1/3 in Symmetrical Components

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Anonymous Poster #1	1/3 in Symmetrical Components 11/16/2016 5:23 AM
	Where does this 1/3 comes from in symmetrical components? Va1=1/3(Va+aVb+a^2VC)
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gutmonarch Guru Join Date: Sep 2016 Posts: 973 Good Answers: 9	#1 Re: 1/3 in symmetrical components 11/16/2016 6:53 AM
	Degrees of freedom, perhaps. Draw a vector in 2D plane and see if 1/2 applies.
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Rixter Guru Join Date: Apr 2010 Location: About 4000 miles from the center of the earth (+/-100 mi) Posts: 9910 Good Answers: 1141	#2 Re: 1/3 in Symmetrical Components 11/16/2016 9:02 AM
	Here is a tutorial.
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67model Guru Join Date: Dec 2010 Posts: 1686 Good Answers: 116	#3 Re: 1/3 in symmetrical components 11/16/2016 9:11 AM
	Sshh...... Questions that look like homework awake the sleeping Gruffalos There are textbooks with the proof of the theory in them with all the algebra. My ancient mind is telling me the theory does not just apply to 3 phases and the "3" is because it is 3 phase.
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Kilowatt0 Guru Join Date: Aug 2007 Location: Earth - I think. Posts: 2143 Good Answers: 165	#4 In reply to #3 Re: 1/3 in symmetrical components 11/16/2016 11:07 AM
	"Gruffalos"? I thought it was the Curmudgeonsaurus! __________________ TANSTAAFL (If you don't know what that means, Google it - yourself)
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Rixter Guru Join Date: Apr 2010 Location: About 4000 miles from the center of the earth (+/-100 mi) Posts: 9910 Good Answers: 1141	#5 Re: 1/3 in Symmetrical Components 11/16/2016 3:59 PM
	Where does this 1/3 comes from in symmetrical components? It comes in when you solve for V1a, V2a, and V3a in terms of Va, Vb, and Vc. This can be done with 3 simultaneous equations of 3 variables, but it is usually done using linear algebra. Linear algebra combines the three variables into a vector. The vector of the known values is multiplied by a 3x3 matrix to get the vector of the unknown quantities. If the matrix is invertible, the inverse operation can be done. If Y=AX, then X=A^-1Y In this case, the magnitude (determinant) of A is 3, so the magnitude of A^-1 is 1/3. The vector of [Va, Vb, Vc]^T is the matrix multiplied by [V1a, V2a, V3a]^T as shown above. To solve for [V1a, V2a, V3a]^T, you need to invert the matrix. The 1/3 comes in when the matrix which has magnitude = 3, is inverted to solve for Va1, Va2, and Va3. http://www.wikihow.com/Find-the-Inverse-of-a-3x3-Matrix http://www.zmuda.ece.ufl.edu/Fall_2013_Power_Systems/6-Symmetrical_Components.pdf
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67model (1); gutmonarch (1); Kilowatt0 (1); Rixter (2)

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