GATE 2015 EE – SET 1 – Question 43

Solution :

$$\delta $$= $$\frac{P_{a}t^{2}}{2M}$$ , where M = $$\frac{H}{f*180} $$

where H is the ratio of the stored KE to machine rating

H = 2 , f = 50 ,t = 0.02 ,P=1 pu

$$\delta $$= $$\frac{P_{a}t^{2}}{2M}$$ =  $$\frac{Pt^{2}}{2*\frac{H}{f*180}}$$

$$\delta $$= $$\frac{1*(0.02)^{2}}{2*\frac{2}{50*180}}$$ = 0.9

$$\delta $$ = 5 +0.9 = $$5.9^{0}$$