Question

An alternator having an induced e.m.f of 1.6
p.u. is connected to an infinite bus of 1.0 p.u.
If the bus bar has reactance of 0.6 p.u. and
alternator has reactance of 0.2 p.u., the maxi-
mum power that can be transferred is given by
A. 2 p.u.
B. 2.67 p.u.
C. 5 p.u.
D. 6 p.u.​

Answer 1

Answer:

A

Step-by-step explanation:

Maximum steady state power = EV/X

Where,

E = Sending side voltage

V = Receiving side voltage

X = Reactance of the system

Maximum steady state power = 1.6*1/(0.6+0.2) = 2 pu

Answer 2

Answer:

Power generated in an alternator is given by the formula

P=(E*V) /X

Where

P=power generated in the alternator

E=emf=1.6

V=voltage=1

X=reactance of the system=(0.6+0.2) =0.8

So power P=(1.6*1) /0.8

=1.6/0.8

=2

So the answer is a. 2 p.u