Kirchoff's law

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K2500

Senior Member
Location
Texas
rattus said:
Well yes, but as Larry Fine says, let's not confuse the OP any more than necessary. For this basic discussion, we should assume perfectly stiff sources. The OP is just starting to crawl, let's not make him run just yet.

OK with you k2500?

Yes, limiting confusion is very OK with me.:)
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Well yes, but as Larry Fine says, let's not confuse the OP any more than necessary. For this basic discussion, we should assume perfectly stiff sources. The OP is just starting to crawl, let's not make him run just yet.

OK with you k2500?
But how do we determine the point where confusion for the OP'er begins?

The OP begins the post with a title of "Kirchoff's law" and then proceeds with "I am trying to gain a better understanding of phase angles..." These aren't the words of someone learning to crawl, but rather the words of someone who wants to but has yet to ride a bicycle. As the analogy goes are we to say here's a tricycle?
 

rattus

Senior Member
Just ask:

Just ask:

Smart $ said:
But how do we determine the point where confusion for the OP'er begins?

The OP begins the post with a title of "Kirchoff's law" and then proceeds with "I am trying to gain a better understanding of phase angles..." These aren't the words of someone learning to crawl, but rather the words of someone who wants to but has yet to ride a bicycle. As the analogy goes are we to say here's a tricycle?

Just ask the OP. I believe he will agree that he is just starting to crawl, and any more detail will be lost on him.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Just ask the OP. I believe he will agree that he is just starting to crawl, and any more detail will be lost on him.
Even if he is starting to crawl, it does not mean his goals are limited to just crawling. So let's test these theories...

K2500, have you reached the level of understanding you desire? If so, congratulations. If not, will you give a basic description of what you wish to understand better?
 

rattus

Senior Member
Smart, I finally see what you are driving at in your post #37. But, you must agree that a separation of exactly 120 degrees is still possible even with unbalanced loads. A little reactance here and there is all that is required. I missed your point because source resistances were not included in the original problem.

The OP has posted this simple problem with ideal conditions in order to learn how to compute In. We should not complicate the issue by introducing non-ideal conditions. In other words, keep it simple.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Smart, I finally see what you are driving at in your post #37. But, you must agree that a separation of exactly 120 degrees is still possible even with unbalanced loads. A little reactance here and there is all that is required.
Yes, I agree... but such reactances were not a part of your response to charlie:
First off Charlie, it is entirely possible that the load currents are separated by exactly 120 degrees, say in the case of incandescent lights.

I missed your point because source resistances were not included in the original problem.
As with most things electrical, you can seldom look at the stated problem with tunnel vision. We must take into account the whole picture, or at least all portions that have or may have an effect. That is where the "keep it simple" premise has its fault.
 

rattus

Senior Member
Smart $ said:
Yes, I agree... but such reactances were not a part of your response to charlie:


As with most things electrical, you can seldom look at the stated problem with tunnel vision. We must take into account the whole picture, or at least all portions that have or may have an effect. That is where the "keep it simple" premise has its fault.

Smart, you miss the point! This was a hypothetical problem posed as an exercise for a student. Even if it were a real problem it is doubtful that anyone would bother to throw in the secondary effects. I would think that two significant figures is good enough for In. If we follow your line of reasoning we could never use the algebraic formula because the angles might be off a little.

I have argued before with engineers who think that each problem should be solved with the greatest possible precision. I would argue though that a seasoned engineer should know when to use approximations and not waste resources with overly complex solutions.

Your statements are technically correct, but in my opinion they are answers to questions not asked.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Smart, you miss the point!
No, I am not missing the point... more precisely, YOUR point, and that is exactly the point.

... This was a hypothetical problem posed as an exercise for a student...
But the student (apprentice) is the one who posed the question, seeking a better understanding of what's involved. He cannot ask questions regarding related issues that he is not aware of.

...Even if it were a real problem it is doubtful that anyone would bother to throw in the secondary effects. I would think that two significant figures is good enough for In. If we follow your line of reasoning we could never use the algebraic formula because the angles might be off a little.
That is not my line of reasoning, but rather your interpretation of it. My line of reasoning is that the result of the algebraic calculation is not one of absolute precision, for other factors not included in the calculation will have a bearing in a real situation. Not armed with that knowledge, the apprentice will not be able to make accurate assessments in such a real life situation, and that situation may well be one where his livelihood, or more drastically his life, is at stake.

I have argued before with engineers who think that each problem should be solved with the greatest possible precision. I would argue though that a seasoned engineer should know when to use approximations and not waste resources with overly complex solutions.
I do not disagree, but I have to ask how many of those engineers, including yourself, do installation and maintenance field work in this area? Say an electrician has to troubleshoot a circuit and has the same exact criteria as the problem posed. In the process, there are two other 4-wire mwbcs running in the same conduit and all three neutral conductors, not bundled or otherwise marked, have a reading in the range of 13A to 17A. Which neutral conductor is the one involved?

Your statements are technically correct, but in my opinion they are answers to questions not asked.
Again, an apprentice cannot ask questions on issues he is not aware of.
 

K2500

Senior Member
Location
Texas
We've been working long hours, sorry for the delay.

Take it easy guys, I think if you take a look back through the thread, that you will find that you are both right from one perspective or another.

That said, this is where I am at:

I have no formal higher education. I metioned previosly that I was begining to study PF, then decided, after reviewing this thread(once again) that an understanding of vector math would be nessasary to do this. The majority of my studying lately has been here http://www.ibiblio.org/obp/electricCircuits/AC/AC_2.html

I study on my own time, when the days work is done, so I apologize for the amount of time it takes for me to reply(where is spell check:) ). I'm still looking for a decent app to apply the vector stuff, once i find it, I will be able to toy with more (hypothetical type) problems. I can only learn so much just reading about it.

To achieve Rms values I'm thinking I would take the product of the waveform peak and .707. Would I be able to take the avarage multiplyed by 1.414 to get the peak, then go from there? Is the practical avarage, as apposed to the technical avarage of 0, the same as nonimal(120, 240...), are these even the correct multiplyers for this application?
How do the reistive loads cause the current to lag the voltage? It seems like peak power occours at the intersection of the V and I waveforms, so I can see why a (unity?)PF would be the most efficient. Why do reactive loads cause a leading PF? Would a three phase induction wound motor be considered a purely reactive load? On a system that is technically a three phase four wire wye, but in practice is a three phase three wire, would load balance be of much greater importance or less? The system in question is 480v with the neutral feeding only the HRG equipment. Belive me when I say, I have more questions than the ability to conjure them. Thank you for all the replies.
 
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rattus

Senior Member
RMS first:

RMS first:

K2500,

There is a formula for computing the RMS value of a sinusoidal wave, but it involves integral calculus, so you must take it on faith that the 0.707 factor is correct for sinusoidal voltages and currents.

FYI though, RMS means the Square root of the Mean value of the voltage or current Squared, but don't sweat that right now.

The mathematical average of a sinusoidal voltage (e.g., 0.636*Vp) is of little use. The RMS value is widely used however, and it applies only to voltages and currents, never to power.

With a resistive load, the average power is given by,

Pavg = Vrms * Irms

RMS values are "effective" values in that they produce the same heating effect in a resistor as a DC voltage or current of the same magnitude.

Example:

What is the power dissipation in a 12 Ohm resistor with 120Vrms across its terminals.

Pavg = Vrms^2/R = 120V^2/12 = 1200W

If you apply 120Vdc to this resistor, the average power is the same.

You must get this solidly in your mind before you proceed.
 

Smart $

Esteemed Member
Location
Ohio
K2500 said:
We've been working long hours, sorry for the delay.
No apology necessary, but appreciated all the same.

Take it easy guys, I think if you take a look back through the thread, that you will find that you are both right from one perspective or another.
Thank you.

That said, this is where I am at:

I have no formal higher education. I metioned previosly that I was begining to study PF, then decided, after reviewing this thread(once again) that an understanding of vector math would be nessasary to do this. The majority of my studying lately has been here http://www.ibiblio.org/obp/electricCircuits/AC/AC_2.html
Are you not participating in a formal apprenticeship program? Recognized apprenticeship programs are a form of formal higher education. I heard some even provide participants with college credits.

The website you linked has some great reference material.

I study on my own time, when the days work is done, so I apologize for the amount of time it takes for me to reply(where is spell check:) ). I'm still looking for a decent app to apply the vector stuff, once i find it, I will be able to toy with more (hypothetical type) problems. I can only learn so much just reading about it.
The initiative you demonstrate is commendable.

To achieve Rms values I'm thinking I would take the product of the waveform peak and .707. Would I be able to take the avarage multiplyed by 1.414 to get the peak, then go from there? Is the practical avarage, as apposed to the technical avarage of 0, the same as nonimal(120, 240...), are these even the correct multiplyers for this application?
I'm not sure what you mean by practical average, but the relationships are:
Effective value = 0.707 ? maximum value
Maximum value = 1.414 ? effective value

Average value = 0.636 ? maximum value
Maximum value = 1.572 ? average value​
The average value is not much use to practical people... [I see in the midst of previewing my message that rattus has covered this and elaborated already. There is no point in my being redundant :wink: ]

How do the reistive loads cause the current to lag the voltage?
You have gotten the wrong impression somewhere along the line. Purely resistive load current neither lead nor lag the voltage. However, there may be some reactance (induction) on the circuit from wiring method. That causes lagging current, but it is usually minimal in effect.

It seems like peak power occours at the intersection of the V and I waveforms, so I can see why a (unity?)PF would be the most efficient. Why do reactive loads cause a leading PF?
Reactive loads are ones which have at least two [edit: which are resistive and one other type, though the resistive component may be very near zero], but possibly all three types of impedence: resistive, capacitive, and inductive (RCL for short). Capacitive reactance results in a leading current and inductive a lagging current.

Would a three phase induction wound motor be considered a purely reactive load?
No. While it achieves its purpose through induction, it also has at least a resistive component to its impedance from the wire used in its windings.

On a system that is technically a three phase four wire wye, but in practice is a three phase three wire, would load balance be of much greater importance or less? The system in question is 480v with the neutral feeding only the HRG equipment.
"...feeding HRG equipment"? High resistance grounding equipment? Help me out here, I must be having a premature senior moment :D

Balancing a system is more critical when the system is heavily loaded. Whether the system is being utilized as 3?, 3-wire or 4-wire makes no difference AFAIK.

Belive me when I say, I have more questions than the ability to conjure them. Thank you for all the replies.
I thought as much... but keep 'em coming :grin:
 
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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
K2500 said:
To achieve Rms values I'm thinking I would take the product of the waveform peak and .707. Would I be able to take the avarage multiplyed by 1.414 to get the peak, then go from there?
In case you haven't tried it, multiply these two numbers together; they're reciprocals.

Since they are approximations, so is the product. It's not exactly 1, but it's real close.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
K2500,

There is a formula for computing the RMS value of a sinusoidal wave, but it involves integral calculus, so you must take it on faith that the 0.707 factor is correct for sinusoidal voltages and currents.
A graphical representation...

rms.gif
 

K2500

Senior Member
Location
Texas
Smart $ said:
Reactive loads are ones which have at least two [edit: which are resistive and one other type, though the resistive component may be very near zero], but possibly all three types of impedence: resistive, capacitive, and inductive (RCL for short). Capacitive reactance results in a leading current and inductive a lagging current.

I see, I took reactive loads to be either capacitive or inductive. Could you explain capsasitive reactance in more detail?

Smart $ said:
While it achieves its purpose through induction, it also has at least a resistive component to its impedance from the wire used in its windings.

It does make sense, now, that most loads would have at least some resistive componants, however negliable.

Smart $ said:
"...feeding HRG equipment"? High resistance grounding equipment?

Yes high resistance grounding.

Smart $ said:
Balancing a system is more critical when the system is heavily loaded.


Do you mean to say that transformers don't run on smoke? :grin:



rattus said:
K2500,

With a resistive load, the average power is given by,

Pavg = Vrms * Irms

With reactive loads, I understand that the power flows through the load, that is, back to the source. By that I mean, when illustrated, it transverses the negative cycle. How would Pavg be with such a load?

rattus said:
RMS values are "effective" values in that they produce the same heating effect in a resistor as a DC voltage or current of the same magnitude.

I belive it was stated earlier that we compare these like this to use DC formulas. Is that the primary function of RMS, to allow us the use of DC formulas?
 

Smart $

Esteemed Member
Location
Ohio
K2500 said:
I see, I took reactive loads to be either capacitive or inductive. Could you explain capsasitive reactance in more detail?
I could... but why rewrite what's already been written?

For example, another lesson from the website you linked above.

With reactive loads, I understand that the power flows through the load, that is, back to the source. By that I mean, when illustrated, it transverses the negative cycle. How would Pavg be with such a load?
Reactive loads have current which lead or lag the voltage. The cosine of the angular difference (theta or θ) between voltage and current waveforms is equal to the power factor (PF or pf)...
cos θ = PF

Pavg = Vrms * Irms * PF​
...where PF will be 1 for purely resistive loads and less than one for reactive loads.
I belive it was stated earlier that we compare these like this to use DC formulas. Is that the primary function of RMS, to allow us the use of DC formulas?
Well, the truth of the matter is that even though we call them DC formulas, they also apply to AC. However, since AC has continuously varying values you'd have to resort to using calculus to compute everything. Instead we use one calculus "averaging" method to simplify the measurements and calculations... and arrive at the same result in the end.
 

rattus

Senior Member
Nice Diagram:

Nice Diagram:

Smart $ said:
A graphical representation...

rms.gif

Nice diagram Smart, but it implies that 0.707 is the average value of a sinusoidal wave, but you know it is 0.636.

Give us a cos^2 plot where the average is 0.5, and the square root is 0.707.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Nice diagram Smart, but it implies that 0.707 is the average value of a sinusoidal wave, but you know it is 0.636.
You misinterpret. Try looking at the diagram from the perspective of area under the curve and not the values along the curve. This is what the RMS value of a sinusoidal curve is... the average height of the area under a sinusoidal waveform. The only difference between this and what you propose, as quoted below...
Give us a cos^2 plot where the average is 0.5, and the square root is 0.707.
...is that it easier for the brain to perceive how the lopped top half of the curve can be inverted and used to fill the valley. But it depicts only part of the process, not the result. Ask yourself, why is it that we use the root mean square method to determine the 0.707 multiplier to begin with?

rms2.gif
 
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rattus

Senior Member
But:

But:

Smart $ said:
You misinterpret. Try looking at the diagram from the perspective of area under the curve and not the values along the curve. This is what the RMS value of a sinusoidal curve is... the average height of the area under a sinusoidal waveform. The only difference between this and what you propose, as quoted below...

...is that it easier for the brain to perceive how the lopped top half of the curve can be inverted and used to fill the valley. But it depicts only part of the process, not the result. Ask yourself, why is it that we use the root mean square method to determine the 0.707 multiplier to begin with?

rms2.gif

But Smart,

You have plotted a cosine wave and by setting the two areas equal you are saying that the average value is 0.707. The average value of the absolute value of a sinusoid is 0.636!

To obtain the value of 0.707, you must perform a definite integral of the sinusoid SQUARED, determine the MEAN value of the sinusoid, and take the square ROOT of the MEAN. That is the reason we must use the cosine squared function in this case.

The average value of the cosine squared is 0.5, and its square root is 0.707.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
But Smart,

You have plotted a cosine wave and by setting the two areas equal you are saying that the average value is 0.707. The average value of the absolute value of a sinusoid is 0.636!
Afraid not!

This plot...

rms2.gif


...is of y = f(x) = cos(x)^2 where x has a range of 0 to 2pi.

The average height of the area under this curve is 0.5. The arithmetic mean value is also 0.5.
 
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