Resonance

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

Maximum voltage accros R is at DC.

Maximun impedence for a parallel tuned circuit is at resonance.

Yes, but here the minimum impedence occurs at resonance.

What does it mean for the capacitor and inductor to have opposite energies?

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

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

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.

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?

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.

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.

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

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

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)

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

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

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.

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.

What I should have said was that there is no CHANGE in the total energy level during the cycle.

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.

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.

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

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.

Well, they're *each* 90 degrees out of phase, so they're in complete antiphase (180') WRT each other. The cap's eneryg is stored in an electric field; the inductor's is stored in a magnetic field and when one's at a maximum, the other's at a minimum and vice versa.

I think we are talking about energy here, and it is true that the time functions of the two energies are sinusoids, 180 out of phase, but the reason is slightly more complicated, Steve. The current in L and C are each 90 degrees out of phase with the reference (voltage), in opposite directions (so to speak), so the two currents are indeed

The detail Rich is missing is that if you plot the energy in L and C separately, you will see that the energy vs. time plot is a *double* frequency function, compared to the voltage or current. This is because the energy involves the *square* of the voltage or current (for C or L), which is always positive regardless of whether the voltage (or current) is in the positive or negative direction. Remember your trigonometry, specifically the formula: SIN^2(x) = (1 - COS(2x))/2 When you square a sinusoid, you get a double frequency sinusoid plus a constant (the constant is the *average* energy). When you look at the squares of two sinusoids that are 90 deg out of phase, you get a couple of double frequency sinusoids that are *180* out of phase with each other. When you add these two double frequency sinusoids (plus their constant terms), the sinusoidal portions cancel and the contants add to give a constant equal to the total energy in L and C.

not go

opposite,

cycle.

---------- There is real energy but it is being shuttled back and forth. The average power over a cycle is 0 so that the total energy input during the cycle is also 0. At any instant in time the power is not 0 nor is the sum of the energies stored in the L and C.

Where does the energy come from? The conditions when operating at steady state are not the same as when the circuit is first energised. There is a transient period in which energy is initially stored in the capacitor and inductor (not necessarily the same in each). You can't handle this period with the concepts of AC steady state analysis (phasors, reactive, etc) but need to consider the differential equations involved.

It's not exponential (though it does look somewhat like it), it's hyperbolic; it's because the reactance of capacitors vary as 1/wC and inductors as wL. The current in a cap is v/(1/wC) which is v*wC; the current in an inductor is v/(wL) which will plot as a hyperbola. If you plot y=1/x you will see a curve like you're describing.