Power Formula

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Hello all,
I have a question, probably a controversial one:

is there any difference between the Formula- P= V*I and P= I*V?
is there any technical difference between the two expressions?
thank you.
 

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  • #2
ENE
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Hello all,
I have a question, probably a controversial one:

is there any difference between the Formula- P= V*I and P= I*V?
is there any technical difference between the two expressions?
thank you.
HELLO,
It is like 2*3=6 or 3*2=6
both are same
 
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  • #3
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in my academic days, I was taught both are not the same in a certain sense,
Power, P, was always V.Icosø, and never I.Vcosø.
 
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  • #4
sophiecentaur
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afaik, a Dot vector product is commutative. The cos of the angle between the vectors is the same whichever one you are measuring relative to: Cos(x) = Cos (-x).
So you can write it either way round.
 
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  • #5
ENE
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I was thing P= V*I as numeric
 
  • #6
ENE
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afaik, a Dot vector product is commutative. The cos of the angle between the vectors is the same whichever one you are measuring relative to: Cos(x) = Cos (-x).
Can you explain again?
 
  • #7
ENE
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Can you explain what you said?
 
  • #8
sophiecentaur
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Can you explain again?
IF you are talking about DC then VI is exactly the same as IV because multiplication is commutative (2X3 is the same as 3X2). If you are talking about AC then you need a vector product between Volts and Current. The Power is given by the Dot Product (not just the RMS Volts times the RMS Current). But that is also commutative so either way round is ok there too.
 
  • #9
ENE
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If you are talking about AC then you need a vector product between Volts and Current.
Thanks for explaining who to do this?
i was not aware as i usually work with DC.
 
  • #10
sophiecentaur
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Your electricity bill charges you for actual Energy delivered. That involves a Dot Product. Google Power Factor to find all about it.
 
  • #11
ENE
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In AC circuits, the power factor is the ratio of the real power that is used to do work and the apparent power that is supplied to the circuit. The power factor can get values in the range from 0 to 1. When all thepower is reactive power with no real power (usually inductive load) - the power factor is 0.
 
  • #12
ENE
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I know about power factor lag and lead of current or voltage due to presence of extra L or C load.
 
  • #13
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OK thanks Ene and Sophiecentaur, but will I be right to contribute from this point of view:
Mathematically the commutative product of Vector proves the product V.I to be equivalent to I.V
but in AC circuits for example, the mean power consumed by a circuit is given by the product of V and that of the CURRENT COMPONENT which is in phase with V, Voltage.
I assume that the fact of possibility that current can be resolved into two components- Isinø and IcosØ- is key in explaining the expressions V.IcosØ and I.VcosØ
the factor to be envisioned here is, cosØ, which is the power factor of the circuit.
current 'chases' voltage, I.e, it always tries to make up a perfect Power (constantly wants to achieve cos0°) hence IcosØ...
so will I be right to attribute I, current, as the variable component in the expression?
 
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  • #14
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I have no intention to teach something, actually. However, as I said somewhere, the vector notion
is not a clear one, today. Once, a vector was a real thing as force [F] or field [E electric or H,B magnetic] intensity, or current density [J] or else. After this a phasor-which represents on a paper a wave- it turned to be a "vector”. No vector product [cross product] it could be done with V or I or other such "vectors”. However, a scalar [dot] product it could be meaningful.
Now, extending the notion, a column of a matrix is named "a vector”. Also, a column of a library
stand... How this "vector" could present gradient , divergence or curl?:frown:
 
  • #15
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hi Babadag,
I'm sorry I don't understand what you stated. Maybe you should give a different perspective to it.
 
  • #16
jim hardy
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I suspect Mr yungboss is not versed in vector calculus.

in my academic days, I was taught both are not the same in a certain sense,
Power, P, was always V.Icosø, and never I.Vcosø.
If all three variables (V I and cosø) are just numbers then it doesn't matter in what order you write them. That's grade school algebra.
If in your prior training ø changes sign depending on which variable you take ,V or I, as starting point for the angle then Sophie's post #4 answers your question for power. But you'll have to pay attention to sign when calculating VARs, VIsinø .

I assume that the fact of possibility that current can be resolved into two components- Isinø and IcosØ- is key in explaining the expressions V.IcosØ and I.VcosØ
Yes, draw your phasor triangle and apply Pythagoras.

current 'chases' voltage, I.e, it always tries to make up a perfect Power (constantly wants to achieve cos0°) hence IcosØ...
I dont understand. Those verbs apply to the observer not to the inanimate charge in motion.
so will I be right to attribute I, current, as the variable component in the expression?
We cannot know what you have in mind. V, I and ∅ are all variables.
Can you draw a phasor diagram in Paint , label the variable(s) and upload it with button in lower right ?
 
  • #17
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please oblige my curiosity and response, not intending labor the point.
ok, what I'm tryingvto express is stated here https://en.m.wikipedia.org/wiki/Leading_and_lagging_current
voltage is generated and this gives rise to a resultant current that is either in phase, lead or lag. because of the above conditions (whereby current is in phase, lead or lag), the expression Icos∅ obtains.
I understand that the three components V, I and cosØ exist as variables but I was of the opinion that cosØ represents the cosine of the angle by which Current departs or conforms with the Voltage phase
sorry if I'm not making much meaning here...
but does the cosine law of vectors have an impact on the expression?
https://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/dotproduct.html
also this link, scroll downwards
https://en.m.wikipedia.org/wiki/Law_of_cosines
 
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  • #18
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but you all say the expression I.VcosØ is correct and technically same (from electricity viewpoint) as v.icos∅, then I must accept it.
many thanks
 
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  • #19
jim hardy
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I understand that the three components V, I and cosØ exist as variables but I was of the opinion that cosØ represents the cosine of the angle by which Current departs or conforms with the Voltage phase
Yes, this time you have stated that very clearly, i understand what you meant.

sorry if I'm not making much meaning here...
No apology necessary; congratulations on your improved clarity.
but does the cosine law of vectors have an impact on the expression?
https://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/dotproduct.html


upload_2016-5-14_16-3-47.png

i guess i don't understand the question.
Trigonometry applies to all triangles, be they phasors or vectors or real geometric shapes like land surveys.


Much of learning is just coming to believe what you already know. One reassures himself by working out with pencil and paper enough examples that the steps become automatic.
Make up three numbers and assign them to A B and cosAOB.
Let me call cosAOB by "C" so it's easier for me to type
There are only two possible sequences because i can repeat the sequence and start anyplace in it.

ABCABCABC is same sequence as BCABCABCA which is same as CABCABCAB just i started at a different place.
Likewise
ACBACBACB is same sequence as CBACBACBA which is same as BACBACBAC ...

does A X B X C equal A X C X B ?

You knew that all along .

old jim
 
  • #20
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got that right on old Jim, thanks...
 
  • #21
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We may represent an “actual” vector in Cartesian or spherical[polar] coordinates.
If we take only three dimensions and these coordinates are mutually perpendicular [ each on other]
then the dot product of two vectors
V1=X1i+y1j+z1k V2=x1i+y2j+z2k where i,j,k are the unit vectors of each coordinate line.
is the sum of the product of same coordinate
[x1*x2+y1*y2+z1*z2) and represents a scalar [a number].
The result of dot product of two phasor-vectors it is another phasor.
V*I*=(Vre+jVim)*(Ire-jIim)=(Vre*Ire+Vim*Iim)+j(-Vre*Iim-Vim*Ire)=S=P-jQ
here I*=I conjugate.
Vre*Ire+Vim*Iim=V*Ire if V is the origin of the angles since Vim=0.
Ire=I*cos(V,I) [I it is the absolute value of I]. I=SQRT(Ire^2+Iim^2)
P=V*I*cos(V,I) [V=Vre]

As one can see no V* it could be employed.
 

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