As per the given in the question,
Xd=1X_d =1Xd=1
Xq=0.6X_q=0.6Xq=0.6
ra=0.02Ωr_a=0.02\Omegara=0.02Ω
Power angle=?
Excitation emf=?
We know that Vcosδ=XdI−−−−(i)V\cos \delta =X_d I----(i)Vcosδ=XdI−−−−(i)
and Vsinδ=XqI−−−(ii)V\sin \delta =X_q I---(ii)Vsinδ=XqI−−−(ii)
From (i) and (ii)
tanδ=XqXd=0.61\tan \delta =\frac{X_q}{X_d}=\frac{0.6}{1}tanδ=XdXq=10.6
δ=tan−1(0.6)=30.96∘\delta =\tan ^{-1}(0.6)=30.96^\circδ=tan−1(0.6)=30.96∘
E=(Ira)2+(IXd)2E=\sqrt{(Ir_a)^2+(IX_d)^2}E=(Ira)2+(IXd)2
⇒E=I2+0.0004I2\Rightarrow E =\sqrt{I^2+0.0004I^2}⇒E=I2+0.0004I2
But here the value of I is not given in the question, so it can not be determined