Question

A motor has an armature resistance of 4 ohm. On a 240 V supply and a light load, the motor speed is 200 rpm and the armature current is 5 A Calculate the motor speed at a full load. When armature current is 20 A.

Answer 1

Here, `R = 4 ohm, V - 240 V, I_(1) = 5 A`,
`V_(1) = 200 rmp, I_(2) = 20 A, v_(2) = ?`
Let `E_(1), E_(2)` be the back emfs in the two cases.
As `I = (V - E)/(R )`
`:. 5 = (240 - E_(1))/(4)` or `E_(1) = 220 V`
and `20 = (240 - E_(2))/(4)` or `E_(2) = 160 V`
As speed of motor is protortinal to back e.m.f., therefore, full load speed
`= (E_(2))/(E_(1)) xx v_(1) = (160)/(220) xx 200 = 145.45 r p m`