Symmetrical Components Example

Symmetrical Components Example - Solving of 3_Phase Systems...

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I c 45.279 amp = arg I c () 34.3deg = I a I b + I c + 9.536 4.254j + amp = Now using Symmetrical Components: V p Z p I p = using V p A V s = V s A 1 V p = I p A I s = I s A 1 I p = we get A V s Z p A I s = multiplying with A -1 we get: A 1 A V s A 1 Z p A I s = which simplifies to: V s A 1 Z p A I s = V s Z s I s = with Z s A 1 Z p A = 1 3 1 1 1 1 e j 2 π 3 e j 4 π 3 1 e j 4 π 3 e j 2 π 3 Z ph Z n + Z n Z n Z n Z ph Z n + Z n Z n Z n Z ph Z n + 1 1 1 1 e j 4 π 3 e j 2 π 3 1 e j 2 π 3 e j 4 π 3 yields Z ph 3Z n + 0 0 0 Z ph 0 0 0 Z ph V s Z ph n + 0 0 0 Z ph 0 0 0 Z ph I s = V 0 V 1 V 2 Z ph n + 0 0 0 Z ph 0 0 0 Z ph I 0
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This note was uploaded on 04/21/2008 for the course EE 4343 taught by Professor Giesselmann during the Spring '08 term at Texas Tech.

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Symmetrical Components Example - Solving of 3_Phase Systems...

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