Understanding of LC resonance

MrCU204

Understanding of LC resonance [Copy link]

 

 

This circuit has neither branches nor power supply, and because there are more than two components, it can be classified as series connection.

Assume that the length of each side is 10cm, R = 1Ω, when resonant, Xι = Xc = 100Ω,

Here, the wires and all components are considered ideal, so the Q values of L and C remain infinite forever, and R is just a pure and linear resistor.

maychang

[Assume that the length of each side is 10cm]

This assumption is redundant.

The characteristic of [circuit] is that it does not matter the length.

MrCU204

The only slot circuit that can oscillate freely in a radio darkroom (C must be charged in advance) and can induce (and rapidly increase) current through radio waves is the LC. High-order RC and LR bandpass networks are also "selective", but they cannot survive without power.

The usual interpretation of LC resonance is series resonance and parallel resonance, but I always feel that what is written in books and taught by teachers cannot reflect the true nature of LC resonance. When resonating, the voltage drop between L and C must be high, and the current in the loop must be large. Regardless of whether the driver is a power supply or an electric wave, the result is actually the same. The effect of field-induced resonance is the true nature of LC resonance.

The difference between series and parallel is not the fault of components or circuits, but the limitation of power supply characteristics. In fact, we should be thankful that there is no ideal reactance in the world, otherwise, the tuning circuit of the receiver or crystal radio will become a "Tesla coil". Haha, on the other hand, if R is 0Ω, then series resonance is strictly prohibited to be matched with a constant voltage source, and parallel resonance should not be paired with a constant current source. Even if the components are safe and sound, it is not fun to cause lightning and thunder!

MrCU204

maychang posted on 2024-8-24 09:40 [Assume that the length of each side is 10cm] This assumption is redundant. The characteristic of [circuit] is that it does not care about the length.

So it is not the closed circuit but the inductor that absorbs the radio wave (electromagnetic field) energy. I understand.

maychang

MrCU204 posted on 2024-8-24 10:18 So, it is not the closed loop that absorbs the energy of radio waves (electromagnetic fields), but the inductor, right? I understand.

[So, it is not the closed circuit that absorbs the radio wave (electromagnetic field) energy, but the inductor, right? I understand.]

You said it yourself, and added brackets (electromagnetic field). Since it is (electromagnetic field), it is out of the scope of "circuit".

maychang

MrCU204 posted on 2024-8-24 09:59 It can oscillate freely in an anechoic chamber (C must be charged in advance), and can be induced by radio waves (and grow rapidly). The current tank circuit is only LC, and high-order RC ...

[But I always feel that what is written in books and taught by teachers cannot reflect the true nature of LC resonance]

The true nature of LC resonance is revealed by you!

MrCU204

Suppose, I want I=10A, what should I do? !

A single passive loop with more than two components can be directly regarded as a series connection. Without the intervention of a power supply, the current will not have branches and can also be regarded as a series connection.

For series resonance, you can use a 10V constant voltage source or a 10A constant current source, or drive it with radio waves (how strong the electromagnetic field must be), and the result is the same: Vι＝Vc＝1kV.

If a constant voltage source is used, R must be present, otherwise it will be short-circuited. If a constant current source is used, R is optional (if not present, the potential of the constant current source will drop to 0V, which is equivalent to no load).

MrCU204

When the LC tank circuit resonates, the current is extremely large, and the voltage drop between L and C is very high, both in series and parallel.

Compared with series resonance, parallel resonance has a shortcoming, that is, it cannot be realized by field-induced resonance. Because, in the parallel resonance circuit, there must be a difference between the currents of L and C (this difference is the bus current, that is, the reactive load of the power supply). The circuit in the figure has no branches and cannot create this difference.

Therefore, parallel resonance can only be driven by a power supply. If the driver is a constant current source, the voltage drop across L and C can be frighteningly high. As for the constant voltage source, the situation where the power factor is 1 is consistent with the parallel resonance (the difference is that the power factor is mixed with the active load).

MrCU204

maychang posted on 2024-8-24 09:40 [Assume that the length of each side is 10cm] This assumption is redundant. The characteristic of [circuit] is that it does not care about the length.

The circuit on the first floor,

When L and C are both present, those wires are equivalent to the secondary side of the transformer, and their function is only to capture electricity into space. L and C form a resonant load.

If L and R are replaced by wires, leaving only capacitance, the entire loop becomes an inductor. The length and layout of the wires will affect the inductance. This is the case with Hertz 's metal bracelet with an air gap (which has the same function as a capacitor).

MrCU204

maychang posted on 2024-8-24 10:44 [But I always feel that what is written in books and taught by teachers cannot reflect the true nature of LC resonance] The true nature of LC resonance depends entirely on you to reveal...

This is not some new insightful discovery, but simply the comprehension of the neglected parts of classical knowledge.

Resonance is not only driven by constant voltage sources, but also by constant current sources and radio waves. If this circuit is driven by a constant current source, the voltage drop of L and C has nothing to do with their own Q value and is not affected by R.

The electromotive force of the constant current source is proportional to the load impedance, but when the circuit is in a resonant state, the voltage drops of L and C are equal but opposite, and the total is zero. If R is 0Ω, the electromotive force of the constant current source will drop to 0V (equivalent to the standby state).

maychang

MrCU204 posted on 2024-8-25 00:08 The circuit on the first floor, when L and C are both present, those wires are equivalent to the secondary side of the transformer, and their function is only to capture electricity into the space, and L and C form a resonance...

But come again.

[Those wires are equivalent to the secondary side of the transformer.] The secondary side of the transformer only needs an alternating magnetic field, not [electromagnetic waves].

maychang

MrCU204 posted on 2024-8-25 01:10 This is not some new insight, but just a realization of the neglected part of classical knowledge. Resonance is not only a constant voltage source, ...

[Resonance is not only driven by constant voltage sources, but also by constant current sources and radio waves]

A power supply that is neither constant voltage nor constant current can also [initiate] resonance.

maychang

MrCU204 posted on 2024-8-25 01:10 This is not some new insight, but just a realization of the neglected part of classical knowledge. Resonance is not only a constant voltage source, ...

[The electromotive force of a constant current source is proportional to the load impedance]

Let’s clearly distinguish between [electromotive force] and [voltage] before we talk about the [ignored parts in classical knowledge].

MrCU204

maychang posted on 2024-8-25 09:07 [The electromotive force of a constant current source is proportional to the load impedance] Let's distinguish between [electromotive force] and [voltage] before we talk about [what is ignored in classical knowledge...

The electromotive force is generated from the power source, and the voltage drop or voltage division is the reaction "force" of the load to the electromotive force.

If the power supply has internal resistance, then the voltage drop across the load will be less than the electromotive force, isn't that right?!

As for voltage, electricians are not physicists. As far as I can see, no matter voltage drop or electromotive force, those electricians call it voltage.

L and C are energy storage elements, which contain reactive power, but they are not power sources after all, so their voltage is still called "voltage drop".

maychang

MrCU204 posted on 2024-8-25 10:18 Electromotive force is generated from the power supply, and voltage drop or voltage division is the reaction "force" of the load to the electromotive force. If the power supply has internal resistance, the load voltage drop will be...

[As for voltage, electricians are not physicists]

Since you are not a physicist, why are you talking about (electromagnetic fields) in the forum?

maychang

MrCU204 posted on 2024-8-25 10:18 Electromotive force is generated from the power supply, and voltage drop or voltage division is the reaction "force" of the load to the electromotive force. If the power supply has internal resistance, the load voltage drop will be...

[As for voltage, electricians are not physicists]

[Voltage] was originally a concept of [field], and was later transplanted into [circuit].

小栩栩

This is related to the frequency. If the frequency is low, you don't need to consider the wire.

MrCU204

The voltage drop V of the resistor always works against the electromotive force EMF of the power supply.

I noticed that Vc also opposes EMF in DC, but in AC it completely matches V only in resonance (from a visual point of view, it is also superposition).

So, if the polarity of Vι is opposite to that of Vc, then it will be in the same direction as the EMF. If they are in the same direction, won’t they be superimposed? !

The reason why Vι and Vc can rise steadily with the alternation of positive and negative power supply at the beginning of resonance establishment is because the capacitor absorbs Vι+EMF each time, until the EMF is completely taken up by R, Vι and Vc stop increasing.

MrCU204

maychang posted on 2024-8-25 10:32 [As for voltage, electricians are not physicists] Since you are not a physicist, why do you talk about (electromagnetic fields) in the forum?

Radio waves are equivalent to the name "person", or my identity MrCU204, while electromagnetic fields are equivalent to my entity;

Radio waves act on antennas, and antennas capture magnetic or electric fields, don't they?!

MrCU204

Xiao Xuxu posted on 2024-8-25 11:06 This is related to the frequency. If the frequency is low, you don’t need to consider the wire

For ideal conductors, there is no need to consider the frequency. For radio waves, the conductor is like the secondary side of a transformer, where the alternating magnetic field induces an electromotive force. L and C are the resonant loads of this so-called "secondary side".

It is different from Hertz's metal bracelet. The entire body of that bracelet is an inductor, and the air gap is a capacitor. There is no part that can be called a conductor at all.