Question

A d.c series motor is connected across a 460 - V supply runs at 500 r/min when the current is 40 A . The total resistance of the armature and field circuits is 0.6 ohms. calculate the torque on the armature in Newton meters.

Answer 1

$Input \space voltage=460V$

$Speed(N) =500rpm\implies \newline\omega=2\pi N/60=52.36 \space rad/sec$

$Current(Ia) =40A$

$Total \space resistance\space of armature \space and \space field=\newline Rse+Ra=0.6\Omega$

$We \space have, V=Eb+Ia(Ra+Rse) \newline Hence , Back \space emf (Eb) =460-40\times0.6=436V$

$We \space have, \newline EbIa=T\omega\implies\newline Torque(T) \newline=EbIa/\omega=436\times40/52.36= 333.08Nm$