\(\begin{array}{l}
{X_d}\; = \;0.8,\phi \; = \;36.86\\
{X_q}\; = \;0.6,{R_a}\; = \;0\\
\tan {\rm{\psi }} = \frac{{Vsin\emptyset + {I_a}.{X_q}}}{{V\;cos\emptyset }}\\
\psi = 56.30
\end{array}\)

for synchronous motor at leading p.f.

\(\begin{array}{l}
\psi \; = \;\phi \; + \;\delta \\
\Rightarrow \delta \; = \;19.70^\circ
\end{array}\)

Now, \(E\; = \;V\;cos\delta \; + \;{I_d}{X_d}\)

\(\begin{array}{l}
{I_d}\; = \;{I_a}\;sin\;\; = \;0.831\\
E\; = \;\left( 1 \right)\;cos\;\left( {19.7} \right)\; + \;\left( {0.831} \right)\;\left( {0.8} \right)\; = \;1.606
\end{array}\)