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May 17th, 2018, 06:14 AM   #1
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3 phase AC signal resultant vector

I want to derive that the resultant of the 3 phase AC signal is a pointer around the circle or in other words the locus is a circle. Am i correct in the above statement and how to arrive at it? Please advise. The 3 phase AC are 3 sine waves with 120 degrees apart.
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May 17th, 2018, 10:47 AM   #2
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Assuming all three phases have the same amplitude the resultant will always be the zero vector.

You can see this geometrically by placing the three vectors tip to tail and noting that they form an equilateral triangle.
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May 17th, 2018, 01:34 PM   #3
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Quote:
Originally Posted by romsek View Post
Assuming all three phases have the same amplitude the resultant will always be the zero vector.

You can see this geometrically by placing the three vectors tip to tail and noting that they form an equilateral triangle.
and I guess more importantly that the tail of the final vector is at the tip of the first.
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May 17th, 2018, 04:01 PM   #4
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At a particular point in time all three waves cannot have the same amplitude.
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May 17th, 2018, 05:39 PM   #5
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One will A*sin(theta) other A*sin(120 + theta) and one more A*sin(240 + theta).
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May 17th, 2018, 06:37 PM   #6
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Quote:
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One will A*sin(theta) other A*sin(120 + theta) and one more A*sin(240 + theta).
gotcha.

Ok just add them

$\begin{align*}
&A\sin(\theta) + A\sin(\theta + 120^\circ) + A\sin(\theta+ 240^\circ) =\\ \\

&A\left(\sin(\theta) + \left(\sin(\theta)\cos(120^\circ)+ \cos(\theta)\sin(120^\circ)\right) +
\left(\sin(\theta)\cos(240^\circ) + \cos(\theta)\sin(240^\circ)\right) \right)= \\ \\

&A\left(\sin(\theta)-\sin(\theta)\dfrac 1 2 + \cos(\theta)\dfrac{\sqrt{3}}{2} - \sin(\theta)\dfrac 1 2 - \cos(\theta)\dfrac{\sqrt{3}}{2}\right) = \\

&A\left(\sin(\theta) - 2\cdot \dfrac 1 2 \sin(\theta)\right) = \\ \\

&0
\end{align*}$
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