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 E_b/_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 E_b = (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.