Calculating Line Current for Wye-Wye for finding Active Power

[BatmanisWatching1987]

Here is what the system looks like below


V_AN = 120∠0°
Z_T = (0.06+j0.12) + (12+j9) = 15.12∠37°

I_Aa = V_AN / Z_T = 7.94∠-37°

To Find Active Power I used the formula P_Load = 3V_pI_pcosθ

But is θ equal to -37° or +37° and why?

 

[mivey]

Here is what the system looks like below
View attachment 22224

V_AN = 120∠0°
Z_T = (0.06+j0.12) + (12+j9) = 15.12∠37°

I_Aa = V_AN / Z_T = 7.94∠-37°

To Find Active Power I used the formula P_Load = 3V_pI_pcosθ

But is θ equal to -37° or +37° and why?

+37.

We use the angle of the voltage minus the angle of the current so we have 0d - -37d = +37d.

Note that S = VI* = P + jQ where the * means we take the complex conjugate.

For inductive loads we have Q as a positive number and for capacitive loads we have Q as a negative number.

edit: typo

 

[BatmanisWatching1987]

Link to Example

If you look at Page 33, they use a positive 37​°, hence way I was confused.

Thanks for the help