# 3-phase power: getting incorrect phase

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!

$$S = 3VI^*$$
Remember that you use the complex conjugate of current. This makes $I_p = 80 \angle -36.86 ^\circ$ as the answer sheet suggests.