# GATE 2015 EE – SET 2 – Q 49

Solution :

3$$\phi$$, 11kV , 50 Hz , P = 2 $$\Rightarrow$$

11kV = ,50 Hz  $$\Rightarrow$$ source

$$X_{s}$$ = 50$$\frac{\Omega}{ph}$$

$$I_{1}$$ = 100 A at cos$$\phi$$ = 1 , $$I_{2}$$= 120A

$$E_{f}$$ =$$V_{t}\angle 0^{0} – jI {a}\bar{X_{s}}$$

$$E_{f}$$ =$$\left [ \frac{11000}{\sqrt{3}} \right ] -j(120)(50)$$

$$E_{f}$$ = 6350.85 – j 6000

$$E_{f}$$ = 8736.89 – $$\angle-47.26$$ (load angle)

$$\delta$$ = -47.26