Notice that when Z_T goes to positive infinity, there is a frequency at which it transitions very sharply to negative infinity. In theory, this occurs instantaneously and there is no frequency in between where it is zero. That zero is an artifact of making a graph that is just connecting dots between closely sampled points. That nearly vertical line should really be perfectly vertical and should be dashed to indicate that it is not actual values.

Okey, so when it goes from infinite to -infinite then this zero like here :



Isn't a zero. But isn't there a frequency very small like 36,00000001 Hz that makes it equal 0 reactance in this point ? From this transition from infinity to - infinity ?

But what does - infinite indicates ? - infinite is also a resonance ?
So basically there are 2 zero ?

Also I asked you what does the circuit sees.
If the total reactance is 0 then it looks like that ?



also how would it look like in equation ?

Z = R +jX right ? so Z = R + j(infinite) ? and Z = R + j(0) ? If yes then how from Z = R + j(infinite) I get Z = R or how from Z = R + j(0) i get Z = 0 because all current goes through this short circuit. Or am I again miss understanding something ? Because this is what the resonance is showing right ?

Edit :

Also in post #35 I've mentioned that some users looked at this complex circuit using series resonance or parallel resonance. But you mentioned that I should be looking at it like that. So how is it exactly ?

EDIT 2:

I tried the same thing do in a simple example (the same way you have calculated things).



I've used some ready to use equations because this is simple parallel resonance. It is not a complex circuit so I could use simple equations.
so the resonance frequency should be 40 Hz. But how do I read it in this diagram which is the resonance frequency ?
And how to I write is as an equation ? Z = R+jX ? = parallel resonance = > Z = R + j*infinite ?

Also when I give it into simulation the voltage source doesn't sees this circuit as a pure resistive https://tinyurl.com/2lrjm72c but the phase is shifted. Even though this is a 40Hz resonance frequency.

 

If you redo you spreadsheet calculations to from, say, 39.5 Hz to 40.5 Hz, and decrease the frequency increments, say, to 0.001 Hz, you will see that the peaks grow huge and that it slams down from positive huge to negative huge right at the ~39.789 Hz resonant frequency. But if one of your same points turns out to be too close to the resonant frequency, you will get a division by zero error. The impedance AT that frequency is not zero, regardless of what drawing a line from the last positive huge value to the first negative huge value makes it look like. There is NO zero crossing there, it is an artifact of how the plot was generated.

The reactance at that point is neither positive positive nor negative infinity, either; it is undefined. But this is only an intellectual curiosity because we can't make pure inductances in parallel with pure capacitances. If we try to get too close to doing so, our model breaks down and we have to model the circuit to take into account parasitic effects that our simple model ignores.

 

The reactance at that point is neither positive positive nor negative infinity, either; it is undefined. But this is only an intellectual curiosity because we can't make pure inductances in parallel with pure capacitances. If we try to get too close to doing so, our model breaks down and we have to model the circuit to take into account parasitic effects that our simple model ignores.

Okey so my job by looking at this diagram is to looking for zero impedance or infinite (infinite is looking like quick transition from positive peak to negative peak).

So basically when I have a circuit and I want to calculate the resonance in this circuit I basically calculate the Z_total including resistance and then in the calculation I just put Z_total = Re{Z_total} + j*Im{Z_total} ? For example I've found that Im{Z_total} is equal infinite so I have Z_total = Re{Z_total} + j*infinity ?

EDIT :

We were using XL and XC which are the module of ZL and ZC.
Can I also do it like that ? :



And then Im{Z_LC} and I don't have this "j".

I also tried with this circuit. Please give it a look because maybe I made a mistake.
I tried using instead of XL and XC I used ZL and ZC.



Those are my calculations.




I should have a Im{Z_Total} = infinity but the problem is that I have something different. It is a simple parallel resonance so for impedance value of imaginary part should be infinite. Where did I make a mistake ? In previous circuit without rezystor in parallel the imaginary was infinite.
It looks like I have resonanse somewhere in 4Hz Because it changes from positive to negative rappidly.



The resonance should be at 39 Hz but here it doesn't look like that.

PS.
In your diagram with X_T, X_A, X_B,X_C.
You said that 37 Hz is correlated with X_A but also 50Hz is correlated with X_C. Is it a coincidence ? X_C min is exactly 50 Hz .
So X_A is in resonance but other branches are not ?

 

I'm a bit confused how to interpret some stuff here.

 

You need to start tracking your units. You messed up well before you got to the final equation you are trying to make sense out of and if you get in the habit of tracking your units you could have caught it and corrected it right then and there.

Look at the following line in your work.


Each of the red boxes has units of impedance. The (ω^2)(LC) factor is dimensionless since it is really (ωL) / (1/(ωC)), which is impedance over impedance.

Look at the middle expression. The numerator has units of impedance-squared and the denominator has units of impedance, so overall the expression has units of impedance. This is good, because we are solve for some Z.

Now look at the rightmost expression. The numerator works out to impedance-cubed (I also added a right-paren that you left off). But look at the denominator. The first term is (ωL)^2, so that's impedance-squared, which is good because if the entire denominator is impedance-squared, that will leave an overall unit of impedance after it divides the numerator. But in second term (in parens) the first term has R and so does the second; but you can't add impedance-squared to impedance. Thus you know that a mistake has been made in going from the middle expression to the right expression. There is NO point going any further, because from this point on you are guaranteed to get a wrong result.

If you get in the habit of checking the units at every step while working the problem, you will catch the overwhelming majority of mistakes you make (which we ALL make) immediately. The same goes for when it comes time to plug values into the equation. Include the units and work the units. Do NOT just do a bunch of arithmetic on a bunch of numbers and then tack on the units to the answer that you expect (really, just hope) that it has. This is your last opportunity to catch mistakes that affected the units (and most, but not all, mistakes mess up the units).
---------------------------------
In the discussion where I mentioned 37 Hz and 50 Hz, I only needed to specify the frequency close enough to make sure it was known which spike I was referring to. They most certainly are NOT exact.
---------------------------------
Look at your circuit and think about the limiting conditions.

At very low frequencies, the capacitor is going to look like an open circuit while the inductor is going to look like a short circuit, so your overall impedance at DC should look like just R (namely the R in series with the source).

At very high frequencies, the inductor is going to look like an open circuit while the capacitor is going to look like a short circuit, so again your overall impedance as the frequency goes toward infinity should look like just R.

At some intermediate frequency, the capacitor and inductor will be in parallel resonance, which means that they will look like an infinite impedance. But from the source's perspective, that infinite impedance is in parallel with R and so that part of the circuit will look like just R, in which case the overall impedance seen by the source will be 2R at resonance.

In the overall impedance, at resonance the reactance will not go to infinity, but rather it will go to zero. So you want to look for where the either the numerator goes to zero or the denominator goes to infinity.

 

You need to start tracking your units. You messed up well before you got to the final equation you are trying to make sense out of and if you get in the habit of tracking your units you could have caught it and corrected it right then and there.

Look at the following line in your work.

Each of the red boxes has units of impedance. The (ω^2)(LC) factor is dimensionless since it is really (ωL) / (1/(ωC)), which is impedance over impedance.

Look at the middle expression. The numerator has units of impedance-squared and the denominator has units of impedance, so overall the expression has units of impedance. This is good, because we are solve for some Z.

Now look at the rightmost expression. The numerator works out to impedance-cubed (I also added a right-paren that you left off). But look at the denominator. The first term is (ωL)^2, so that's impedance-squared, which is good because if the entire denominator is impedance-squared, that will leave an overall unit of impedance after it divides the numerator. But in second term (in parens) the first term has R and so does the second; but you can't add impedance-squared to impedance. Thus you know that a mistake has been made in going from the middle expression to the right expression. There is NO point going any further, because from this point on you are guaranteed to get a wrong result.

If you get in the habit of checking the units at every step while working the problem, you will catch the overwhelming majority of mistakes you make (which we ALL make) immediately. The same goes for when it comes time to plug values into the equation. Include the units and work the units. Do NOT just do a bunch of arithmetic on a bunch of numbers and then tack on the units to the answer that you expect (really, just hope) that it has. This is your last opportunity to catch mistakes that affected the units (and most, but not all, mistakes mess up the units).

I see the error.
I forgot to increase one value, here is correction.




Now it really looks like there is resonance :




Is it correct ?
This is a parallel resonance, but here the resonance is min which is 0 or is it infinite ? Because it is going from positive value to negative value like in the infinite examples.
So the Im{Z_Total} = 0 ? Even if it's parallel resonance ?

At some intermediate frequency, the capacitor and inductor will be in parallel resonance, which means that they will look like an infinite impedance. But from the source's perspective, that infinite impedance is in parallel with R and so that part of the circuit will look like just R, in which case the overall impedance seen by the source will be 2R at resonance.

In the overall impedance, at resonance the reactance will not go to infinity, but rather it will go to zero. So you want to look for where the either the numerator goes to zero or the denominator goes to infinity.

That's why I don't get it. It is simple parallel resonance but still Im{Z_total} is 0 and not infinite.

If I add in parallel R and X. It would look something like that : 1/Z_RLC (note this is only parallel R||XC||XL) = 1/R + 1/XC + 1/XL.
It is parallel resonance so 1/Z_RLC =1/R + 1/j*infinite. So how from this I get Z_RLC = R ?
And then Z_total = Z_RLC + R = 2R.

I don't get it.


So how I get Z = R or like in my example Z = 2R when the imaginary value is infinite ? In this picture it is said that LC is infinite. 1/Z = 1/R +1/j*infinite? and from this I get Z = R and the imaginary part disappears ? I can understand if the imaginary part value was 0. Because Z = R +j*0 then it is Z = R. But I don't undestand why infinite is disappearing.

For example here :


If instead of 100 ohm I give infinite then the value isn't 0.

 

The overall circuit you have is NEITHER a series circuit nor a parallel circuit. It is a combination of the two. So you can't expect it to behave like either a series circuit or a parallel circuit.

Let's tear it apart. I'm going to call the resistor in series with the source Rs and will call the other one R.

The overall circuit is a series circuit in which Rs is in series with an impedance Z1, consisting of an RLC parallel resonant circuit.

The overall impedance is

Z_T = Rs + Z1

Z1 is a simple, parallel RLC circuit so we could look up the equations for the impedance for Z1, but we can also break it down further and look at it as the parallel combination of R and the LC

Z1 = R || Z2

where Z2 is the impedance of the parallel LC components.

What happens to Z2 as its resonant frequency is reached? Answer: It goes to infinity. Coming from lower frequencies, the reactance is positive (inductive), and at higher frequencies it is negative (capacitive).

Now, think back to DC circuits when you had two resistors in parallel. What happened as one of them went to infinity? Answer: It disappeared, leaving only the other one.

The same thing happens here. As Z2 goes to infinity, it has less and less effect on Z1. At resonance, it has no effect. But in order for it to have no effect on Z1, that means that the reactance of Z1 has to go to zero.

Note the HUGE distinction here. As the reactance of Z2 goes to infinity, the reactance of Z1 goes to zero.

You MUST keep in mind WHICH impedance you are considering.




, .

 

What did I mean about minimum and maximum in your diagram is that X_T minimum is the same as X_A minimum at 39 Hz and X_T min is the same as X_C min which is at 50 Hz (minimum here I mean that the reactance is 0), is it a coincidance that individual branches that has a min resonance are the same as X_T minimum ?

You've mentioned that it should be and overall resonance that not individual branches are in resonance.

I mentioning it because Jony also mentioned in his post that those are series resonance. And he calculated those individual branches that are in resonance makes the whole circuit also resonance. Maybe this is a coincidence but I'm curiouse if Jony was right. Because you mentioned that there should be a combination of all branches not an individual resonance.
So if the branches as a combination acts like 0 or infinite reactance does it mean that all branches exchange the energy ? Even if they are not in resonance individualy and they still don't drain energy from the source ? For simple RLC it is easy to see that LC doesn't drain energy from source and it is in resonance. But when there are 2 branches that are not in resonance but as a whole are a resonance it is wierd. How does the energy goes fluently from one branch to another. Or maybe a good question is, how 2 branches that are not in individual resonance doesn't drain energy from source and remains in overall resonance.

Z1 = R || Z2

where Z2 is the impedance of the parallel LC components.

What happens to Z2 as its resonant frequency is reached? Answer: It goes to infinity. Coming from lower frequencies, the reactance is positive (inductive), and at higher frequencies it is negative (capacitive).

Okey so Z2 goes to infinity. But as you mentioned that Z2 has no effect on R in parallel that Z1 end up with only R.

But Z2 is an imaginary value and R is a real value. Like in the example I gave, there was 1 ohm in parallel with j*100 in the end I had an equation for Z2 which was = (100+j*10)/101. The impedance consist now the real number and imaginary number.
For infinity example the imaginary disappears and only real number is avaible.
Shouldn't be it like this :



Or If I didn't have R in parallel with Z2.
Then I have only Z = R + j*ZC||ZL it goes to this point : Z = R + j*infinity. In practice I know that this is a open circuit. But in this equation it doesn't look like that? What do I mean that module from this equation : Z = R + j*infinity is really a Z = infinite. But this is a module what about the phase shift ? Because there is still a imaginary value right ? So calculating the current would look like that : I = Vin/Z. It ends up with current with a phase shift because Z still consist imaginary value.

The same thing happens here. As Z2 goes to infinity, it has less and less effect on Z1. At resonance, it has no effect. But in order for it to have no effect on Z1, that means that the reactance of Z1 has to go to zero.

So my diagram which shows the resonance of Z1 is correct as I can assume ?

 

What did I mean about minimum and maximum in your diagram is that X_T minimum is the same as X_A minimum at 39 Hz and X_T min is the same as X_C min which is at 50 Hz (minimum here I mean that the reactance is 0), is it a coincidance that individual branches that has a min resonance are the same as X_T minimum ?

Look back at that circuit. As seen by the source, there are four parallel branches, R and what I called X_A, X_B, X_C. What do you know about the impedance of a bunch of parallel branches if any one of them goes to zero?

You've mentioned that it should be and overall resonance that not individual branches are in resonance.

In general, yes. But in THAT circuit, because the source sees a bunch of parallel branches, if any branch goes to zero impedance, the overall circuit goes to zero.

Look at the peaks. They do NOT occur where peaks of X_A or X_C occur. Why not? It's because, in a parallel circuit, if a branch goes to infinite impedance, it effectively removes itself from the circuit and thus has no effect. For THIS circuit, this is where the three branches have to work together in order to reach an overall peak impedance.

 

In general, yes. But in THAT circuit, because the source sees a bunch of parallel branches, if any branch goes to zero impedance, the overall circuit goes to zero.

Okey.

I have a question more ideological.
LC components while in resonance doesn't take any energy from voltage source. But If I had RLC without voltage source and with charged C, then the voltage will drop pretty fast but it is oscilating in resonance frequence why ?

Also if RLC in resonance the LC component doesn't need energy from voltage source then why if I take of voltage source and short circuit it then the voltages drops. So the LC components need energy from voltage source even if they are resonating.

And last thing, when I see this this branch being in resonance and it's impedance is equal 0. Then it means that it doesn't need any energy from the voltage source, but the other branches aren't in resonance and they need additional energy from voltage source. So how does it work that overall it is 0 impedance while other branches are not in resonance but one branch is in resonance (0 impedance). Does this resonant branch provides extra energy to other branches (because they need additional energy from voltage source) so if they take energy from resonant branch then they don't need energy from the current source/voltage source so thanks to that resonant branch the other branches takes energy from the resonant one hence the voltage source don't provide any extra energy to any branches so overall it is in 0 impedance ?

 

Okey.

I have a question more ideological.
LC components while in resonance doesn't take any energy from voltage source. But If I had RLC without voltage source and with charged C, then the voltage will drop pretty fast but it is oscilating in resonance frequence why ?

I assume we are talking about a parallel RLC circuit?

Forget about the R for the moment and just consider an ideal LC circuit that has some energy in it. The energy is transformed back and forth between between the capacitor (as a voltage across it) and the inductor (as a current through it). At any given moment the total energy between the two must be a constant value, since there is no mechanism for it to be transformed to anything else. Furthermore, the voltage across both must be the same and the current through each must be the same. Crank through the math while applying these constraints and you discover that the only way for it to work is for the voltage and current to oscillate at the a specific frequency.

Now put a resistor across them. Any time there is a voltage across the resistor, a current will flow in it and energy will be dissipated as heat. Since the LC has a sinusoidal voltage appearing across it, on each half cycle some of the energy stored in the LC will be transferred to the R and dissipated as heat, thus reducing the amplitude of the oscillations exponentially.

Also if RLC in resonance the LC component doesn't need energy from voltage source then why if I take of voltage source and short circuit it then the voltages drops. So the LC components need energy from voltage source even if they are resonating.

I'm again going to assume that you are talking about a circuit in which everything is in parallel. It would be really nice if you wouldn't make your readers make basic assumptions about what you are talking about.

If short circuit the voltage source, then you FORCE the voltage across each of the RLC components (as well as the source) to be zero. That will immediately dump whatever energy is on the capacitor, but the energy in the inductor will be trapped as a persistent DC current.

And last thing, when I see this this branch being in resonance and it's impedance is equal 0. Then it means that it doesn't need any energy from the voltage source, but the other branches aren't in resonance and they need additional energy from voltage source. So how does it work that overall it is 0 impedance while other branches are not in resonance but one branch is in resonance (0 impedance). Does this resonant branch provides extra energy to other branches (because they need additional energy from voltage source) so if they take energy from resonant branch then they don't need energy from the current source/voltage source so thanks to that resonant branch the other branches takes energy from the resonant one hence the voltage source don't provide any extra energy to any branches so overall it is in 0 impedance ?

If we are talking about parallel LC branches, when the impedance of one branch goes to zero, the other branches have no energy in them because they are shorted out by the one whose impedance went to zero. It is clamping zero voltage across them.

 

Forget about the R for the moment and just consider an ideal LC circuit that has some energy in it. The energy is transformed back and forth between between the capacitor (as a voltage across it) and the inductor (as a current through it). At any given moment the total energy between the two must be a constant value, since there is no mechanism for it to be transformed to anything else. Furthermore, the voltage across both must be the same and the current through each must be the same. Crank through the math while applying these constraints and you discover that the only way for it to work is for the voltage and current to oscillate at the a specific frequency.

Now put a resistor across them. Any time there is a voltage across the resistor, a current will flow in it and energy will be dissipated as heat. Since the LC has a sinusoidal voltage appearing across it, on each half cycle some of the energy stored in the LC will be transferred to the R and dissipated as heat, thus reducing the amplitude of the oscillations exponentially.

So the voltage source provides energy to resonating RLC (parallel one as you assumed).

I'm again going to assume that you are talking about a circuit in which everything is in parallel. It would be really nice if you wouldn't make your readers make basic assumptions about what you are talking about.

If short circuit the voltage source, then you FORCE the voltage across each of the RLC components (as well as the source) to be zero. That will immediately dump whatever energy is on the capacitor, but the energy in the inductor will be trapped as a persistent DC current.

So resonating RLC (parallel) needs a voltage source so the LC components take energy ? Even if resonating LC doesn't take energy from voltage source ? Is it the same with series RLC ? Does the Capacitor and Inductor take energy from voltage source even if it's LC component resonating ? Because if the voltage source sees the circuit as purely resistive then LC doesn't take energy from voltage source, but If I shorten the voltage source then the energy from RLC (parallel or series) is going to 0.

If we are talking about parallel LC branches, when the impedance of one branch goes to zero, the other branches have no energy in them because they are shorted out by the one whose impedance went to zero. It is clamping zero voltage across them.

I mean Capacitor and Inductors have some voltage while having 0 impedancje :



Here L and C have voltage even if their impedance is 0.
So I thought that if one of the branches has 0 impedance but still has voltage on them then the other branch also needs to have voltage on them. So is the energy from resonating LC being partly transfered to not resonating branch LC, so the non resonating branch doesn't take additional energy from voltage source ?

 

So the voltage source provides energy to resonating RLC (parallel one as you assumed).

For a parallel RLC circuit, the source must provide the energy that is dissipated in the resistor. Then the energy in the inductor and capacitor is shuttled between them making them invisible to the source, which only sees the R.

I mean Capacitor and Inductors have some voltage while having 0 impedancje :

View attachment 274471

This is NOT a parallel RLC circuit!!!!

The R is in SERIES with the parallel combination of L and C.

That is DIFFERENT and it is going to behave DIFFERENTLY from a parallel RLC circuit.

Here L and C have voltage even if their impedance is 0.
So I thought that if one of the branches has 0 impedance but still has voltage on them then the other branch also needs to have voltage on them. So is the energy from resonating LC being partly transfered to not resonating branch LC, so the non resonating branch doesn't take additional energy from voltage source ?

Go back and look at the basic rules for voltage and current in a circuit. If one branch of a parallel network has zero impedance, then the voltage across that branch is zero. But the definition of parallel branches requires that all branches have the same voltage across them. So it is impossible for one branch to have zero impedance and the other branches to have a voltage across them.

The non-resonating branches are just sitting there looking stupid because the resonating branch is forcing the voltage across them to be zero. They have no energy in them because the voltage across them is being forced to be zero. They are not taking any energy from the voltage source because they have no energy in them because the voltage across them is being forced to be zero.

In your original circuit, Z_A and Z_B are series parallel combinations of capacitors and inductors. The top of the circuit is a parallel LC. There is a frequency at which the reactance of this LC becomes infinite. That makes the reactance of that entire branch go to infinity because the reactance of the inductor at the bottom is added to the inductance at the top. As the frequency goes above this point (the frequency at which the top resonates and goes to infinite impedance), the top of the circuit looks like a capacitor (below that it looked like an inductor). At some point, the effective capacitance of the top is just the right value to resonate with the inductor on the bottom. THIS then behaves like a series resonant circuit and the impedance of this branch goes to zero.

 

For a parallel RLC circuit, the source must provide the energy that is dissipated in the resistor. Then the energy in the inductor and capacitor is shuttled between them making them invisible to the source, which only sees the R.

Okey so this voltage source is only for R to not use the energy from LC.

This is NOT a parallel RLC circuit!!!!

The R is in SERIES with the parallel combination of L and C.

That is DIFFERENT and it is going to behave DIFFERENTLY from a parallel RLC circuit.

I know but this series RLC where LC has 0 impedance.
So I thought that maybe they act similar. Because it is a short circuit, and a short circuit should have 0V.

https://tinyurl.com/2ma9f697 - something like here ?



One branch has impedance 0 and the other one is not.

 

You are looking at the voltage across the individual components (and I can't tell to what degree phase information is being taken into account or if these are instantaneous values at one particular moment in time, or what).

Look at the voltage across the entire branch. In both cases it is ~20 mV.

But I don't understand the values there. The resonant frequency of the 15uF cap and 1 H coil should be around 41 Hz, not the 1 Hz shown for the source (the right branch should resonate at about 29 Hz).

A series LC branch can have significant voltage across the L and the C, corresponding to significant energy stored in them, and yet have zero net voltage across the LC combination.

What is the amplitude of that source? 1 V? Let's assume it is. Since the impedance of the circuit to the right of the resistor is zero, that means that the current in the 10 Ω resistor is 100 mA. This current has to be flowing through the right hand components. For branches not in resonance, the current's will be small because it is a small voltage across a not-so-small impedance. So nearly all of the current will be flowing in the resonant branch. That will result in significant voltage across the L and the C individually, but not across the combination of them.

 

But I don't understand the values there. The resonant frequency of the 15uF cap and 1 H coil should be around 41 Hz, not the 1 Hz shown for the source (the right branch should resonate at about 29 Hz).

O my god I cut it wrong, there is 41 Hz.

What is the amplitude of that source? 1 V? Let's assume it is. Since the impedance of the circuit to the right of the resistor is zero, that means that the current in the 10 Ω resistor is 100 mA. This current has to be flowing through the right hand components. For branches not in resonance, the current's will be small because it is a small voltage across a not-so-small impedance. So nearly all of the current will be flowing in the resonant branch. That will result in significant voltage across the L and the C individually, but not across the combination of them.

Oh so the L and C will have a voltage in this resonant branch but the combination is equal close to 0V (for ideal resonance it is 0V).
And the other branches that are not in resonance will get a small amount of current.
So the same goes with the example with many branches and with the circuit RLC with 2 series branches (the one I did a wrong cut)

 

Correct.

At 41 Hz the capacitor has a reactance of -258.8 Ω while the inductor has a reactance of +257.6 Ω, so this is very close to resonance. Let's call it 258 Ω each.

For a 1 V AC source at 41 Hz, the current will have an amplitude of about 100 mA. The amplitude of the voltages across the capacitor and inductor will each be

|V_C| ~= |V+L| = (100 mA)(258 Ω) = 25.8 V

Because both components have the same current through them, and because the voltage across a capacitor lags the current by 90° while the voltage across an inductor leads the current by 90°, the voltage across the capacitor will always be 180° out of phase with the voltage across the inductor. If they are the same magnitude, they therefore cancel out, even though each by itself significantly greater than the voltage of the source.

Assuming you captured your voltage values near the max, that would mean that your voltage source probably had about a 5 V amplitude?

 

Assuming you captured your voltage values near the max, that would mean that your voltage source probably had about a 5 V amplitude?

Something like that.

May I ask one more question ?

In the circuit with many branches we talked about.
This one branch was in resonance and it had 0 impadance. So it acted like series resonance this branch with 0 impedance.

Is it possible that combination of 2 branches can have 0 impedance ? Or only single branches can have 0 impedance and parallel elements can have infinite impedance ?

 

WBahn
Can 2 branches combined have 0 impedance ?
If not then only single branches (with LC) can have 0 impedance, because this single branch with L can't have 0 impedance.
But I wondered if 2 branches combined can have 0 impedance like in series resonance. I mean more like Parallel connecting 2 branches can create new frequency for 0 impedance in which it results with new series resonance? Or is it impossible because those branches are in parallel?

This is the last question.

EDIT:

What is the frequency for infinite impedance ? I want to test in here : https://tinyurl.com/2n2yg2gz

 

WBahn
Can 2 branches combined have 0 impedance ?
If not then only single branches (with LC) can have 0 impedance, because this single branch with L can't have 0 impedance.
But I wondered if 2 branches combined can have 0 impedance like in series resonance. I mean more like Parallel connecting 2 branches can create new frequency for 0 impedance in which it results with new series resonance? Or is it impossible because those branches are in parallel?

This is the last question.

You are asking some good questions -- the kinds of questions that far too many students don't bother to even ponder. So that's good.

The next step is to start trying to answer your questions for yourself to the degree that you can. You've been doing that somewhat in trying to think things through. So that's good.

The next step is to delve into the math -- the math is what it is all about.

So let's see what the math tells us.

Imagine we have two impedances, Z1 and Z2, in parallel. What is the combined impedance?

Z_T = (Z1)(Z2) / (Z1+Z2)

So under what conditions can Z_T go to zero. Keep in mind that each impedance is generic, meaning that it could have both a real part and an imaginary part. The imaginary part can be positive or negative or zero. In some circuits, the real part can also be negative, but let's ignore those and assume that the real part is always nonnegative.
These are the kinds of questions that you need to start trying to answer yourself.

The total impedance can go to zero if either the numerator goes to zero or the denominator goes to infinity. This is assuming the other states finite and nonzero. If both go to zero or both go to infinity, we have to look at the specifics much more closely.

The numerator is the product of two impedances, and this can only go to zero if one of the impedances goes to zero. There is no way for two nonzero impedances to make the numerator go to zero.

The denominator, on the other hand, can only go to infinity if one of the impedances goes to infinity, but this gives us the infinity/infinity problem and, in this case, the result is that the infinite impedance goes away leaving us with just the other one (exactly as you would expect given two impedances in parallel when one of them is open-circuited).

So multiple parallel branches can't combine to give a net zero impedance unless at least one of the branches has zero impedance.

But what about the total impedance going to infinity? This can happen if either the numerator goes to infinity or the denominator goes to zero. The numerator can only go to infinity if one of the two impedances goes to infinity and we are back to the infinity/infinity case which we know reduces to just the other impedance. But the denominator CAN go to zero while both impedances remain non-zero and finite, so multiple parallel branches CAN combine to create an infinite impedance even while the individual impedances each have finite values.