- #36

anorlunda

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- #36

anorlunda

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- #37

hutchphd

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It would be like the marvellous medicines that are advertised on the net to sort out your arthritis, impotence, shortness of breath and dyspepsia, all in the same bottle.

💊 Can I get a citation here?

- #38

artis

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Ok to be honest I haven't put all of this together and maybe I'm just making a fool of myself here , but at least in theory shouldn't that be possible?

Also was I right in saying that since I'm driving my load with an amplifier even under varying load conditions where the output voltage increases/decreases the frequency should stay fixed as it it set by the input signal?

- #39

sophiecentaur

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It doesn't "start to lag" because the word Lag can only apply to the steady state. You connect the load and then you wait until it's all settled down. Only then can you measure a meaningful Lag.say I attach a very inductive load and the current starts to lag

One way (the only one) that you can cause the output frequency to be different from the input frequency (say, lower) is to separate the source from the output with a steadily increasing distance (increasing time lag as the ambulance speeds away from you). Then you have the Doppler effect at work. There is no way that the output frequency of a static setup can be different from the input frequency. Ask yourself at which point in the circuit would the voltage be going up and down at a different rate from the input voltage? Amplitude continuity is always maintained across an interface.the frequency should stay fixed as it it set by the input signal?

There are signal processors that can achieve this, though - the easiest to describe is the old tape loop with several read heads on a wheel which rotates faster of slower than the tape speed, the outputs from the heads are switched to select the head that happens to be scanning over the tape. This is really just the Doppler shift at work. Other (digital) methods are available but none of this is relevant to the simple source / load system that's been suggested.

- #40

sophiecentaur

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You can actually write your own and someone, somewhere will read and believe it.💊 Can I get a citation here?

- #41

anorlunda

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No. In theory it is not possible. That's what we have been trying to tell you.The phase shift I simply thought to be the current either lagging or leading , say I attach a very inductive load and the current starts to lag , then the idea is to simply adjust the signal waveform driving the amplifier so that the phase shift is minimized .

OK to be honest I haven't put all of this together and maybe I'm just making a fool of myself here , but at least in theory shouldn't that be possible?

If you connect a pure resistive load, the phase between V and I will be zero.

If you connect a pure inductive load, the phase between V and I will by 90 degrees.

If you connect a pure capacitive load, the phase between V and I will be 90 degrees the other direction.

The voltage source can do nothing to change that. There is nothing to adjust. The signal source can't make an inductor into a resistor.

- #42

artis

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@anorlunda hey, I messed up terribly and only now I understood that when I said phase shift I was actually thinking just controlling the shape of the output waveform itself not the waveform of the voltage vs current as it would change under a different type of load (inductive/capacitive or reactive)

when you said that an inductor would never perform as a resistor I though damn sure makes sense ,

so surely I would not be able to undo the phase shift present within an inductor , what I wanted to is to simply simulate a capacitor added to an inductor , although I assume that would need to be done the old fashioned way as to connect a physical cap in the output much like power stations switching on capacitor banks to compensate for overly lagging current , am I now thinking correctly?

So I guess what I am asking is , how could one simulate the added capacitor effect to an inductive load simply by having an amplifier and an inductive load without a real capacitor , is that even possible ?

IIRC a capacitor added to an inductor like an induction motor doesn't actually decrease the total lag of current or reactive power it just forms a loop where these reactive currents circulate between the cap and motor instead of them circulating from the motor back to grid etc ?

- #43

anorlunda

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I'm afraid your misconception is more basic. Consider the following circuit.

Let Z be an impedance constructed of any combination of R, L and C you please, series and/or parallel connections, as many passive R, L, C components as you please. The voltage measured across Z is V, and the current through Z is I.

Let S be a source. It could be anything, a mains plug, a signal generator, a battery , or whatever. We'll exclude DC, but S is a periodic signal of any shape.

Z and S are connected only via the two terminals (black dots).

Now, if you allow V, I and Z to be complex numbers, Ohm's Law determines the relationship ##\frac{V}{I}=Z## between voltage and current. Nothing that S does can change that.

If S is not a sine wave, then it can be represented as a Fourier Series which has several harmonic frequencies. ##\frac{V}{I}=Z## applies separately to each harmonic, and the total is the sum of all harmonics. Changing the fundamental frequency of S can change Z, but still ##\frac{V}{I}=Z## . The relationship between V and I depends on Z, not S. Changing the waveform of S, changing the phase angle of S relative to some external reference does not change ##\frac{V}{I}=Z##.

So, what you say your objective is, to change the relationship between I and V using adjustments in S is theoretically impossible. That can be done only by changing the contents of the box Z. I hope that I'm making the message clear.

Let Z be an impedance constructed of any combination of R, L and C you please, series and/or parallel connections, as many passive R, L, C components as you please. The voltage measured across Z is V, and the current through Z is I.

Let S be a source. It could be anything, a mains plug, a signal generator, a battery , or whatever. We'll exclude DC, but S is a periodic signal of any shape.

Z and S are connected only via the two terminals (black dots).

Now, if you allow V, I and Z to be complex numbers, Ohm's Law determines the relationship ##\frac{V}{I}=Z## between voltage and current. Nothing that S does can change that.

If S is not a sine wave, then it can be represented as a Fourier Series which has several harmonic frequencies. ##\frac{V}{I}=Z## applies separately to each harmonic, and the total is the sum of all harmonics. Changing the fundamental frequency of S can change Z, but still ##\frac{V}{I}=Z## . The relationship between V and I depends on Z, not S. Changing the waveform of S, changing the phase angle of S relative to some external reference does not change ##\frac{V}{I}=Z##.

So, what you say your objective is, to change the relationship between I and V using adjustments in S is theoretically impossible. That can be done only by changing the contents of the box Z. I hope that I'm making the message clear.

- #44

artis

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but my question about the classical case of an capacitor connected to an induction motor is true isn't it ? where you basically control how far in the grid the "Z" imposed phase shift goes , that's like PFC correction in smps , one cannot change the phase shift of the mains input rectifier or the power of the switches being taken mostly from the peaks of the rectified sine due to stored charge in capacitors but the PFC circuit changes the properties of this connected Z load as to make it more resistive in nature ?

maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

- #45

Baluncore

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Phase is independent of voltage in a linear system.maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

Now you are just confusing things by writing one long misinterpretable sentence.

Sit down with a book on AC theory and write out the equations.

You will be able to ask more sensible questions once you understand the concepts.

- #46

sophiecentaur

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I was contemplating an answer to this when I read @Baluncore 's post, above - which I totally endorse. The above quote is a great demonstration of how 'home brewed' personal theories can lead you into more nonsensical questions and answers.maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

All elementary 'AC Theory' is based on linear generators and Loads so changing the supply volts cannot do anything to change the IMPEDANCES in the circuit.

You really should read some basic texts and do some graft to get a hold of this topic, instead of trying to get there by Q and A which, as PF often points out, is not a way to learn anything technical. Just being apologetic about asking bad questions is not excuse, either, if you really want to find out stuff.

- #47

anorlunda

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@artis , take the advice given by @Baluncore and @sophiecentaur

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