Answer / tarak

the flux becomes very less dat is only residual
magnetism.as speedinversly proportional to flux d speed
will become very high.

Answer / ashokreddy r

When field ckt is opened field current will be zero,flux
will be zero,as speed is inversly proportional to speed
becoz of zero flux motor tries to rotate at maximum
(infinite),dangerous speed .motor will be damaged(un rooted
etc..)

Answer / gajanan

when fied is open then field current will be zero and so
field flux also tends to zero,as speed is inversaly
proportional to flux, motor will rotate at (infinite)high
speed then winding will burn

if as always protection is use or drive used for motor then
it will trip and stops motor and protect it from burning

Answer / taraka mohan

The speed of the shuntmotor becomes high(uncontrollable)
and the windings are burn

Answer / tommy

When field ckt is opened field current will be zero,flux
will be zero,as speed is inversly proportional to speed
becoz of zero flux motor tries to rotate at maximum
(infinite),dangerous speed .motor will be damaged(un rooted
etc..)

Answer / nitin panwar

when a dc motor field is suddenly opened then flux dies out or flux decreases and as speed is inversely proportional to the flux therefore speed increases than the normal speed

Answer / menu

For a dc shunt motor if field is suddenly opened, no
current flow so no flux and no magnetic field..so according
to MOTOT PRINCIPLE...conductor experiences torque only if
it is placed in magnetic field...hence the speed of motor
decreases slowly due to inertia and finally it stops.

Answer / nayan mondal

when the shunt field is open the flux becomes near zero, not
reaches the flux zero due to there have some residual
magnetism. so it will runs at very high speed(SPEED
INVERSELY PROPORTIONAL TO FLUX)

Answer / manoj

when field circuit is open sudden high torque wii develop
due to this as speed is directly proportional to speed
motor runs very fastly than its rated speed

Answer / anusha

when rield circuit is suddenly opened, as the flux becomes
zero, back emf would become zero, hence high current flows
through armature circuit and the armature winding may
burnout.

We know that primary transmission is done to reduce current so that i^2*r loss could be minimised but according to ohm's law v is directly proportional to i,which indirectly means higher potential difference means larger current?so how the current is minimised in high power transmission line??

2 Answers  

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