Answer to Question #203301 in Electrical Engineering for Sonwabile

Question #203301

In a three-phase, four-wire system, the currents in lines A, B and C under abnormal conditions of loading were as follows:

 Ia = 100∠ 30°  A;            Ib=50∠ 300° A ;    and       Ic=30∠ 180 ° A.

 Determine the zero-phase sequence current  (Iao) in line A in polar form.  


1
Expert's answer
2021-06-07T05:46:46-0400

Write the components matrix

"\\begin{bmatrix}\n I_{a_{o} } \\\\\n I_{a_{1} } \\\\\n I_{a_{2} }\n\\end{bmatrix}=\\begin{bmatrix}\n 1 &1&1\\\\\n 1&k&k^2 \\\\\n 1&k^2&k\n\\end{bmatrix}\\begin{bmatrix}\n I_A \\\\\n I_B \\\\\n I_C\n\\end{bmatrix}"

"k=1\\angle120^o; k^2=1\\angle120^o"

"I_A= I_{a_{o} } + I_{a_{1} } + I_{a_{2} }"

"I_{a_{o} }=\\frac{1}{3} [I_a + I_b+I_c]= 81.777\\angle4.693=\\frac{81.777}{3}=27.259\\angle4.693^o"

"I_{a_{o} }=\\frac{1}{3} [I_A +k I_B+k^2I_C]=\\frac{1}{3} [100\\angle30^o+\\angle120^o+50\\angle300^o+30\\angle180^o]=\\frac{173.944\\angle43.29^oA}{3} =57.98\\angle43.29^oA"

"I_{a_{2} }=\\frac{1}{3} [I_A +k^2 I_B+kI_C]=\\frac{1}{3} [100\\angle30^o+\\angle-120^o+50\\angle300^o+\\angle120^o +30\\angle180^o]=\\frac{56.929\\angle24.96^oA} =18.973\\angle24.96"




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