A small d.c. motor operating at 200 V draws a current of 5.0 amp.at its full speed of 3000 r.p.m. The resistance of the armature of the motor is `8.5 Omega`. Determine the back e.m.f. of the motor. Obtain the power input, power output and efficiency of motor.

Explanation

Here, `V = 200 V, I = 5.0` amp.
`n = 3000//60 = 50 r.p.s., R = 8.5 Omega`
If E is the back e.m.f. then
`I = (V - E)/(R )` or `V - E = IR`
`:. E = V - IR = 200 - 5 xx 8.5 = 157.5` volt
Input power `= VI = 200 xx 5 = 1000` watt
Output power `= VI - I^(2) R = I (V -IR) = IE`
`= 5 xx 157.5 = 787.5 W`
Efficiency = Output power/Input power
`eta = (787.5)/(1000) = 0.7875 = 78.75%`