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Here's my working:

$$ V_P = \frac{V_L}{\sqrt{3}}$$ $$ \therefore V_P = \frac{21.651kV}{\sqrt{3}} = 12.5kV$$ $$ S = 3V_PI_P$$ $$ \therefore I_p = \frac{S}{3V_p} = \frac{2400+j1800kVa}{(3)(12500)} = 80\angle36.86^\circ$$

The solution given in the answer sheet is $$80\angle-36.86^\circ$$

Is there a mistake I made? because this messes up the answers for the rest of the tut. Thanks!


Your equation for complex power isn't correct:

$$S = 3VI^* $$

Remember that you use the complex conjugate of current. This makes \$ I_p = 80 \angle -36.86 ^\circ \$ as the answer sheet suggests.

  • \$\begingroup\$ Thanks! Just scoured the notes for that equation and frustratingly found it hidden quite deeply \$\endgroup\$ – Makoto Jan 9 '17 at 15:41
  • \$\begingroup\$ That sounds like 3-phase power classes to me! Electric machines courses can be pretty bad, too, with such subtleties. \$\endgroup\$ – calcium3000 Jan 9 '17 at 15:46
  • \$\begingroup\$ haha :P ill probably be posting questions about machines in a couple days... \$\endgroup\$ – Makoto Jan 9 '17 at 15:52

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