 The term torque means ‘Turning movement of the force about an axis.’
T = F × r Newton
– meter
Where T = Torque
F = Force in Newton
 Consider an armature of radius r meter and force F newton acts on it.
 Let us assume that the armature rotate at speed of N rpm.
 When the armature rotates one revolution, it cuts distance 2πr in time of 60 / N second. Therefore the work done per revolution
= Force ×
distance
= F × 2πr
But F × r = T
 So the work – done / revolution = 2πT Newton – meter
 Now the Power developed
= 2πT / ( 60 / N
)
= 2πNT / 60
= Tω
Where ω =
Angular velocity in radian / second
= 2πN / 60
 The electrical equivalent to mechanical power developed by the armature is given by
T = ( 60 / 2πN )
E_{b}I_{a} …………….( 1 )
T = 9.55 ( E_{b}I_{a }/ N ) 
If the speed is given in revolution per second ( rps )
T = ( 9.55 / 60 ) ( E_{b}I_{a }/ N )

 T = 0.159 ( E_{b}I_{a }/ N )
 As the back emf E_{b} = ФZNP / 60A
T = ( 60 / 2πN )
( ФZNP / 60A ) I_{a}
= ( 1 / 2π ) ( ФZNP / A ) I_{a} N –
m
= [ 1
/ ( 2π × 9.81 ) ] ( ФZNP / A ) I_{a} Kg – m
 The number of conductor Z, number of poles P and number of parallel paths A is constant in the DC motor therefore
T α ФI_{a}
_{}
_{}
Shaft
Torque
 The shaft torque T_{sh} always less than the armature torque due to small amount of friction losses in the motor.
Shaft torque =
Armature torque – Friction and windage losses
T_{sh} =
T_{a} – Friction and windage losses
Output
power
 Output power = Power developed in the armature
 The mechanical power develops at the shaft of the DC motor is always less than the armature power due to friction and windage losses.
P_{sh} =
T_{sh} × ( 2πNT / 60 ) Watt
 The mechanical power developed at the shaft is called as brake horse power ( BHP ).
One HP = 735.5
watt
P_{sh} =
( T_{sh} × 2πN / 60 )( 1 / 735.5
) HP
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