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Resonance

D

Deniz

Jan 1, 1970
0
-->------L--
I | |
C R
| |
= =

Circuit seen above (series RL and a parallel C) is supposed to
resonate somehow. There is a current source driving the circuit (since
I thought that a voltage source driving the circuit would be useless).

Overall impedance (Zo) is calculated as: 1/(R/(w^2*L^2 +
R^2)+j*w*(R^2*C - L + w^2*L^2*C)/(w*L^2 + R^2))

Now in order to maximize abs(Zo) we let imaginary part to be 0, by
letting wo = [1/LC - (R/L)^2]^0.5 (here i assume that abs(Zo) is
maximum when imaginary part is 0). Since Zo is maximized, I*Zo
(voltage across the capacitor (Vc)) will reach the maximum, and at wo
the circuit is said to be resonating?

My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

2)If we want voltage across R to reach its maximum, we calculate
complex expression for Vr and if we let Vr's imaginary part to be 0,
we come up with a new resonance frequency w1 = (-R/L)^0.5, which is
meaningless. So can we immediately say or predict that voltage accross
R will reach its maximum when we set w = wo = [1/LC - (R/L)^2]^0.5
(which will reveal that finding resonance frequency has nothing to do
with letting the imaginary parts equal to 0)? If wo is making both the
voltage across Zo and R maximum, what is the reason for this?

(calculating the frequency which will make abs(Vr) maximum seemed to
be impossible, so i made a prediction)

3) Is there a series RLC equivalence of this above circuit (for Vr or
Vc)? If there is no such equivalence, how is parallel RLC equivalent
of this circuit calculated (for Vc and Vr) ?
 
C

CBarn24050

Jan 1, 1970
0
Subject: Resonance
From: [email protected] (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <[email protected]>
My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.

2)If we want voltage across R to reach its maximum, we calculate
complex expression for Vr and if we let Vr's imaginary part to be 0,
we come up with a new resonance frequency w1 = (-R/L)^0.5, which is
meaningless. So can we immediately say or predict that voltage accross
R will reach its maximum when we set w = wo = [1/LC - (R/L)^2]^0.5
(which will reveal that finding resonance frequency has nothing to do
with letting the imaginary parts equal to 0)? If wo is making both the
voltage across Zo and R maximum, what is the reason for this?

Maximum voltage accros R is at DC.

(calculating the frequency which will make abs(Vr) maximum seemed to
be impossible, so i made a prediction)

Maximun impedence for a parallel tuned circuit is at resonance.

Yes, but here the minimum impedence occurs at resonance.
 
S

Spajky

Jan 1, 1970
0
-->------L--
I | |
C R
| |
= =

Circuit seen above (series RL and a parallel C) is supposed to
resonate somehow. There is a current source driving the circuit (since
I thought that a voltage source driving the circuit would be useless).

could be driven also with voltage IMHO & rappresents a impedance step
down transformation @ resonant frequency; R is just a load ...
 
T

The Phantom

Jan 1, 1970
0
NO, the enrgy in both components are equal but opposite.

What does it mean for the capacitor and inductor to have opposite
energies?
2)If we want voltage across R to reach its maximum, we calculate
complex expression for Vr and if we let Vr's imaginary part to be 0,
we come up with a new resonance frequency w1 = (-R/L)^0.5, which is
meaningless. So can we immediately say or predict that voltage accross
R will reach its maximum when we set w = wo = [1/LC - (R/L)^2]^0.5
(which will reveal that finding resonance frequency has nothing to do
with letting the imaginary parts equal to 0)? If wo is making both the
voltage across Zo and R maximum, what is the reason for this?

Maximum voltage accros R is at DC.

(calculating the frequency which will make abs(Vr) maximum seemed to
be impossible, so i made a prediction)

Maximun impedence for a parallel tuned circuit is at resonance.

Yes, but here the minimum impedence occurs at resonance.
 
R

Robert Monsen

Jan 1, 1970
0
The said:
What does it mean for the capacitor and inductor to have opposite
energies?

At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
T

The Phantom

Jan 1, 1970
0
At any given time, they don't have 'opposite' energy

CBarn24050 says they do. My question was directed at him; I was
hoping he would explain about these 'opposite' energies.

He also denies that the energy in the cap is max when the energy in
the inductor is 0. I was hoping he could give some details showing
just how this comes about.

, because energy is
 
S

Steve Evans

Jan 1, 1970
0
At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

It is a fact that in a parallel resonant circuit, the impedance is at
a maximum. Is this due to the fact that - at resonance - the energy
flows between the cap and coil are so large as to be able to repell
any current from the external energy source, thereby rendering its
path effectively blocked?
 
C

CBarn24050

Jan 1, 1970
0
Subject: Re: Resonance
From: Robert Monsen [email protected]
Date: 21/11/04 19:50 GMT Standard Time
Message-id: <[email protected]_s04>



At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does not go
from 1 to the other and back again. When I said that they equal but opposite,
what I meant was that they allways add up to zero at any time in the cycle.
 
S

Steve Evans

Jan 1, 1970
0
Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does not go
from 1 to the other and back again. When I said that they equal but opposite,
what I meant was that they allways add up to zero at any time in the cycle.

Zero over an aeveage *whole* cycle, I think you mean.
 
R

Robert Monsen

Jan 1, 1970
0
CBarn24050 said:
Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does not go
from 1 to the other and back again. When I said that they equal but opposite,
what I meant was that they allways add up to zero at any time in the cycle.

I dont' want to put you on the spot, but I think there is energy stored
in resonant systems. Look at the film of the tacoma narrows bridge being
torn apart due to resonant oscillations. Each little nudge from the wind
stores more energy in the resonant system, until it collapses.

And, in electrical systems, the energy does go from the inductor to the
capacitor and back again. At any point, the energy stored in a capacitor is

Uc = 1/2 C * V^2

the energy stored in an inductor is

Ul = 1/2 L * I^2

Thus, the energy is maximum in the inductor when the current in the
system is maximum, and the energy in the capacitor is maximum when the
voltage across it is maximum. However,

Uc + Ul = U

which is the total energy at a given time, which is what I think you
were trying to say.

This is exactly analogous to a mass and spring system in basic physics,
where the energy goes from kinetic energy to potential energy and back
again.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
T

The Phantom

Jan 1, 1970
0
Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy.

When the voltage across a capacitor is greater than zero, is there
not an energy of .5*C*V^2 associated with that capacitor? And
likewise given a current greater than zero in an inductor, is there
not an energy of .5*L*I^2 stored in the inductor? Are these not
*real* energies?

I notice on some of your other posts that you seem to be familiar
with circuit simulators such as Spice. If you simulate the circuit in
the OP's post (I chose C=.01uF, L=25mH and R=1 ohm, for a resonance
freq of 10.066 KHz, and a current source of 1 amp at that freq), and
plot the current in the C and L for about 10 milliseconds, you will
see that as the currents in C and L increase, they stay 180 degrees
out of phase with each other. If you plot .5*C*v^2 and .5*L*i^2 (the
instantaneous power in C and in L), you will see that these quantities
are double frequency sinusoids with a DC component. Let the current
source drive be applied for about 8 milliseconds, then reduced to
zero; plot the *sum* of those instantaneous powers. You will see that
after the 8 milliseconds, the result is very nearly a constant (if you
make R=0 so the Q is infinite, then the sum of the two instantaneous
powers will be dead constant after the current source drive is reduced
to zero). If the current source drive stays on with R=1, it
superimposes a little ripple on the sum of the powers as the stored
energy in C and L ramps up.

(In the low-Q case, where for example, I made R=10k, the sum of the
instantaneous powers is not constant. But that is not the interesting
case.)

So, in fact, (in the infinite-Q case, where R=0) after the current
source drive is turned off, the capacitor and inductor continue to
exchange energy indefinitely. The voltage and current in each are
sinusoids, and the sum of the instantaneous powers is constant. The
law of Conservation of Energy requires this, because without loss
components in the circuit, there is no mechanism for energy to be
lost. The energy is sloshed back and forth between the capacitor and
inductor forever (in the simulator, at least, where it is possible to
have a circuit with R=0, including the wiring and parasitics).

It does not go
 
T

The Phantom

Jan 1, 1970
0
It is a fact that in a parallel resonant circuit, the impedance is at
a maximum. Is this due to the fact that - at resonance - the energy
flows between the cap and coil are so large as to be able to repell
any current from the external energy source, thereby rendering its
path effectively blocked?

Steve, I've been watching your postings as you strive to understand
a somewhat difficult subject--AC circuit theory. I can see that you
are struggling with it, but I admire your perseverance! Keep at it,
and you'll eventually get it. I'll add my input to the help others
have been giving you.

I think it's helpful to realize that what are called two-terminal
circuit elements (R, L, and C are the fundamental components)
*enforce* a relationship between voltage and current. The voltage
*across* a component and the current *through* it are not independent.
It is what two-terminal components do; they establish a relationship
between voltage and current, *for that component only*.

So, if some two-terminal circuit elements are in series, the
current in each of them *must* be identical. If they are in parallel,
the voltage seen (applied across) by each of them *must* be identical.

Thus, if they are in series, the currents must be the same and
*only* the voltages across each can be different. If they are in
parallel, only the *currents* in each can be different; the voltage
seen by each is the same.

For components in series, since the current in all of them is the
same, it makes sense to use current as a reference, and speak of the
phase of the voltages *across* (not to ground) each component with
respect to the current through all of them.

For components in parallel, it is appropriate to use the voltage
*across* them as the reference, and speak of the phase of the current
in each with respect to the voltage across all of them.

Now, since for a C and L in parallel the voltage across the two
components is the same, only the phase of the currents can differ.
The current in one is 180 degrees out of phase with the other, and
when those currents are added by the parallel connection, they tend to
cancel. If the magnitude of the currents is identical, which is what
happens at a frequency such that the reactance of each is the same
(this is resonance), then we get complete cancellation of the currents
(for ideal L and C). Thus the current into the parallel combination
of the L and C is zero, even though we have applied some non-zero
voltage to the two of them. When we have a circuit that has the
property that no current (or very little) is produced with a finite
applied voltage, we say that the impedance of that circuit is high.
It's not that the parallel combination of L and C at resonance repel
the applied voltage. In fact, a current does exist in both the L and
C, but the two currents are 180 out of phase, and completely add to
zero at the connection of the L and C.

The same thing happens in a series resonant L and C circuit, but
with the roles of current and voltage reversed. For a given current
through the L and C, if the applied current is at a frequency where
the reactance of the L and C is the same, then the magnitude of the
voltage *across* the inductor and *across* the capacitor is the same
and since these voltages are 180 degrees out of phase (for ideal
components), they completely cancel (add to zero). Remember that the
current in the L and in the C is the same, since they are in series.
Thus we have a circuit with (almost) no voltage across it even though
a current is passed through it. We say that such a circuit has a low
impedance. This circuit doesn't repel the applied current; it's just
that the voltage *across* one component cancels the voltage *across*
the other, giving a resultant of zero *across* the series combination.

I hope this helps.
 
C

CBarn24050

Jan 1, 1970
0
Subject: Resonance
From: The Phantom [email protected]
Date: 22/11/2004 13:04 GMT Standard Time
Message-id: <[email protected]>
When the voltage across a capacitor is greater than zero, is there
not an energy of .5*C*V^2 associated with that capacitor? And
likewise given a current greater than zero in an inductor, is there
not an energy of .5*L*I^2 stored in the inductor? Are these not
*real* energies?

Just because the energy is not real doesn't mean that it does not exist, it's
just the confusing terms (real & imaginary) that apply to complex (but not
complicated) maths. You would think that they would have come up with something
a bit better by now.

It does not go
What I should have said was that there is no CHANGE in the total energy level
during the cycle.
 
S

Steve Evans

Jan 1, 1970
0
tnx, phantom! Thats the clearest explanation i've come across so far.
there are still a couple of outstandingpoints i need to clear up....
I'll get back in a shrot while!

Steve
 
S

Steve Evans

Jan 1, 1970
0
I dont' want to put you on the spot, but I think there is energy stored
in resonant systems. Look at the film of the tacoma narrows bridge being
torn apart due to resonant oscillations. Each little nudge from the wind
stores more energy in the resonant system, until it collapses.

Yeah, but5 the energy has to come at the *right* moment each time.
it's a bit like pushing a child on a swing. you have to impart the
force at the rpecise time in each arc to get the swing moving with
minimal effort. i think the equivalent term in elctronics is the
'flywheel effect'; get the timing spot-on and you can keep the
oscillation moving with minimal energy input/maximal efficitency.
 
T

The Phantom

Jan 1, 1970
0
Just because the energy is not real doesn't mean that it does not exist, it's
just the confusing terms (real & imaginary) that apply to complex (but not
complicated) maths. You would think that they would have come up with something
a bit better by now.

I think the problem here is that you are trying to analyze the
problem with the *phasor* representation of the voltages on the L and
C. But the question the OP asked about stored energy cannot be
appropriately dealt with from that point of view. You must analyze
the instantaneous time response of the circuit.
What I should have said was that there is no CHANGE in the total energy level
during the cycle.

This is better than saying they add up to zero, because it allows
for the possibility that the sum is non-zero. But, as I explained in
my long post, it is only when the Q of the circuit is infinite (and
there is no drive from the current source) that there is no change in
the total energy level during one cycle. If the current source drive
is in operation, or if R is non-zero, then the total energy level
varies somewhat during a cycle. This cannot be determined from a
phasor point of view, however.
 
R

Rich Grise

Jan 1, 1970
0
At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

They're 90 degrees out of phase, because the capacitor stores energy as
charge, and the inductor stores energy as current. The current hits its
maximum as charge is going through zero, and vice-versa.

But yes, it's the same energy, just being sloshed back and forth. Think of
a spring and a weight. The spring's compression is the voltage, its
spring constant is the capacitance, the weight is inductance, and its
movement is current.

And somebody said something about a series resonant circuit and a parallel
resonant circuit - if you just look at the inner loop, a parallel resonant
circuit is just a series resonant circuit with its ends tied together. So
the difference really depends on your point of view. :)

Hope This Helps!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Just because the energy is not real doesn't mean that it does not exist, it's
just the confusing terms (real & imaginary) that apply to complex (but not
complicated) maths. You would think that they would have come up with something
a bit better by now.

It's definitely real - it's just that it stays in the resonant circuit,
rather than doing work or being thrown away as heat, which is the part
that you pay for. :) This is why power companies like power factor
correction.
What I should have said was that there is no CHANGE in the total energy
level
during the cycle.

Yeah. :)

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
CBarn24050 wrote:
I dont' want to put you on the spot, but I think there is energy stored
in resonant systems. Look at the film of the tacoma narrows bridge being
torn apart due to resonant oscillations. Each little nudge from the wind
stores more energy in the resonant system, until it collapses.

They've determined that the bridge wasn't resonating - it just got blown
down:

http://www.math.umbc.edu/~gobbert/teaching/math101.s2003/reports/Group1Tacoma.doc

Cheers!
Rich
 
T

The Phantom

Jan 1, 1970
0
They're 90 degrees out of phase, because the capacitor stores energy as
charge, and the inductor stores energy as current. The current hits its
maximum as charge is going through zero, and vice-versa.

But yes, it's the same energy, just being sloshed back and forth. Think of
a spring and a weight. The spring's compression is the voltage, its
spring constant is the capacitance, the weight is inductance, and its
movement is current.

And somebody said something about a series resonant circuit and a parallel
resonant circuit - if you just look at the inner loop, a parallel resonant
circuit is just a series resonant circuit with its ends tied together. So
the difference really depends on your point of view. :)

If the L and C are all there is, and you don't make any connection
to the pair, then the series/parallel distinction doesn't exist. See
E. A. Guillemin's classic book on circuit theory. He talked about
what he called a "soldering iron entry" or a "pliers entry" into a
(sub) circuit. The L and C combination looks different to the
observer depending on whether you open the loop (pliers entry) and
make the connection, or leave the loop intact and tack on in parallel
(soldering iron entry).
 
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