Actually... we can.
You can join us if you wish, but it appears you wish to be excluded. Your choice. Don't say we didn't try.
I choose understanding
not making things fit my lack there of
enjoy your 'alternative facts'
:lol:
Is this calc right?
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I choose understanding
not making things fit my lack there of
enjoy your 'alternative facts'
:lol:
Well if you choose understanding over not I would think you'd be more open-minded. Apparently not and contradictions abound...
Well if you choose understanding over not I would think you'd be more open-minded. Apparently not and contradictions abound...
saying something that is incorrect, is correct, is not open minded
it's stupid
save your 'open mind' for things like tolerance of others and introspection of your shortcomings
not manipulation of fact to fit fiction
it's very difficult for you to accept your faults
let the reader form their own opinion on the matter
wwhitney
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Great. So far we have two conventions:
Every textbook Ingenieur has ever seen: Ia + Ib + Ic = In
Several posters here: Ia + Ib + Ic + In = 0.
Both are valid conventions. Either choice may be made, and the resulting physics is the same. All that differs is the mathematical representation, as far as the sign of In.
Ingenieur, if you have never seen the second convention before, consider this an opportunity to broaden your horizons.
Cheers, Wayne
it's not convention, nor valid
into node +, out of node -, that is convention
opposite in sign, not the same
I have seen it
it is incorrect, since when is +60 + (-92) = 32 ?
show me a text with Ia, Ib, Ic and In ALL going INTO the node/junction and ALL positively signed
let the reader decide
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wwhitney
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Exactly, that is the convention under which Ia + Ib + Ic + In = 0 is the statement of Kirchoff's current law. Which is the convention Smart$ used in the spreadsheet in the OP. And the numbers in the OP satisfy Ia + Ib + Ic + In = 0 (up to a decimal place). So what's the problem? What would you change in the OP's numbers?
Cheers, Wayne
no it isn't
KCL Ia + Ib + Ic - In = 0
actually Ia + Ib + Ic = In
60 - 92 not equal 32
show me a text using that convention
one
broadening my horizons does not involve incorrect statements
let the reader decide
wwhitney
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These two statements are in contradiction. If you choose the convention in your first statement, then KCL is Ia + Ib + Ic + In = 0. That's the mathematics.
Edit: Just look at an example with DC (or AC without any phase shifts, so scalar math), a 4 way node with Ia, Ib, Ic each 1 amp going into the node (so +1 amp), and In 3 amps going out of node (so -3 amps). Then Ia + Ib + Ic + In = 0.
Cheers, Wayne
are you serious?
Wrong
Summing at the node
Iabc + entering
In - leaving
summing
Ia + Ib + Ic + (-In) = 0
or Ia + Ib + Ic = In
one text, just one
let the reader decide
wwhitney
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You still have a sign error, see the example in my last post.
How about wikipedia? https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
Or if you want dead trees, Feynman's Lectures on Physics Volume II is on my shelf, page 22-8 equation 22.17.
Cheers, Wayne
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I cannot speak for engineering texts, but physics texts tend to leave out any prior "understanding" of how things work when setting up conventions.
All wires connected to the node are equal and equivalent. Either you look at all currents as positive when they enter the node or all are positive when they leave the node. The neutral is in no way "special".
If you have only one line hot, the current in that hot line and the neutral are equal and *opposite* at any moment in time. So their sum is zero.
mobile
All wires connected to the node are equal and equivalent. Either you look at all currents as positive when they enter the node or all are positive when they leave the node. The neutral is in no way "special".
If you have only one line hot, the current in that hot line and the neutral are equal and *opposite* at any moment in time. So their sum is zero.
mobile
mivey
Senior Member
Then you have been exposed to a limited range of texts. I'm telling you they are stated both ways.
The shorthand is what is confusing you. If we use a sending/receiving convention we get
INsr + IAsr + IBsr + ICsr = 0
If you mix sending and receiving you can get:
INrs = IAsr + IBsr + ICsr
The shorthand notation is making you incorrectly conclude that IN should always mean INrs but that is simply wrong.
You are in error.
mivey
Senior Member
Absolutely.:thumbsup:
mivey
Senior Member
I may see if I can post a few so it may help him expand his knowledge.
you are just making stuff up now
I am not the one confused or wrong
Show me one, just one textbook
let the reader decide
that is what you need to do
a page out of an ee power textbook
it is not my knowledge base that needs expansion
let the reader decide
simple
mivey
Senior Member
So we go from EVERY textbook to a EE power textbook? Already backing up/hedging your bets? How amusing.
But I'm bored at the moment so after I get a snack I'll grab a few off the shelf to see if I can't help you open your eyes. May take a little bit to scan and reference so you might have to wait.
wwhitney
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So, the way I've been reading this, and perhaps others too, is that currents into the node have positive values and currents out of the node have negative values. That convention yields Ia + Ib + Ic + In = 0.
But it occurs to me that you could mean instead that all current values are positive, and currents into the node get a positive coefficient, and currents out of the node get a negative coefficient. That convention could yield Ia + Ib + Ic - In = 0, as you have been writing.
Either convention works for DC, but for AC does this "into the node" or "out of the node" make any sense? Making a choice of 0 phase angle and simply writing Σ I = 0 seems simpler to me.
Cheers, Wayne
mivey
Senior Member
If you really think that is what Smart$ has calculated then you are not getting it.
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