Question

In a PMDC motor, when the armature is rotated at 100rpm, an emf of 2V is developed across its terminals. For this motor, calculate the armature current flow required to generate a torque of 0.4Nm

Answer 1

Required torque($T$ ) = 0.4 Nm

$Eb$ =2V

Mechanical power delivered =$P = Eb*Ia =T * \omega$

Where $Eb$ = Back emf and , $Ia$ = Armature current, $\omega$ = anguar frequency.

$\omega = 2\pi*n/60$ $= 2\pi*100/60$ = 10.472 rad/sec

Therefore Ia = $T*\omega/Eb$ = 0.4*10.472/2 = 2.0944 A