Electrical Engineering ⇒ Topic : Armature Torque of a Motor
Consider Equation and multiply both the sides by /a ,we get
In this equation, V Ia is the total electrical power input to the motor, is the copper loss in the armature, and Eb/a is the gross mechanical power developed. All this mechanical power not available at the load because some of it is lost as mechanical loss and some more in core losses If T is the torque in Nm, which is defined as the turning moment of a force about an axis,then the gross mechanical output power developed can be written as
where,ω , the regular velocity in rad/sec 2 πN/60, in which N is the speed of the motor in rpm, i.e. mechanical power developed (P) = (2πT.N)/60 Watts
Substituting Eb = (ΦN.Z/50) (2P/2a) ,we get
For a given machine, Z and P are fixed and therefore
i.e. the torque in a dc motor is proportional to the product of armature current and flux per pole.
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