Resonance state - Alternating current

[Guillem_dlc]

TL;DR Summary

Resonance state in alternating current

In alternating current, is the resonance state that the phase angle of the current is 0?

 

[Baluncore]

In alternating current, is the resonance state that the phase angle of the current is 0?

If the resonance is of an LC circuit, then the energy is circulating between the electric field of the capacitor, (E = ½ C V²), and the magnetic field in the inductor, (E = ½ L I²). The current in the resonant circuit will therefore be 90° out of phase with the voltage across the resonant circuit.

 

[Guillem_dlc]

If the resonance is of an LC circuit, then the energy is circulating between the electric field of the capacitor, (E = ½ C V²), and the magnetic field in the inductor, (E = ½ L I²). The current in the resonant circuit will therefore be 90° out of phase with the voltage across the resonant circuit.

Then is $\dfrac{\pi}{2}$, right?

 

[Baluncore]

That is a possible approximation to "90° out of phase".
But is the phase of the current to the voltage +90° or -90° ?
The answer will depend on the actual question, and I am still not sure what the resonator really is, how polarity is defined, and how the resonator is driven.

 

[vanhees71]

Which resonance do you mean? Amplitude resonance or energy resonance? For an undamped oscillator you have rather a "resonance catastrophe". Take a pure LC series. Then you have
$$LC \ddot{u} + u=U_0 \exp(\mathrm{i} \omega t).$$
Here $u$ is the "voltage" across the capacitor, and of course only the real part is the physical $u$.

It's clear that the eigenfrequence is $\omega_0=1/(LC)$, and if $\omega \neq \omega_0$ you can make the ansatz $u=u_0 \exp(\mathrm{i} \omega t)$ to solve the in-homogeneous equation since then you simply get
$$u_0 (1-L C \omega^2)=U_0 \; \Rightarrow \; u_0 = \frac{U_0}{1-\omega^2/\omega_0^2} = \frac{U_0 \omega_0^2}{\omega_0^2-\omega^2},$$
which diverges for $\omega=\omega_0$.

For $\omega=\omega_0$ you get as a particular solution for the inhomogeneous equation
$$u(t)=\frac{\omega_0 t}{2 \mathrm{i}} U_0 \exp(\mathrm{i} \omega_0 t),$$
which goes to $\infty$ for $t \rightarrow \infty$. That's the said "resonance catastrophe".

 

[Delta2]

If the circuit has some ohmic resistance yes in the resonance state the current is in phase with the voltage that is applied in the circuit. That is because by definition of resonance, the total complex impedance of a circuit in resonance has only real part and imaginary part zero, hence the phase angle defined as $$\phi=\arctan \frac {Im(Z)}{Re(Z)}=\arctan \frac {0}{Re(Z)}=\arctan 0=0$$
The case where Re(Z)=0 and Im(Z)=0 (like in a pure LC circuit) leads to the undetermined $\phi=\arctan \frac{0}{0}$. I believe post #5 explains very well what happens in this case.

 

[nasu]

Summary: Resonance state in alternating current

In alternating current, is the resonance state that the phase angle of the current is 0?

"In alternating current" is not enough to specify the conditions. There is no general rule for all possible AC circuits.You need to focus on a specific circuit configuration and then you can look at the resonance.

 

[Delta2]

"In alternating current" is not enough to specify the conditions. There is no general rule for all possible AC circuits.You need to focus on a specific circuit configuration and then you can look at the resonance.

I think for a "random" AC circuit made of components R,L,C resonance is defined as the condition where the total complex impedance of the circuit has only real part. With this definition it is true then that the total current is in phase with the voltage source.

EDIT: Except in the case where also the real part is zero...

 

[nasu]

Except for these cases when it si not.

 

[tech99]

If the L, C and generator are connected in series, then first of all consider a known current, such as 1 Amp, through the circuit. The voltage across the L will lead this current by 90 deg and the voltage across the C will lag the current by 90 deg. As the two voltages are in series and out of phase, they subtract from each other. The generator finds itself working into a load which is reactive (voltage 90 deg to the current) and has smaller impedance than either L or C, so it can produce 1 Amp with a smaller voltage. At resonance, when Xc is numerically equal to Lc, the 1 Amp current develops zero voltage across the LC circuit.

Now connect L, C and the generator in parallel, and consider a known voltage, such as 1 Volt across all three items. Now the current in L will lag the voltage by 90 deg and current in C will lead the voltage by 90 deg. The total current will be reactive (voltage 90 deg to the current) and is the difference between these two. At resonance, Xl is numerically equal to Xc, so the current drawn from the 1 volt generator is zero. The generator is seeing an impedance which is infinitely high.

If there is resistance in the circuit, then for the series connection the generator sees only this resistance at resonance, and for the parallel connection it sees a very high resistance, being a magnified version of the actual resistance. So for both cases, at resonance the load seen by the generator is resistive and has zero degrees phase angle between V and I.
[For accuracy I need to mention that for very low Q operation, for a parallel circuit where all the R is in the inductor, zero phase angle and max impedance do not quite coincide].

 

[DaveE]

If the resonance is of an LC circuit, then the energy is circulating between the electric field of the capacitor, (E = ½ C V²), and the magnetic field in the inductor, (E = ½ L I²). The current in the resonant circuit will therefore be 90° out of phase with the voltage across the resonant circuit.

Except at the resonant frequency where it will be in phase. Really, for a theoretically losses circuit, as you sweep the frequency, it will instantly switch from +90^o to -90^o*, and undefined right at the resonance frequency with 0 or ∞ amplitude. Of course in the real world this can't happen, it will smoothly switch from +90^o to -90^o*.
Here's an example:

* edit - or from -90^o to +90^o for the other type.

 

[Baluncore]

Except at the resonant frequency where it will be in phase.

I am sorry, but I think you are looking at the drive to the resonator, not the much greater circulating energy stored within the resonator.

If you look at a parallel LC resonator, with no external connections, stored energy circulates between L and C. The current through the inductor must be in quadrature with the inductor voltage. Obviously, the same holds for the parallel capacitor. V=L·di/dt necessitates quadrature, as does the integral of charge in the capacitor. At resonance, those two equations work together in quadrature.

If the resonator was to be driven by an energy source at the resonant frequency, the drive current to the resonator would be in phase with the drive voltage. But that drive is not part of the resonator. It simply makes up for energy lost in the resistance of the resonator. For drive at frequencies away from resonance, the phase of the drive current will approach quadrature with the drive voltage.

 

[Delta2]

@Baluncore I think the OP implies that there is external drive to the circuit. We usually talk about the phase of the current with respect to the drive voltage and not with respect to the individual voltages in the capacitors or inductors of the circuit.

 

[Baluncore]

@Delta2 We can all guess at what the OP might have been thinking.
There are several possibilities.
I am simply explaining why there are multiple answers.

 

[Delta2]

@Delta2 We can all guess at what the OP might have been thinking.
There are several possibilities.
I am simply explaining why there are multiple answers.

In my opinion you answers regarding what's happening in a self resonating LC circuit is slightly off topic. But ok that is just my opinion :D.

 

[Baluncore]

In my opinion you answers regarding what's happening in a self resonating LC circuit is slightly off topic.

Likewise, I consider your focus on the missing drive circuit to be off topic, but so long as we acknowledge our different viewpoints, we have no disagreement, just a poorly written, incomplete question.

 

[DaveE]

I am sorry, but I think you are looking at the drive to the resonator, not the much greater circulating energy stored within the resonator.

If you look at a parallel LC resonator, with no external connections, stored energy circulates between L and C. The current through the inductor must be in quadrature with the inductor voltage. Obviously, the same holds for the parallel capacitor. V=L·di/dt necessitates quadrature, as does the integral of charge in the capacitor. At resonance, those two equations work together in quadrature.

If the resonator was to be driven by an energy source at the resonant frequency, the drive current to the resonator would be in phase with the drive voltage. But that drive is not part of the resonator. It simply makes up for energy lost in the resistance of the resonator. For drive at frequencies away from resonance, the phase of the drive current will approach quadrature with the drive voltage.

Don't be sorry, I totally agree with you. If you restrict the analysis to the natural (transient) response for an undriven resonator with non-zero ICs, you are correct. But that isn't a common interpretation of the "phase shift" question IMO.

Summary: Resonance state in alternating current

In alternating current, is the resonance state that the phase angle of the current is 0?

In an undriven series resonant circuit there is only one current, you must assume some external reference for the phase question to make sense. Because the only internal variables to reference are the voltages of the components, which is a trivial question; i.e. "what is the phase shift between the capacitor voltage and its current?"

In an undriven parallel resonant circuit (with losses), there are three currents (two independent currents) and a single common voltage. In that case you are correct, but again, the phase shift question for currents is trivial. They share a common voltage so it is almost the same; i.e. "what is the phase shift between the capacitor current and the inductor current when they have the same voltage across them?" It is also poorly phrased for the parallel case since they didn't think it important to specify which current.

This sort of question is rarely raised unless there is a driving source so it can either make sense or be non-trivial. I think the rest of us may have been guilty of answering the more interesting and common interpretation here.

Finally, the last few post are a clear illustration of why I think all EE questions should include a schematic. But, we all know that's not likely to happen.

 

[DaveE]

For drive at frequencies away from resonance, the phase of the drive current will approach quadrature with the drive voltage.

You mean kind of like in the graph I posted?

 

[Baluncore]

You mean kind of like in the graph I posted?

Yes. You are looking at the drive to a resonator.
Off-frequency, the resonator will appear to be reactive, and so will be circulating energy between the driver and the resonator.
At resonance, the resonator will appear to be resistive, so energy will flow only towards the resonator, to satisfy the internal real resistive losses.

 

[Guillem_dlc]

Okay, I get it.

The exercise in question was as follows: an ac generator of emf $\varepsilon (t)=150\cos (2\pi ft)$, feeds a series RLC circuit with $R=40\, \Omega$, $L=50\, \textrm{mH}$ and $C=50\, \mu \textrm{F}$. Determine at what frequency $f$ the generator will have to work at for the circuit to be at resonance.

Ressonance $\rightarrow X_L=X_C$
$$L\omega =\dfrac{1}{C\omega}$$
$$\omega =\dfrac{1}{\sqrt{LC}}\rightarrow f=\dfrac{\omega}{2\pi}=\dfrac{1}{2\pi \sqrt{LC}}=100\, \textrm{Hz}$$

 

[Baluncore]

XL cannot equal XC, since XC is negative.
The condition for resonance is; XL + XC = 0 ;

 

[Guillem_dlc]

XL cannot equal XC, since XC is negative.
The condition for resonance is; XL + XC = 0 ;

Thanks!

 

[vanhees71]

For the un-driven oscillator we have
$$u(t)=u_0 \cos(\omega_0 t) + \frac{i_0}{\omega_0 C} \sin(\omega_0 t)$$
and
$$i(t)=C \dot{u}(t) = -C u_0 \omega \sin(\omega_0 t) + i_0 \cos(\omega_0 t)= C u_0 \cos(\omega_0 t-\pi/2) + i_0 \sin(\omega_0 t-\pi/2),$$
i.e., the current is a phase of $\pi/2$ behind the phase of the voltage on the capacitor.

That's intuitive, because suppose in the beginning there's no current and the capacitor at a voltage $u_0$. When closing the switch the current needs time to be built up against the inductance of the coil, because the magnetic field building up in the coil is such as to hindering this current to build up ("Lenz's rule"). So it takes time for the current to get its maximum value, and thus the current is behind the voltage on the capacitor. The calculation shows it's behind by a quarter of the period, corresponding to the phase shift $(-\pi/2)$.

 

[Guillem_dlc]

Okay, okay, I understand.

Thank you all very much!