Answer / ravi693

If field of DC Shunt generator will opened then generated
voltage will reduced to 2 or 3 volts.

Answer / v.karthikeyan

If the sunt field of the Generator got opened the Back EMF
will be drastically reduced and there will not be flux
linkage between the armature field and sunt field.Hence
output voltage will not be as sufficient as stipulated.

Answer / j.d

if the field winding of the shunt generator gets open then the generating voltage gets reduced to very low level finally it gets zero output from the generator, because field winding is the main element to produce flux. The flux is directly proportional to the output voltage.

Answer / hskk

if in dc generator field is open the flux will be zero and
as speed is inversly proportional ti flux the speed will be
infinite theoretically. it means practically the speed will
so high that it will damage the machine

Answer / vilas

In dc shunt generator if the field is open in running
ondition the generators speed will increase,definately,but
in practical there some protection, as a field failure
relay,it will operate and stop the generator.

Answer / lavakumar

As the speed is inversely proportional to flux,as field is
removed,speed is said to increase theoritically but
practically the speed enourmously increases which may
damage the generator.

Answer / v.selvakumar

n=emf/flux according to this formula field open means flux
will be zero.flux zero means speed will be infinity,but we
know that no machine can't reach infinite speed.so that
therotically ok but practically starting speed will
increase then it will reduced zero.

Answer / nimish

if the field of dc shunt generator is open the speed of the
generator will be infinite and it will damage.

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