$$V_4^{(3)} = \frac{1}{-j21.02564} \left[ \frac{3.18 - j0.90}{0.98634 - j0.14656} - (j13.33333(0.95818 + j0.00600) + j7.69231(0.95046 - j0.03616)) \right]$$

asked by guest
on Mar 05, 2026 at 1:13 am



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