Back e.m.f.:
When the motor armature
rotates,the conductors also rotate and hence cut the flux.In accordance with
the laws of electromagnetic induction, e.m.f.is induced in them whose
direction,as found by Fleming,s Right-hand Rule,is in opposition to
the applied voltage. Because of its opposite direction, it is referred to as
counter e.m.f. or back e.m.f.Eb.
Fig:1
Prove that the gross power developed by a DC motor is
maximum when the back e.m.f. (Eb) equals to the half of the applied
voltage(V):
Back e.m.f.Eb=Vt-IaRa
Where Vt=applied terminal voltage, Ia=armature current, Ra=armature resistance
The gross mechanical power
developed by a motor is Pm=VIa-I2aRa
Differentiating both sides with
respect to Ia and equating the result to zero,we get
dPm/dIa=V-2IaRa=0
Or, IaRa=V/2
As V=Eb+IaRa
and IaRa=V/2 So Eb=V/2
Thus gross mechanical power
developed by a motor is maximum when back e.m.f. is equal to half the applied
voltage.
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