Excitation emf of motor drawing 160 A

Reading of data: "3.3 kVA" is a misprint for 3.3 kV (a 3.3 kVA motor cannot take 160 A). Star connection. The stray loss does not affect the excitation emf, since $E_f$ depends only on $V$, $I_a$ and $Z_s$.

$$V = \frac{3300}{\sqrt3} = 1905.26\ \text{V per phase}$$

Motor equation: $\vec E_f = \vec V - \vec I_a(R_a + jX_s)$, with $V$ as reference.

$$\begin{aligned} \vec I_a &= 160\angle -36.87^\circ = 128 - j96\ \text{A} \\ \vec I_aZ_s &= (128 - j96)(0.8 + j5.5) \\ &= 102.4 + 528 + j(704 - 76.8) = 630.4 + j627.2\ \text{V} \\ \vec E_f &= 1905.26 - (630.4 + j627.2) = 1274.86 - j627.2 \\ &= 1420.8\angle -26.20^\circ\ \text{V} \end{aligned}$$

Line value $= \sqrt3 \times 1420.8 = 2461$ V.

Since $E_f < V$ the motor is under-excited, which agrees with its lagging power factor.

(For reference: input $= \sqrt3 \times 3300 \times 160 \times 0.8 = 731.6$ kW, copper loss $= 3 \times 160^2 \times 0.8 = 61.44$ kW; the 30 kW stray loss would be used only for output or efficiency.)

Answer: Excitation emf $E_f = 1421$ V per phase (2.46 kV line), lagging $V$ by $26.20^\circ$.