Exploring LRC AC Series Circuits

In summary, an inductive circuit stores more of its internal energy in electric fields than in magnetic fields. This can be seen by the phase angle of the current passing through the circuit.
  • #1
Albertgauss
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TL;DR Summary
LRC AC inductors capacitor magnetic fields electric fields at which frequencies are fields created
Hi all,

In an LRC AC series circuit, at which frequencies are where you are mainly dumping your generator/current energy into capacitor to create electric fields or into the inductor to create magnetic fields? So, for example, at low frequencies, f --> 0, the impedance of the inductor goes to 0 and the phase angle is +90 degrees, yet the impedance of the capacitor (1/wC) is ∞ . Is my circuit creating and destroying magnetic fields for the inductor, or is it creating and destroying electric fields in the capacitor? On the one hand, the inductor is acting like a short, but the capacitor is blocking all the current, so I'm not sure how to reason this out. I also need help for the same case for high freqs at phase angle -90 degrees. Which of these cases is called an "inductive circuit" and which case is a "capacitve circuit"?
 
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  • #2
Hi,

You are -- in the usual analysis cases -- not 'dumping' energy in L and C, only in R.
For precisely the 90##^\circ## reason in both L and C: ##\vec V(t)## and ##\vec i(t)## are perpendicular, so ##\vec V(t)\cdot\vec i(t) = 0##.

LC: video1 and video2
 
  • #3
Yes, I understand that energy only actually leaves the circuit through the resistor.

But, within the circuit, it takes energy to create magnetic fields for an inductor and electric fields in a capacitor. Depending on what time, t, of the cycle you are in, there are some magnetic fields that exist in the inductor and some electric fields between the capacitor. Both kinds of fields require some of the circuit's energy. I wrote the equations relating to this on the attached powerpoint.

I think I can rephrase this question better as: In an AC, LRC circuit, that is inductive (low freqs, phase angle +90 degrees), does this correlate to more of the circuit's internal energy being devoted to the electric fields of the capacitor rather than the magnetic fields of the inductor?

The other jpeg I attached is from a textbook of a slide of an LC circuit showing the energy transferred between magnetic and electric fields. I am trying to relate this energy transfer to the impedance or phase angle of an LRC circuit, if that is possible. Something like, as simple as "an inductive circuit stores internal circuit energy more in electric fields than in magnetic fields", but I realize that such a statement might not be as simple as I thought.

I did watch the videos you mentioned, but they didn't seem to answer my question.

EnergyAC_LRC_Circuits.jpg


P4260773.JPG
 
  • #4
I'm a bit confused by your question. Are you asking about the instantaneous energy storage on L and C? If so the answer is that the peak energy stored in each is the same value (this is neglecting R, for now). When the energy stored in one is at the peak, the energy stored in the other is zero. You can compare this to a pendulum shifting energy from kinetic to potential. This is also assuming we are talking about the natural response, not a driven response from a connected source.

edit: Sorry, I didn't answer your question. This resonant behavior doesn't depend on what the resonant frequency is. Just as the behavior of a pendulum doesn't depend on it's length. The numbers may change but the behavior doesn't.
 
  • #5
OK. First thing, forget about fields. There is no need to consider electric or magnetic fields to analyze a circuit.

Second, the key concept here is impedance. For a series RLC circuit, the impedance is:

##R+j2\pi{f}L-j\frac{1}{2\pi{f}C}## where ##f## is the frequency.

The tricky part if you don't have the math is ##j##. That implies a phase shift of 90 degrees.

If ##2\pi{f}L## is greater than ##\frac{1}{2\pi{f}C}## then the circuit is primarily inductive. If it is less, then the circuit is primarily capacitive.

Next consider the phase angle. If the circuit is series, then the same current passes through the R and the L and the C. Use this schematic.

schemeit-project (1).png

The current I is ##I = \frac{V0}{R+j2\pi{f}L-j\frac{1}{2\pi{f}C}}##

The voltage measured between points 0 and 3 is V03. It is the same as V0.

The voltage measured between points 2 and 3 is V23. It is in phase with I.

The voltage measured between points 1 and 2 is V23. It is 90 degrees ahead the current I.

The voltage measured between points 0 and 1 is V23. It is 90 degrees behind the current I.

When averaged over an entire cycle, no matter what the frequency, the real power is all dissipated in R, and there will be zero net energy change in L or in C.

As @DaveE mentioned, you must be careful to specify if you are asking about instantaneous values, or average values (averaged over an entire cycle). You can't mix the two. At the instantaneous level, yes the L and the C can store some energy, but the voltage and current both reverse direction during a cycle, so averaged over an entire cycle, both the L and the C store zero energy (assuming zero initial conditions).
 
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  • #6
I see. I realize that what I was asking didn't quite make sense, I being a victim of my own misunderstanding. Yes, instantaneous values of energy on the components is what I am talking about here. What I thought is that; somehow, erroneously, the impedances could relate to the magnetic and electric fields stored in the inductor and capacitor; that is, I thought maybe that a large impedance on the capacitor at low function generator frequencies below resonance would indirectly mean that more internal energy of the circuit was somehow manifest in the electric fields of the capacitor. I realize this is not true, though, as mentioned above, the L and C together combine for total zero energy. The pendulum analogy above makes sense--I'd forgotten about that, but I'm glad Dave let me know about it.

I think I'm actually okay for now, and consider this question answered. @Dave, they were still good videos, actually, really good and I will use them later. So thanks for letting me know about them.
 
  • #7
I think you may be confusing the natural (transient) response of the circuit at the resonant frequency with the driven response at the signal generator frequency. If an LCR circuit is driven in steady state at an arbitrary frequency (this allows the transients at the resonant frequency to dissipate to zero). Then the average energy stored in the L and C are not equal. At very low frequencies (compared to the resonant frequency) the inductor can be modeled as a short circuit and the voltage, and energy, from the source will all be across the capacitor. The opposite will happen at very high frequencies, where the capacitor can be replaced with a short circuit. At the resonant frequency the average energies of L and C will be equal.
 
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  • #8
anorlunda said:
but the voltage and current both reverse direction during a cycle, so averaged over an entire cycle, both the L and the C store zero energy (assuming zero initial conditions).
The average energy isn't zero since it is Li2/2 or Cv2/2. Yes the steady state energy is dissipated in R, but there is typically average energy stored in L and C too.

edit: although it isn't immediately clear to me why we would care much about average stored energy in a steady state AC circuit.
 
  • #9
Oh, that does change something. Yes, I am off resonance, driving the circuit at a frequency much less than the resonant frequency. So, then, if the energy stored in L and C are not equal, and at low freqs all the energy is across the capacitor as mentioned in the previous post, does that mean, indeed, that all of the energy is manifest as electric fields in the capacitor?

I thought of something else, the power factor, the cosine of the phase angle. At resonant frequency, the phase angle would be zero in the power factor, the interpretation being all of the generator energy is dissipated through the resistor. In that case, the L and C cancel out, as mentioned above.

But if you drive the circuit where the phase angle is +90 degrees, at a frequency much, much less than resonance, and the cosine in the power factor is zero, then could you interpret that situation that all of your generator energy is not getting out through the resistor, but is being turned into electric fields stored in the capacitor? This was something else I heard but am not sure if it is correct.
 
  • #10
Are you familiar with the concept of complex impedances (i.e. impedances that have a magnitude and phase)? The answer to your question is in calculating the impedances in your circuit at the frequency you care about. Of course you have to add them up like vectors. The only simple answers are at f→0 and f→∞.

It may help if you could model this circuit in a simulator (I like LTSpice). Although in general, I hate teaching fundamentals only with simulators, for very simple circuits they can help you visualize what is happening.
 
  • #11
DaveE said:
Then the average energy stored in the L and C are not equal.
Come on, let's not confuse the OP even more. They are equal unless the resistance in the circuit is nonzero.

When we teach circuit analysis, we do so using ideal components. Ideal wires and L and C have zero resistance. Only L has inductance and only C has capacitance. Kirchoff's Laws apply perfectly. There is no propagation speed. There is no Poynting vector. There are no fields. We might also add ideal transformers and ideal diodes to the mix. Those are sufficient to study elementary circuit analysis(CA) including energy transfer and energy losses. Series, parallel, node and mesh methods can be taught.

Once CA skills are mastered, we can introduce more advanced concepts like E fields, and B fields, and Maxwell's equations, and RF propagation, and atoms and electrons :nb), and semiconductors, and even Quantum Electrodynamics.

DaveE said:
The average energy isn't zero since it is Li2/2 or Cv2/2. Yes the steady state energy is dissipated in R, but there is typically average energy stored in L and C too.
Then how would you propose preserving energy conservation?

Make it simpler. Take an AC voltage source V and a L (no R no C). Write the equation for current I as a function of time, and power P=VI as a function of time. Then integrate power over an entire cycle and what is the answer?

You are getting confused by mixing AC and DC expressions.
 
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  • #12
I almost forgot. We have a PF Insights article on this topic.

https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/

In particular, this plot showing V and I and P for the case where V and I are 90 degrees out of phase. In an RLC circuit, that should be applied separately to the voltage across the L and the voltage across the C. In both cases, we have that 90 degree shift.

1588016908508.png
 
  • #13
I've attached a calculation for an arbitrary Series LCR circuit driven by a sinewave. This is based on the phasor method and is only applicable for a single frequency source after any transients have died away.

Notice the difference between the values when the circuit is driven at resonance compared to away from the resonant frequency.

The issue of "average" stored energy is confusing because, as far as I know, no one ever uses it in a staedy stae analysis. The instantaneous energy in L and C are well defined and positive (wrt the 0 voltage, 0 current rest state). You could average it over one cycle and call that the average energy. However, if all of the voltages and currents are sine waves at a single frequency, then the instantaneous energy will be given by a sin2 function. I think everyone knows this and rarely does the calculation (if you care, it's 1/2 the peak value).

Edit: In this calculation I am defining the voltages and currents as peak values. Every one else (including me, LOL) would use RMS values. The difference comes into play when you calculate the peak energy. You can be off by a factor of 2 if you do that part wrong.
LCR1.jpg
 
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  • #14
anorlunda said:
Then how would you propose preserving energy conservation?
So a mass oscillating on a spring doesn't store energy? And it is confusing to teach that it does?
There are transients required to reach steady state and energy is added. I strongly disagree with your position here.
 
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  • #15
DaveE said:
The average energy isn't zero since it is Li2/2 or Cv2/2.
I think he meant average "power" associated with L and C is zero.

If I understood the OP correctly, he wants to know what it means when the net reactance of a series RLC circuit is capacitive or inductive in terms of energy stored in L and C.
It can be shown that
peak energy stored in L= (1/2)(XL/ω)(ipeak)2 and
peak energy stored in C= (1/2)(Xc/ω)(ipeak)2,
where ω is the angular frequency of the ac source.

So, at a given ω, if XL < Xc, the net reactance is capacitive and peak energy stored in C is more than that in L. This happens at frequencies lower than the resonant frequency.
For ω > ωresonant, XL > Xc and the net reactance is inductive. Here, the peak energy stored in L is more than that in C.
 
  • #16
Your circuit is more or less equal to the deflection coil and driver in a CRT monitor (yes, I have been there).

Way back (I think 1993) we were trying to develop a multi-frequency monitor, and my lead analog developer was very upset over the fact that the R component would distort the picture in various ways depending on the horizontal frequency. I told him that the solution was obvious - insert a negative resistance in the circuit (I knew that it was possible, but I did not know if he knew the circuit). The next day he had solved the problem in the way I had suggested!
 
  • #17
[Retracted. See #23]

hutchphd said:
I strongly disagree with your position here.
I think the source of our disagreement is that you are thinking of the energy as an unsigned quantity ##\frac{CV^2}{2}## whereas I am thinking of the signed quantity ##\int_0^{\frac{N}{f}} V(t)I(t) dt##. Physics and engineering teachers should always back up the words they use with math. Natural language is so ambiguous; in this case "energy."

For the beginning student, let's reduce it to the DC case with two ideal capacitors with capacitance C and initially charged to the same voltage V. Each holds energy ##\frac{CV^2}{2}##, so the total energy is ##CV^2##. Now we connect the two + to + and - to -; nothing happens. Next we connect the two + to - and - to +; they discharge each other leaving zero net charge, zero total energy at the end. Did that violate conservation of energy? Not at all. It demonstrates that in a circuit, the potential energy must be considered a signed quantity.

When we connect a pre-charged capacitor to a circuit, the energy it adds to the circuit is + or - depending on how we connect the leads. Reverse the leads and you reverse the sign of energy added.

Conservation of energy is a very simple concept, never violated here on Earth. But accounting for energy can be devilishly complicated. Students mess it up regularly. Using math rather than words helps avoid some misunderstandings.
 
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  • #18
I just posted this in another thread.

anorlunda said:
I think the source of our disagreement is that you are thinking of the energy as an unsigned quantity ##\frac{CV^2}{2}## whereas I am thinking of the signed quantity ##\int_0^{\frac{N}{f}} V(t)I(t) dt##. Physics and engineering teachers should always back up the words they use with math. Natural language is so ambiguous; in this case "energy."

For the beginning student, let's reduce it to the DC case with two ideal capacitors with capacitance C and initially charged to the same voltage V. Each holds energy ##\frac{CV^2}{2}##, so the total energy is ##CV^2##. Now we connect the two + to + and - to -; nothing happens. Next we connect the two + to - and - to +; they discharge each other leaving zero net charge, zero total energy at the end. Did that violate conservation of energy? Not at all. It demonstrates that in a circuit, the potential energy must be considered a signed quantity.

When we connect a pre-charged capacitor to a circuit, the energy it adds to the circuit is + or - depending on how we connect the leads. Reverse the leads and you reverse the sign of energy added.

Conservation of energy is a very simple concept, never violated here on Earth. But accounting for energy can be devilishly complicated. Students mess it up regularly. Using math rather than words helps avoid some misunderstandings.
 
  • #19
Yes, as Dave mentioned, the following is what I was after:

If I understood the OP correctly, he wants to know what it means when the net reactance of a series RLC circuit is capacitive or inductive in terms of energy stored in L and C.
It can be shown that
peak energy stored in L= (1/2)(XL/ω)(ipeak)2 and
peak energy stored in C= (1/2)(Xc/ω)(ipeak)2,
where ω is the angular frequency of the ac source.

So, at a given ω, if XL < Xc, the net reactance is capacitive and peak energy stored in C is more than that in L. This happens at frequencies lower than the resonant frequency.
For ω > ωresonant, XL > Xc and the net reactance is inductive. Here, the peak energy stored in L is more than that in C.

Also, it has been a long time since I have added reactances with complex numbers, so I didn't do that. I also admit my question was more about trying to understand if/what the impedence values meant in terms of what was happening in terms of the E,B fields of the capacitor and inductor in LRC at non-res freq. I realize the final picture is more complicated than what I initially guessed when I posted this question.

I waited awhile to respond, just to see what other posts people would put up.

I certainly appreciate everyone's help in trying to understand these concepts.
 
  • #20
anorlunda said:
Next we connect the two + to - and - to +; they discharge each other leaving zero net charge, zero total energy at the end. Did that violate conservation of energy? Not at all. It demonstrates that in a circuit, the potential energy must be considered a signed quantity.
They also produce a very large pulse of heat somewhere in the circuit of size +CV2. Where did that come from? We agree that energy is conserved but I simply do not understand what you are trying to accomplish with this scheme. You have positive energy and negative energy and apparently "free" energy heating the wires...
Is the energy in the inductor also sometimes "negative"?
Electric potential difference already is a signed quantity...
 
  • #21
anorlunda said:
signed quantity ∫Nf0V(t)I(t)dt
Like this?
Energy_Calc.jpg

The only way to make this a signed quantity is by how you define it initially, for example in the constant of integration, or the definition of what zero energy is. This is a trivial definition. Any quantity can be negative if you redefine what zero is.

That's ok to do if you are clear about your definitions, but you can't justify it in the math.
 
  • #22
anorlunda said:
For the beginning student, let's reduce it to the DC case with two ideal capacitors with capacitance C and initially charged to the same voltage V. Each holds energy ##\frac{CV^2}{2}##, so the total energy is ##CV^2##. Now we connect the two + to + and - to -; nothing happens. Next we connect the two + to - and - to +; they discharge each other leaving zero net charge, zero total energy at the end. Did that violate conservation of energy? Not at all. It demonstrates that in a circuit, the potential energy must be considered a signed quantity.
misunderstandings.
This is slightly off-topic, I understand your point.

However, be careful with the transients from ideal capacitors connected in parallel. This is a case where the ideal lumped elements allow us to create a physically unrealizable circuit. For example, we know you won't have infinite current flowing in zero time.

On way out is to recognize that there is usually some resistance in the circuit that both limits the current and absorbs half of the transferred energy.

I think the more interesting case is if you assume superconducting everything so resistance is zero. This is often presented as the capacitor energy paradox: Assume 2 1uF capacitors, with one charged to 2V the other at 0V. Then the total energy is 2uJ. Then you connect them and conservation of charge results in each capacitor having 1V, with the total energy now equal to 1uJ. So where did the other 1uJ go?

The resolution is to recognize that the question wasn't realistic to start with. While you can have zero resistance and perfect capacitors, you can't move charge without creating a magnetic field. In the lumped element circuit world, this means that you should have included an inductor. The real solution is an LC harmonic oscillator with energy oscillating between the capacitors and inductor.
 
  • #23
OK. After confirmation by @mfb, I see that @DaveE is right and I was wrong. The key math in #21 is the identity ##\int \sin{x}\cos{x} = \sin^2 x##. I apologize @DaveE .

Now I have to go back and rethink that example with two reversed capacitors discharging. It could be done with superconducting wires, so residual resistance can't be the key. Could the energy go into the spark as we try to close the switch? Any scenario which predicts infinite V or I when using ideal components is invalid, and must be analyzed using real-life non-ideal components.
 
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  • #24
anorlunda said:
OK. After confirmation by @mfb, I see that @DaveE is right and I was wrong. The key math in #21 is the identity ##\int \sin{x}\cos{x} = \sin^2 x##. I apologize @DaveE .

Now I have to go back and rethink that example with two reversed capacitors discharging. It could be done with superconducting wires, so residual resistance can't be the key. Could the energy go into the spark as we try to close the switch? Any scenario which predicts infinite V or I when using ideal components is invalid, and must be analyzed using real-life non-ideal components.

.
I think absent resistance any circuit inductance would come into play (there will always be some) to limit the current to finite values. Of course with that inductance and no resistance the circuit will oscillate forever!
 
  • #25
hutchphd said:
.
I think absent resistance any circuit inductance would come into play (there will always be some) to limit the current to finite values. Of course with that inductance and no resistance the circuit will oscillate forever!
Yes. The peak current is determined by the characteristic impedance of the LC tank.
 
  • #26
I'm not sure if we really answered the OP question. Of course the B field is proportional to the loop current, and the E field is proportional to the capacitor voltage. At low frequencies, the circuit looks like a capacitor; at high frequencies, it looks like an inductor. Here are some normalized plots for the series LCR response.
LCR2p2.jpg
 
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Related to Exploring LRC AC Series Circuits

1. What is an LRC AC series circuit?

An LRC AC series circuit is a type of electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C) connected in series with an alternating current (AC) power source. This type of circuit is commonly used in electronic devices to control the flow of electricity and can exhibit unique properties such as resonance and phase shifts.

2. How does an LRC AC series circuit differ from other types of circuits?

An LRC AC series circuit differs from other types of circuits in that it contains all three basic circuit elements (resistor, inductor, and capacitor) connected in series. This combination of components allows for the control and manipulation of AC power in a more complex and precise manner compared to other types of circuits.

3. What is the significance of resonance in an LRC AC series circuit?

Resonance is a phenomenon that occurs in an LRC AC series circuit when the inductive and capacitive reactances cancel each other out, resulting in a high amplitude of current flow. This can be beneficial in certain applications, such as radio tuning, but can also cause issues if not properly controlled.

4. How do you calculate the impedance of an LRC AC series circuit?

The impedance of an LRC AC series circuit can be calculated using the formula Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. This formula takes into account the resistance and reactances of the circuit to determine the total impedance.

5. What are some common applications of LRC AC series circuits?

LRC AC series circuits are commonly used in electronic devices such as radios, televisions, and computers. They are also used in power transmission and distribution systems to control the flow of electricity. Additionally, LRC AC series circuits are used in filters, oscillators, and other electronic circuits to manipulate and control AC power in various ways.

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