Question

Suppose the no load voltage V = 90V, Ra + R1 = 4 ohms, Rsh + RF = 150 ohms, and the motor runs at a rated speed of 1000 RPM when IL = 4 amps. What is the armature current IA in amps at the rated speed

Answer 1

Explanation:

Field current= 400200=2A

So, no load armature current= 5–2=3A

So,

ENL=V−IaRa=400–3×0.5=398.5V

and full load armature current

= 50–2=48A

So,

EFL=400–48×0.5=376V

E is the back emf and for a dc motor

E=kφN

So, E is directly proportional to speed (N) as φ for a dc shunt motor remains constant for constant field current and k is constant.

So, EFLENL=NFLNNL=376398.5=0.94

Or, it can be said that full load speed is 94% of no-load speed.

Answer 2

Answer:Ia=3.4[A] at 1000 rpm

Explanation:

If=90/150=0.6[A]

Ia=IL-If=4-0.6=3.4[A]

E=V-(Ra+R1)*Ia=90-4*3.4=76.4[V]