Answer / rajeshkumar k

The power factor at no load is low because the magnetizing component of input cur rent is a large part of the total input current of the motor. When the load on the motor is increased, the in-phase current supplied to the motor increases, but the magnetizing component of current remains the same. This means that the resultant line current is more nearly in phase with the voltage and the power factor is improved when the motor is loaded, compared with an unloaded motor which has its magnetizing current as a major component of the input current.

Answer / sree

At the starting condition, the induction motor draws higher magnetisation component of current in order to overcome the reluctance offered by air gap between stator and rotor. Then the total current increases.  In the phasor representation, the angle between coreloss component and net current increases. As the power factor is the cos of angle between net current and core loss component, therefore the power factor decreases.

Answer / suman pakhrin

cos∅=R∕√((R2)²+(s.X)²)

During starting,reactance of rotor is maximum Xr=X since slip=1
as load increases slip decreases and so does the reactance
as Xr=S.X therefore its obvious from above that during starting the impedance offered by rotor is more inductive as that during loading conditions!and hence power factor at start low compared to loadin conditions

Answer / kiran kumar c

At the starting condition, the induction motor draws higher magnetisation component of current in order to overcome the reluctance offered by air gap between stator and rotor. Then the total current increases.A motor which exhibits a high starting current, i.e. 850% will generally produce a low starting torque, The induction motor operates due to the torque developed by the interaction of the stator field and the rotor field. Both of these fields are due to currents which have resistive or in phase components and reactive or out of phase components. The torque developed is dependant on the interaction of this phase components "A low rotor resistance will result in the current being controlled by the inductive component of the circuit, yielding a high out of phase current and a low torque".

Answer / xyz

Because electrical current is inversly proportional to cos phi or vice versa at starting there is high current so pf is low

Answer / electrical usa

because if the power factor would have been high at the starting, you morons have asked
"why induction motor having high power factor at starting time and high at full load?

so, here is the answer

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