How does an equivalent impedance consume a real power?

In summary, the two circuits have different real power consumption due to the different reactance of the elements in the circuits. The circuit on the right consumes more power due to the reactance of Zeq itself. To impedance match the circuits, the source resistance in the right circuit should be different from the resistance of the load RL.
  • #1
goodphy
216
8
Hello.

Please looks at the attached image first.

Equivalent impedance in LCC network.jpg


The left image is LCC impedance matching network (one of T-type impedance matching networks) and the right is the corresponding equivalent circuit when CT and CL are adjusted such that an equivalent impedance of the sub-circuit surrounded by dashed-line becomes RS.

You don't need to take care of what ZL really is, that was not important for my question I would like to ask you here. In the left circuit, there is only one resistance, RL, located in ZL so that's the only point that a real power (active power) is consumed. But in the right, Zeq itself is a purely resistive. In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

How can I solve this apparent contradiction? I think reactive powers for reactive components have some role to match these two power. For example, reactive components receive power over a quarter cycle and release it to the RL for another quarter cycle, not to the source. But I don't know how such an energy flow occurs.

Could you please help to clarify this?
 
Engineering news on Phys.org
  • #2
That diagram may be more complicated than necessary to resolve this dilemma.
You can think in terms of a simple, ideal transformer., which avoids the need to do complex calculations and the operative word here is 'transformation'. The load RL is presented with a voltage V2 (the secondary volts). The secondary current will be V2/ RL. and the power delivered will be (V2)2/RL. The power that flows into the transformer primary will be V1I1, which must be the same as the power dissipated in the load RL. So the resistance presented to the source is V1/I1, which will be RL T2 where T is the turns ratio.
The voltage across the load is not the same as the voltage across the primary winding and, if you want to impedance match, the source resistance Rs will not be the same as RL
The different matching network you use in your diagram will still behave in the same way, with RL being presented as a different value to the source. It's what you would expect. If you are bothered about the "Energy Flow" then you can consider the energy that is not dissipated in the load as being stored within the reactive components. There is stored energy but no Power dissipation. (Perhaps that last sentence may be enough explanation)
 
  • #3
goodphy said:
You don't need to take care of what ZL really is, that was not important for my question I would like to ask you here. In the left circuit, there is only one resistance, RL, located in ZL so that's the only point that a real power (active power) is consumed. But in the right, Zeq itself is a purely resistive. In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

How can I solve this apparent contradiction?
You can solve this by not jumping to conclusions as you appear to do. Check the proper expressions including the phases. For complex numbers the > and < aren't directly applicable or useful if you forget about the phases.
 
  • Like
Likes sophiecentaur
  • #4
sophiecentaur said:
That diagram may be more complicated than necessary to resolve this dilemma.
You can think in terms of a simple, ideal transformer., which avoids the need to do complex calculations and the operative word here is 'transformation'. The load RL is presented with a voltage V2 (the secondary volts). The secondary current will be V2/ RL. and the power delivered will be (V2)2/RL. The power that flows into the transformer primary will be V1I1, which must be the same as the power dissipated in the load RL. So the resistance presented to the source is V1/I1, which will be RL T2 where T is the turns ratio.
The voltage across the load is not the same as the voltage across the primary winding and, if you want to impedance match, the source resistance Rs will not be the same as RL
The different matching network you use in your diagram will still behave in the same way, with RL being presented as a different value to the source. It's what you would expect. If you are bothered about the "Energy Flow" then you can consider the energy that is not dissipated in the load as being stored within the reactive components. There is stored energy but no Power dissipation. (Perhaps that last sentence may be enough explanation)

Hello. Thanks for giving a quick answer:)

I'm sorry but I'm afraid I couldn't get your point yet. I know how the transformer works and it seems that you're equating any complex network with a load ZL = RL + iXL to some ideal transformer with a load RL in its secondary side. I don't know how this correspondence is made.

You use an expression of power as voltage squared divided by an impedance while I use it as current squared multiplied by an impedance, they are both equal. With sticking on my power expression, I have to say these: the power consumed by the sub-circuit surrounded by the dashed-line in left diagram must be P1 = I2RS , according to the right equivalent diagram and it is an active power as the load RS is purely resistive (the active power is the power consumed by a resistance, not reactance). Obviously, the active power of P1 must be consumed by the sub-circuit somehow. I think P1 should be consumed by RL in the left diagram as it is the only resistance in the circuit, so I12RL = I2RS where I1 is the current going through RL. If RL is lower than RS , then I1 is higher than I. However..how this is possible? I is split into I1 and I2 so I1 is lower than I.

I'm sorry not to understand your comment, but I need to get a little more help to follow your sense. What did I miss?
 
  • #5
BvU said:
You can solve this by not jumping to conclusions as you appear to do. Check the proper expressions including the phases. For complex numbers the > and < aren't directly applicable or useful if you forget about the phases.

Hello.

I'm sorry but phase? Do you want me to write phase of ..impedance? I'm talking about general case so a specific phase is not relevant. And what do you mean > and < ?
 
  • #6
goodphy said:
I know how the transformer works
I think that statement should read "I am familiar with what a basic ideal transformer does". A transformer that you would recognise as such is only one of many ways to achieve its function of Impedance Transformation.
A transformer can be regarded as a set of impedances, connected in such a way as to change volts. current and resistances. Yes, there is Mutual Impedance involved in a common or garden transformer but, only when the transformer is 'ideal' (where the various capacitances and self inductances play a minor part in its operation) can we assume it's the only element involved. Then we can make do with the simple 'transformer equations' that we start off with. My example of a familiar thing like a transformer was one which you (I mean you and not "one") felt you 'understand'. But many other arrangements of impedances can produce the same effect as a transformer with an iron core and two windings. In all of them, there is EM energy stored, temporarily, in reactances and that energy can influence the power being dissipated in the resistive elements. That's the philosophy behind reactive circuits and how they can change the apparent resistance values.
It's a pain in the butt to do the full analysis of such circuits but you could choose appropriate reactances such that the V and I referred to the point where the circuit emerges through that box such that V/I = RS. You could actually produce any resistance you wanted.
Is this the first complex circuit that you have considered? If so, it would be better to start a bit further back in the learning process and do some easier examples.
goodphy said:
Do you want me to write phase of ..impedance?
If you plot X and R on an Argand diagram, the angle between them and the resultant Z can be referred to as Phase.
A "specific phase" is relevant when it's referred to the input signal.
I think you need to go back to basics for this.
 
  • Like
Likes goodphy
  • #7
goodphy said:
In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

maybe i misunderstand the question... but here's how i would answer the question as i understand it..

Of course RL can be lower than RS and the two consumed(dissipated?) powers therefore unequal. But then you have mismatched impedances, and the only downside to that is you have not optimal transfer of power.

Observe that with matched impedance your efficiency is 50% because half the power gets dissipated inside the source.
I assure you that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
But it's fine for a low power RF signal line, and even desirable because it suppresses reflections.

old jim
 
  • #8
goodphy said:
How can I solve this apparent contradiction? I think reactive powers for reactive components have some role to match these two power. For example, reactive components receive power over a quarter cycle and release it to the RL for another quarter cycle, not to the source. But I don't know how such an energy flow occurs.
I well remember struggling with these concepts.

Yes reactive components have some role. You can deduce it intuitively.

Reactance opposes flow of current, so minimizing reactance will maximize current flow. When reactances of load and source have opposite sign and equal magnitude they'll add to zero and that's pretty well minimized, eh ?
Therefore for optimum power transfer you want Zload to be complex conjugate of Zsource.

That logic is easy enough to remember ?
 
  • #9
jim hardy said:
maybe i misunderstand the question... but here's how i would answer the question as i understand it..

Of course, RL can be lower than RS and the two consumed(dissipated?) powers therefore unequal. But then you have mismatched impedances, and the only downside to that is you have not optimal transfer of power.

Observe that with matched impedance your efficiency is 50% because half the power gets dissipated inside the source.
I assure you that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
But it's fine for a low power RF signal line, and even desirable because it suppresses reflections.

old jim

Hello. Thanks for giving comments.

I think I need to think this problem more carefully, before asking further and further.

However, in order to start right reasoning, I need to confirm one thing: In impedance matching by adjusting CT and CL so that Zeq = RS, a power dissipated in Zeq is P1 = I2RS, which is a pure active power (non-zero net power transfer). As Zeq is an equivalent impedance of the sub-circuit (presented by the dashed-line in my figure) and the power on Zeq is the active power, P1 must be consumed by the sub-circuit without going back to the power source (unlikely to the active power, a reactive power goes back to the source after being received by a reactive circuit or component so net power transfer is zero). In the sub-circuits, RL is the only one resistive component so whole amount of power P1 is dissipated on RL, eventually.

Could you tell me whether these statements I made are correct or not ? If these are correct, I'm on a right track. If not, there is something big I missed so I need to check up what I was totally wrong with.
 
  • #10
goodphy said:
Could you tell me whether these statements I made are correct or not ?
I don't know. Still trying to parse them. :smile:
Also your power flows right to left,, unconventional but i finally figured it out.

equivalent-impedance-in-lcc-network-jpg.112245.jpg


goodphy said:
In impedance matching by adjusting CT and CL so that Zeq = RS, a power dissipated in Zeq is P1 = I2RS, which is a pure active power (non-zero net power transfer).
hmm latex doesn't quote well ?
Regardless of that, yes that'll be the power transferred when impedances are matched.
I'm not sure what your term "pure active" means, though. Are you suggesting the currents I and I1 are in phase ? I don't see that as a condition , yet.

goodphy said:
As Zeq is an equivalent impedance of the sub-circuit (presented by the dashed-line in my figure) and the power on Zeq is the active power, P1 must be consumed by the sub-circuit without going back to the power source
I think i agree with that one, because in order for Zeq to match a resistive source impedance Zeq must appear resistive also. Complex conjugate of R +j0 is R-j0, and that's resistive.

goodphy said:
(unlikely to the active power, a reactive power goes back to the source after being received by a reactive circuit or component so net power transfer is zero).
did you mean "(unlikely to the REactive power,.." ?
Yes that's what reactance does, shuttles energy back and forth between components that store it.

goodphy said:
In the sub-circuits, RL is the only resistive component so whole amount of power P1 is dissipated on RL, eventually.
Sounds right to me. P1 = I12RL and RL is the only circuit element that dissipates power.
Current through RL ≠ current through RS though; as Sophie observed your matching circuit behaves like a transformer.

We're speaking of course about average not instantaneous power. That's what you meant when you said "net power" , Correct ?
goodphy said:
If these are correct, I'm on a right track. If not, there is something big I missed so I need to check up what I was totally wrong with.
I think you're on the right track. I hope above was some help. Sophie and the radio guys here are better with the maths than i am.

A word on style:

When writing technical stuff i try to break it up into single stepwise thoughts and place one thought per sentence, starting each sentence on a fresh line..
It requires a lot of editing because I'm prone to run-on sentences..
But the payoff is i can read it a week later and probably figure out what i was talking about.
I think also it makes it easier for others to follow my thoughts.
Try it, you might like it. After all, we do our maths that way don't we ?

old jim
 
  • Like
Likes goodphy
  • #11
jim hardy said:
I'm not sure what your term "pure active" means, though. Are you suggesting the currents I and I1 are in phase ? I don't see that as a condition , yet.
Please never mind. "Pure active" just means that the power on Zeq has no reactive power, it only has an active power.

jim hardy said:
did you mean "(unlikely to the REactive power,.." ?
The original sentence I made was right. I was trying to emphasize a difference between an active power and a reactive power.

jim hardy said:
A word on style:

When writing technical stuff i try to break it up into single stepwise thoughts and place one thought per sentence, starting each sentence on a fresh line..
It requires a lot of editing because I'm prone to run-on sentences..
But the payoff is i can read it a week later and probably figure out what i was talking about.
I think also it makes it easier for others to follow my thoughts.
Try it, you might like it. After all, we do our maths that way don't we ?

old jim
Thanks for giving me some advice. I'll consider your suggestion of writing style when I have a new post.

All right. It seems my original reasoning is right, but I still have to figure out how I1 can be larger than I. Instanteously, I1 can be smaller than I but on average, it must be higher than I in order to match up powers, P1,net = <I2RS> and P2,net = <I12RL>. Maybe I1 can be larger in other instantaneous time when currents from reactive components reinforce it.

I appreciate your kind help. You made me felt more clear view.
 
  • #12
goodphy said:
but I still have to figure out how I1 can be larger than I. Instanteously, I1 can be smaller than I but on average, it must be higher than I in order to match up powers, P1,net = <I2RS> and P2,net = <I12RL>
Hmmm i was thinking the other way, your matching network acts as a step up transformer not step down .

goodphy said:
Maybe I1 can be larger in other instantaneous time when currents from reactive components reinforce it.
or maybe you'd need a different circuit topology.
Wikipedia suggests that step-down uses reactance in parallel not series with the source

L-section
220px-Matching_L_Pad.png


Basic schematic for matching R1 to R2 with an L pad. R1 > R2, however, either R1 or R2 may be the source and the other the load. One of X1 or X2 must be an inductor and the other must be a capacitor.
220px-LMatchingNetworks.svg.png


L networks for narrowband matching a source or load impedance Z to a transmission line with characteristic impedance Z0. X and B may each be either positive (inductor) or negative (capacitor). If Z/Z0 is inside the 1+jx circle on the Smith chart (i.e. if Re(Z/Z0)>1), network (a) can be used; otherwise network (b) can be used.[3]
A simple electrical impedance-matching network requires one capacitor and one inductor. In the figure to the right, R1 > R2, however, either R1 or R2 may be the source and the other the load. One of X1 or X2 must be an inductor and the other must be a capacitor. One reactance is in parallel with the source (or load), and the other is in series with the load (or source). If a reactance is in parallel with the source, the effective network matches from high to low impedance.
The amateur radio buffs here are versed in practical side of impedance matching.http://www.hamuniverse.com/kb1lqc41balun.html
 
  • #13
jim hardy said:
that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
Yes. That point was made to me the first time the maximum power theorem was introduced.
But the problem of reflections is solved with a good match into the load, at the other end of the feeder line. There are then, no reflected signals hitting the source (=transmitter). This applies to high powered RF transmitters. When appropriate, they are made with high efficiency / very non-linear output stages and they, again, cannot afford to be only 50% efficient. So impedance matching is only vital at the antenna end of the transmission line. For a high power valve transmitter, the output stage needs to be 'tuned' to present the right impedance for the efficient operation of the valve rather than to match the Z0 of the line
Audio amplifiers seldom have an 8Ω output impedance - often, they are 'Voltage sources" with an output impedance of less than 1Ω. You can expect a (basically) emitter follower style of output stage.
There are many occasions when matching at the source end is important - particularly with measuring equipment, where a standing wave can be a huge embarrassment.
 

Related to How does an equivalent impedance consume a real power?

1. What is an equivalent impedance?

An equivalent impedance is a single impedance value that represents the combined effect of multiple impedances in a circuit. It takes into account both the resistance and reactance of each individual impedance.

2. How does an equivalent impedance differ from a regular impedance?

An equivalent impedance is a simplified representation of a more complex circuit, whereas a regular impedance refers to the individual impedance of a single component in a circuit.

3. Can an equivalent impedance consume real power?

Yes, an equivalent impedance can consume real power just like any other impedance in a circuit. The real power consumed by an equivalent impedance is equal to the voltage squared divided by the equivalent impedance.

4. How does an equivalent impedance affect the power consumption in a circuit?

An equivalent impedance can affect the power consumption in a circuit by altering the amount of real power that is consumed. This can be seen through changes in the voltage and/or current in the circuit.

5. Why is it important to understand how an equivalent impedance consumes real power?

Understanding how an equivalent impedance consumes real power is important for analyzing and designing circuits. It allows for more accurate calculations of power consumption and can help identify potential issues or inefficiencies in a circuit.

Similar threads

Replies
1
Views
979
Replies
33
Views
2K
  • Electrical Engineering
Replies
7
Views
2K
Replies
8
Views
2K
Replies
8
Views
2K
  • Electrical Engineering
Replies
9
Views
2K
Replies
15
Views
1K
Replies
19
Views
3K
  • Electrical Engineering
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
884
Back
Top