Answer to Question #203301 in Electrical Engineering for Sonwabile

Question

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

 I_a = 100∠ 30°  A;            I_b=50∠ 300° A ;    and       I_c=30∠ 180 ^° A.

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

Answer 1

Write the components matrix

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

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

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

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

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

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