nicholaaaas
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PF= Cos(Tan^-1(VARS/True Power))
Power factor and VA vs Watts
- Thread starter cyriousn
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Cold Fusion
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Smart -
You have certainly straightened me out about this thread.
later
cf
You have certainly straightened me out about this thread.
I agree
Using the current english definition of "instantaneous", this statement is wrong. Please note that I am certainly not trying to convinince you of this fact.
Yor're right. In fact, for this thread I vote for pig latin
Is this a pun on "Dirk Gently's Holistic Detective Agency". If it is, it's a good one.
For this I am horrible apologetic. In my defence, I can only say that adhering to the known body of the laws of God and physics that have been built upon for the last 350 years, would never have occured to me as asking you to lessen yourself. I would never want you to do that.
Iyay inkthay Iyay dooday.:grin:
later
cf
What about it is wrong????...
Not a pun at all... it's interesting though that you found it entertaining.
I do believe you would not do it... knowingly. But sometimes people's eyes have to be opened for them. And your defense is pretty weak considering my offense has essentially the same background
Cold Fusion
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It isn't going anywhere. You made the assertion. You bugged out. Where do think its going to go? Perhaps you just prefer that it go away
Hameedulla-Ekhlas
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Cold Fusion
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No I'm still here. I don't know where the thread will go. I don't have a preference.It isn't going anywhere. You made the assertion. You bugged out. Where do think its going to go? Perhaps you just prefer that it go away
Why is this wrong? Well, it is a trick question - I can't tell you because there isn't any. All of the necessary discussion detailing why this is true has already been posted. Repeating won't help. So, I am following your lead.
gar
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Ham:
See
http://www.karlscalculus.org/trig_id.html
Go to equations 7.1b-10a and 10b
In post 155 I clearly indicated what sin^2 t was.
In post 179 you provided a plot of instantaneous power to a resistor without a specified relation to the voltage, and your assumption was for a cosine voltage curve without my knowing that. I had to assume you were referring to a sine wave because that was what I had defined in post 155, and the post to which you were questioning.
.
Ham:
See
http://www.karlscalculus.org/trig_id.html
Go to equations 7.1b-10a and 10b
In post 155 I clearly indicated what sin^2 t was.
In post 179 you provided a plot of instantaneous power to a resistor without a specified relation to the voltage, and your assumption was for a cosine voltage curve without my knowing that. I had to assume you were referring to a sine wave because that was what I had defined in post 155, and the post to which you were questioning.
.
Hameedulla-Ekhlas
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gar
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100417-0941 EST
Ham:
It is wrong for a sine wave and correct for a cosine wave.
.
Ham:
It is wrong for a sine wave and correct for a cosine wave.
.
Hameedulla-Ekhlas
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Cold Fusion
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Ham -
I'm not jumping in. I just have a question to help me follow along.
Can I rewrite p = p + pcos2wt as:
p(t) = VI + VIcos2wt or:
p(t) = VI (1 + cos2wt)
If I can, that will help me understand what your are saying. If not, then I am still lost - but that's okay, I'll catch up eventually.
cf
gar
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Ham:
The reason I prefer sine waves is that I grew up with them, and it is virtually impossible to synchronize on the peak of a waveform on an oscilloscope. Too much amplitude variation and dv/dt is small.
I have no idea what current books do when discussing sinusoidal waveforms. In general the books I have discuss sine waves as the reference.
Except for special reasons, on most arbitrary waveforms I synchronize on a zero crossing. But for TTL logic it is usually about 2 V. Zero crossing sync for sinusoidal waveforms means I am looking at a sine wave.
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Ham:
The reason I prefer sine waves is that I grew up with them, and it is virtually impossible to synchronize on the peak of a waveform on an oscilloscope. Too much amplitude variation and dv/dt is small.
I have no idea what current books do when discussing sinusoidal waveforms. In general the books I have discuss sine waves as the reference.
Except for special reasons, on most arbitrary waveforms I synchronize on a zero crossing. But for TTL logic it is usually about 2 V. Zero crossing sync for sinusoidal waveforms means I am looking at a sine wave.
.
Hameedulla-Ekhlas
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Ok wait please and there is a confusion for me too. I know you, smart and gar are more experience than me. But I want to make it clear for me.
Trying to catch up here...
I followed your "proof" in post # 240.
p = P + Pcos2wt is correct when θv=θi, or pf=1.
Otherwise, the third term in your proof does not cancel out (i.e. where θv<>θi or pf<1).
p = P + Pcos2wt - Vm*Im*sin(0v - 0i)sin2wt/2
Hameedulla-Ekhlas
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Ummm... forgot to mention your "P" is not an accurate representation of P_real. You reduced your eq. 8 by P=Vm*Im cos(0v-0i)/2. The waveform of P_real is Vm*cos(wt)*Im*cos(0v-0i)
Unless you are computer language scripting, rewriting the equation with lower case p on both sides of the equation as cf did does not make any mathematical sense unless for all instances of t, cos2wt=0
Hameedulla-Ekhlas
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My only concern, p = p + pcos2wt should be ploted by consin not by sine. Because
1- According to Electric Circuits 7th Edition Nilsson-Riedel 2004 book it says.
We are operating in the sinusoidal steady state, so we may choose any convenient reference for zero time. Engineers concerned with operating systems involving the transfer of large blocks of power have found that a zero time corresponding to the instant when the current is passing through a positive maximum is convenient.
2- According to Transforms and Applications Handbook, Third Edition by Alexander D. Poularikas.page-51
p = cos(0i-0v) = P/S
"is called the power factor. The power factor may be regarded as a normalized correlation coefficient of the voltage and current signals while sin(0i - 0v) = SQR(1 - sqr(p)) may be called the anticorrelation of coefficient.
Now I need your help and I might be wrong but I want to clear my self thanks
Hameedulla-Ekhlas
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Smart$ I have to go out and will reply you later.
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