A small dc motor operating at 200 V draws a current of 5.0 A at its full speed of `100 pi rad s^(-1)`. The resistance of the armature of the motor is `10 Omega` . Determine the back emf of the motor. Obtain the power input, power output and efficiency of the motor.

Explanation

Given : V=200 ,I=5.0,R= `10 Omega,omega=100 pi rad s^(-1)`.
if `varepsilon` is the back emf then ,
`V-varepsilon=IR` (from KVL)
`therefore varepsilon=V-IR=200 -(5)(10)=150V`
Input power , `VI =200xx5=1000W`
Output power , `varepsilonI=150xx5=750W`
Efficiency of the motor , `eta=("output power")/("input power")xx100`
i.e., `eta=(750)/(1000)xx100`
`eta=75% `