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Does current have a frequency

Merry Christmas

RichB

Senior Member
Location
Tacoma, Wa
Occupation
Electrician/Electrical Inspector
This fits in several topics so I put it here

This came up in as discussion the other day;

My thoughts are:

No-Current does not have an inherent frequency because:
1. In an open AC circuit you can measure Voltage and Frequency but no current
2. You can change the frequency of the applied EMF at any time as it is a function of # of poles and RPM of the source
3. Current is a measurement of electrons past a given point
4. You can not change the "frequency" of current without changing the frequency of the applied EMF at the source to cause said change

With a 'scope placed on a circuit you would read the value of the current changing value in a sinusoidal manner--following the frequency of the applied EMF.
Therefore, current has no inherent frequency but its apparent frequency is driven by the frequency of the applied EMF


So, now tell me where I messed this up

Thanks All!!
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
This fits in several topics so I put it here

This came up in as discussion the other day;

My thoughts are:

No-Current does not have an inherent frequency because:
1. In an open AC circuit you can measure Voltage and Frequency but no current
2. You can change the frequency of the applied EMF at any time as it is a function of # of poles and RPM of the source
3. Current is a measurement of electrons past a given point
4. You can not change the "frequency" of current without changing the frequency of the applied EMF at the source to cause said change

With a 'scope placed on a circuit you would read the value of the current changing value in a sinusoidal manner--following the frequency of the applied EMF.
Therefore, current has no inherent frequency but its apparent frequency is driven by the frequency of the applied EMF


So, now tell me where I messed this up

Thanks All!!
Well, yes. The frequency of the current in a conductor will always be the same as that of the voltage no matter what its magnitude, which is zero for an open circuit.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Semi-Retired Electrical Engineer
I think you are over thinking this. The notion of current does not exist in an open circuit. It is nonsense to discuss the frequency of a non-existent quantity.

In a closed AC circuit, current has frequency that can be measured independently from the frequency of the voltage source. The two will be the same, as the voltage generally calls the current into existence. But that last part is not always true. In an inductor (e.g., a motor), a voltage is called into existence by changes in current. Here again, the frequency of the current and the voltage will be the same, but this time it is current that is driving the bus.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
And let us not forget about the effects of capacitance and inductance that cause shifts in current timing.

So, yes, if there is a current, it has the same frequency as the driving voltage, even if leading or lagging.
 

RichB

Senior Member
Location
Tacoma, Wa
Occupation
Electrician/Electrical Inspector
Thanks for the replies and the edification!!:)
I was going way to deep into the weeds like you said Charlie---All of these brought me back to reality!!
You can all go home now with pay as "larnin' has tooken place!" Send the bill to Mike:LOL:
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
And let us not forget about the effects of capacitance and inductance that cause shifts in current timing.

So, yes, if there is a current, it has the same frequency as the driving voltage, even if leading or lagging.

As long as the resistive and/or reactive loads placed across a source of AC voltage are linear, then the current they draw will be at a single frequency that is the same as the frequency of the source. But if the load is non-linear, the current can also include harmonics at integral multiples of the source frequency (the fundamental frequency).

One thing you learn early in electrical engineering courses is that a source of power can be represented as an ideal (zero impedance) voltage source with an impedance in series with it. This is known as the Thevenin equivalent source.
Alternatively, it can also be represented as a current source with a shunt impedance across it (the Norton equivalent source).

In electric power applications the Thevenin voltage source representation is much more appropriate because the series source impedance is usually much less than the load impedance, and so it can often be ignored. In voltage drop and short circuit calculations of course the impedance must be known.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Source impedance discussions remind me of audio amplifier damping factor. Not only does it affect the amplifier's ability to accelerate a speaker diaphragm in response to an audio signal, but also its ability to stop the diaphragm when the signal stops.

The amplifier's internal feedback circuitry let's the amp know what the output terminal's voltage at any given moment, so it effectively knows what the diaphragm's movement is, and if it is different from the audio signal, it can put out an opposite voltage.

Sort of an active dynamic braking. (It's actually active all the time)

Sorry for the
hijacked.gif
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
As long as the resistive and/or reactive loads placed across a source of AC voltage are linear, then the current they draw will be at a single frequency that is the same as the frequency of the source. But if the load is non-linear, the current can also include harmonics at integral multiples of the source frequency (the fundamental frequency).
With solid state switching (I'm not super knowledgeable about electronics), and a power source/sink, couldn't we create an arbitrary current waveform drawn from the AC voltage source?

One thing you learn early in electrical engineering courses is that a source of power can be represented as an ideal (zero impedance) voltage source with an impedance in series with it. This is known as the Thevenin equivalent source.
For real world power sources (utility transformers, large generators, large inverters), how close do they come to matching this idealization with a fixed linear source impedance, across various orders of magnitude of current draw? E.g. 0.1A, 10A, 1,000A.

Cheers, Wayne
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
LOL...
We have an opinion section in our local paper... Often just reading the heading and the 1st sentence I can accurately predict the author without continuing the article.

I read the header and accurately predicted the responders :)
 

Carultch

Senior Member
Location
Massachusetts
This fits in several topics so I put it here

This came up in as discussion the other day;

My thoughts are:

No-Current does not have an inherent frequency because:
1. In an open AC circuit you can measure Voltage and Frequency but no current
2. You can change the frequency of the applied EMF at any time as it is a function of # of poles and RPM of the source
3. Current is a measurement of electrons past a given point
4. You can not change the "frequency" of current without changing the frequency of the applied EMF at the source to cause said change

With a 'scope placed on a circuit you would read the value of the current changing value in a sinusoidal manner--following the frequency of the applied EMF.
Therefore, current has no inherent frequency but its apparent frequency is driven by the frequency of the applied EMF


So, now tell me where I messed this up

Thanks All!!

It is possible in concept for the frequency of current to be different than the frequency of the voltage. You would need a kind of load that distorts the shape of the voltage waveform for this to happen. Usually, there is at least some kind of relationship between the current and voltage waveform frequencies. A circuit may also attenuate, or amplify a certain range of frequencies, or it may produce harmonics of the original fundamental. I can't think of any example that would produce completely different frequencies, that aren't directly determined by the frequency of the voltage waveform.

As an example, an ideal full wave rectifier will turn a simple sine wave of voltage, V(t)=sin(ω*t), into the following Fourier series, when applied to a unit resistance load:
1713402887317.png

This is what the circuit looks like, with the load resistor being 1 ohm:
full_rectifier.png


Here's a plot that shows how this evolves as more cosine waves are added (I've deliberately drawn offsets, so you can see them separately):
1713403286842.png

For an ~0.16 Hz voltage source, the original ω=1 rad/s. Constructing the first few terms of this sum:
I(t) = 0.6366 - 0.4244*cos(2*t) - 0.08488*cos(4*t) - 0.03638*cos(6*t) - 0.02021*cos(8*t)

This produces all kinds of different frequencies from the original 0.16 Hz. First being 2ω, then 4ω, then 6ω, and so forth. But they are all whole multiples of the fundamental (ω), and are all related to it. The current has frequencies of 0.32 Hz, 0.64 Hz, 0.96 Hz, etc, while the voltage only has an 0.16 Hz frequency.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
I don't understand what you mean by adding waves or terms. What is actually changing? :unsure:
A Fourier series is a way to approximate any sufficiently well behaved periodic function by an infinite sum of cosine (or sine) functions of increasing frequency. The lowest frequency cosine function in the sum will be the frequency implied by the statement the function is periodic; the higher frequency cosine terms in the sum will be the integer multiples of that fundamental frequency. The cosine term with frequency k times the fundamental frequency is called the kth harmonic.

So the dashed functions in post #12, which are shown offset vertically to separate them for exposition, are what you get as approximations to the blue function if instead of taking the infinite sum, you just add up the first so many non-zero terms of the Fourier expansion of I(t) for the circuit described. I say non-zero terms, as apparently (from the algebraic formula given) all the odd harmonics of I(t), including the fundamental frequency, are zero, so only the even harmonics show up in the Fourier series of I(t). As you add more and more terms to your approximation, the shape of the approximation gets closer and closer to I(t), the function being approximated, which you can visually see from top to bottom among the dashed functions.

Cheers, Wayne
 
Last edited:
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
A Fourier series is a way to approximate any sufficiently well behaved periodic function by an infinite sum of cosine (or sine) functions of increasing frequency. The lowest frequency cosine function in the sum will be the frequency implied by the statement the function is periodic; the higher frequency cosine terms in the sum will be the integer multiples of that fundamental frequency. The cosine term with frequency k times the fundamental frequency is called the kth harmonic.

So the dashed functions in post #12, which are shown offset vertically to separate them for exposition, are what you get as approximations to the blue function if instead of taking the infinite sum, you just add up the first so many non-zero terms of the Fourier expansion of I(t) for the circuit described. I say non-zero terms, as apparently (from the algebraic formula given) all the odd harmonics of I(t), including the fundamental frequency, are zero, so only the even harmonics show up in the Fourier series of I(t). As you add more and more terms to your approximation, the shape of the approximation gets closer and closer to I(t), the function being approximated, which you can visually see from top to bottom among the dashed functions.

Cheers, Wayne
Well that certainly explains it, says the guy who had to retake 9th grade algebra!

🤔
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Well that certainly explains it, says the guy who had to retake 9th grade algebra!
How about this:

Let's say we want to write out 1/3 as a decimal. We can think of decimals as a situation where the only primitives we have available are the powers of 10 (in this case negative powers of 10, so 1/10, 1/100, etc) and their multiples by a single digit (0-9). So our first approximation to 1/3 would be using just one of these primitives, 3 * (1/10), or 0.3. Our second approximation would be to use two of these primitives, 3 * (1/10) + 3 * (1/100) = 0.33. That's closer to 1/3 than our first approximation. The third approximation would be 0.333, closer still. As we add more and more terms to our finite sum, the answer will get closer and closer to 1/3. But it never equals 1/3 until we pass to the infinite sum.

With Fourier series we can do the same thing for a function f that is periodic with cyclic frequency ω. We can approximate it as an infinite sum of primitive functions, meaning that the longer and longer finite sums get closer and closer to our function f. And the primitives we use are the functions cos(k * ωt) and sin(k * ωt), where k is an integer at least 0. We get to multiply these function by an arbitrary real number constant, rather than just the digit 0-9.

So in post #12, we can do this for V(t). That one is easy, as it already a sine wave; and ω has been chosen to be 1 for simplicity. We only need one term, sin(t), and the multiple looks to be about 0.9.

But we can also do this for I(t). (I didn't check the expansion that is provided in the post but am just explicating it). The first term would be when k=0, in which case cos(0*t) = 1, so we just get a constant function. The coefficient for this term is 2/pi = 0.6366. So our first approximation for I(t) would be a constant current I(t) = 0.6366A. Not a great approximation, as the raw output of the full bridge rectifier has a lot of ripple.

So to better approximate I(t) we can use more terms. It turns out we don't need any sin terms, and all the cos terms are of the form cos(kt) where k is even. We say that I(t) has only even harmonics. The second harmonic gets a coefficient of 4/(-3pi) = -0.4244. So this approximation would be I(t) = 0.6366 - 0.4244 cos(2t). This approximation is graphed as the dashed line labeled "1 term" meaning we just used one cosine function. And it's graphed offset upwards, just so the functions don't pile up on each other.

We can keep doing this with more and more terms. The second approximation add a cos(4t) term or fourth harmonic; the coefficient is 4/(-15pi) = -0.08488. This approximation is I(t) = 0.6366 - 0.4244 cos(2t) - 0.08488 cos(4t) and is the dashed graph labeled "2 terms". It looks a little closer to the blue graph we are approximating.

Etc.

Cheers, Wayne
 
Last edited:

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
LOL...
We have an opinion section in our local paper... Often just reading the heading and the 1st sentence I can accurately predict the author without continuing the article.

I read the header and accurately predicted the responders :)
I also correctly predicted how quickly the complexity of the discussion would outstrip the intent of the OP's question by several orders of magnitude, but hey, it's what we do, innit? :D
 

RichB

Senior Member
Location
Tacoma, Wa
Occupation
Electrician/Electrical Inspector
I also correctly predicted how quickly the complexity of the discussion would outstrip the intent of the OP's question by several orders of magnitude, but hey, it's what we do, innit? :D
It is and as usual I got my answer and a soft smack to remind me to stop chasing things down Alice's rabbit hole:cautious:
And then on to the learning part and having theory re taught to me as well as going waaaay above and beyond--ILOVE IT:love:
 

RichB

Senior Member
Location
Tacoma, Wa
Occupation
Electrician/Electrical Inspector
If you are correct then maybe they should have called it alternating voltage and direct voltage instead of alternating current and direct current?
kwired--true --but here i go down the rabbit hole again--we probably could-as we are "Alternating" the voltage from Positive to negative in an AC circuit and the voltage is "straight or direct",i.e., in DC
 
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