Calculating Excitation Voltage Ef for Synchronous Motor: 350V, Xs=80Ω, 1A, 0.8 PF Lagging

Hi, I'm doing some homework, and was wondering if anyone would help me on track with this problem.
(I'm translating it, so it might be a bit incorrect)

It says:A synchronous motor with synchronousreactance (X_s) of 80 ohms is coupled to 350 V line to line voltage.
The motor absorbs 1 (A) current with 0.8 lagging.

Find the machines magnetizing voltage/excitation voltage (Ef) (as line to neutral voltage).

I was thinking of finding Ef like this: Ef = Vln (Vll/sqr3) + I_a * XsEf = 202.073 + 1*80 = 282.073VAm i completly wrong on this or? 
Thank you in advance

Hi Alba king,the excitation voltage Ef is given by the following expression:Ef = Va -j*Xs*Iawhere Va and Ia are the voltage and current of the synchronous machine equivalent circuit (you can find it on any textbook)In your specific case, we have:Va = 350V/sqrt(3) = 202.07VA lagging power factor of 0.8 corresponds to a power factor angle theta given by:
theta = -cos-1 (0.8) = -36.87°The phase a current is then given by:Ia = 1 * e-j36.87° APutting all together:Ef = 202.07 -j*80* (1 * e-j36.87°) V = 217.15 V

That makes sense.Thank you Paul!