LTS I guess...
Perhaps this is one way to think of it.

Imagine a series connection of a resistor, inductor, and capacitor. If all you have is a resistor, the current will be in phase with the voltage. If all you have is a capacitor, the current will lead the voltage. If all you have is an inductor, the current will lag the voltage. So, in essence, a resistive element tries to bring the current and voltage into phase, a capacitor tries to move the phase of the current ahead of the voltage, and an inductor tries to move the phase of the current behind the voltage. Since a series circuit has but a single current in it, the final phase will be the result of these three competing influences. If the capacitor has greater influence than the inductor, the total phase will be such that the current leads the voltage, but if the inductor has greater influence than the capacitor, the total phase will be such that the current lags the voltage. The influence of a capacitor is greatest at DC and gets less and less as the frequency increases, while the influence of an inductor is nothing at DC and gets greater and greater as the frequency increases. At low frequencies the capacitor thus dominates and the circuit looks a resistor in series with a capacitor (the impact of the inductor is just to make the capacitor look like it is larger than it really is). At high frequencies the inductor dominates and the effect of the capacitor is just to make the inductor look smaller than it really is). At some point we cross from looking capacitive to looking inductive and that point is the resonant frequency at with the two influences just cancel each other out.
 

crutschow

Joined Mar 14, 2008
23,365
It boils down to: when you go above resonance the inductive reactance increases while the capacitive reactance decreases, and vice-versa when you go below resonance.
So the further the frequency deviates from resonance, the more the one reactance will dominate, and the more the phase will shift towards that dominate reactance.
 

Thread Starter

SamR

Joined Mar 19, 2019
1,171
Thx guys! Let me digest this a bit. I'm sure it will gel with some practice. I understand reactance and impedance and their phasors and I have been through resonance and oscillators previously but somehow the nuance of f1 and f2 being slightly capacitive or inductive slipped by me or I just plain forgot it. I much appreciate the input.

The influence of a capacitor is greatest at DC and gets less and less as the frequency increases, while the influence of an inductor is nothing at DC and gets greater and greater as the frequency increases.
That is the exact piece of the puzzle I was overlooking. I knew that and I just wasn't adding it into my thinking!
 

MrAl

Joined Jun 17, 2014
6,500
It boils down to: when you go above resonance the inductive reactance increases while the capacitive reactance decreases, and vice-versa when you go below resonance.
So the further the frequency deviates from resonance, the more the one reactance will dominate, and the more the phase will shift towards that dominate reactance.
Hi,

I am not sure i like that line of reasoning because the individual deviations dont tell the story it's the sum (or difference) that tells what is happening. In short, 2+1/2=1/2+2 even though only 1=1 is correct, so how do we know if the impedance went up or even down, or neither, from the point of resonance.
 

Papabravo

Joined Feb 24, 2006
12,405
Hi,

I am not sure i like that line of reasoning because the individual deviations dont tell the story it's the sum (or difference) that tells what is happening. In short, 2+1/2=1/2+2 even though only 1=1 is correct, so how do we know if the impedance went up or even down, or neither, from the point of resonance.
Because imepdance only goes one way as a function of frequency. It is in fact monotonic!
 

Thread Starter

SamR

Joined Mar 19, 2019
1,171
Al, the question I was having trouble with was to predict whether at f1 would the resonant serial circuit be reactive (no, it was off of fr resonance), capacitive or inductive? I could do the calculations for theta and find out and I knew verbatim it was capacitive but I was having problems grasping the concept until I made the connections between what actually happened to the capacitor or coil as the frequency changed. Once Crut jogged my memory with his brief note it all fell into place for me. It was just the very basic reaction of a cap or coil to AC frequency. Simple to predict once that missing piece of the puzzle fell into place for me. Thx again guys I really do greatly appreciate your help!
 

danadak

Joined Mar 10, 2018
3,578
A way to think of it as F drops the Xc rises, at some point it is >> Xl,
so the dominant current is in the L. The opposite is true, if F rises
the Xl Z rises, and the Xc keeps dropping, more and more current
is C related.



Same logic, inverted, applies to series resonant. As F rises
Xc drops to the pouint it is essentially zero. But at same time Xl
has been rising so it becomes the dominant element controlling
waht happens in the network. Invert this for F dropping.



Regards, Dana.
 

crutschow

Joined Mar 14, 2008
23,365
I am not sure i like that line of reasoning because the individual deviations dont tell the story it's the sum (or difference)
Liking it or not is your choice, but you can do the calculations for the impedance change with frequency if you don't believe what I said.
 

MrAl

Joined Jun 17, 2014
6,500
Aren't those labels backwards?!?! If
Xc > Xl it is capacative not inductive
Hi,

Here is probably the best way to look at it because we also like to know what changes when we increase C or decrease C, or increase or decrease L. Note the graphic "Z" is really "|Z|" which is the magnitude of Z.
Note that as C increases, the graph opens up wider, while when L increases, the graph closes more.
In all cases when we reach resonance (w0) we see only resistance as the impedance, and that is the lowest point on the graph.

Another interesting point is that if we look at the current and voltage in both C and L we see it changing. But how can it change even at resonance when we have just an impedance of R? C and L continuously exchange energy, where L has it for some time then C gets it, then back again, over and over again, with no loss. The only energy loss occurs in R. That's something to really think about.

Note also the graph is kind of old, and the x axis goes from w0/2 to w0 then to 2*w0.
See the other graphic for a better drawing.

SeriesResonancePhysical-1.jpg
 

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