My Math Forum 3 phase AC signal resultant vector

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 May 17th, 2018, 05:14 AM #1 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 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.
 May 17th, 2018, 09:47 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281 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.
May 17th, 2018, 12:34 PM   #3
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Quote:
 Originally Posted by romsek 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.

 May 17th, 2018, 03:01 PM #4 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 At a particular point in time all three waves cannot have the same amplitude.
 May 17th, 2018, 04:39 PM #5 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 One will A*sin(theta) other A*sin(120 + theta) and one more A*sin(240 + theta).
May 17th, 2018, 05:37 PM   #6
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 Originally Posted by MathsLearner123 One will A*sin(theta) other A*sin(120 + theta) and one more A*sin(240 + theta).
gotcha.

\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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