A separately excited DC motor has an armature resistance of 0.5 W. It runs from a 200 V DC supply drawing an armature current of 20 A at 1500 rpm. For the same field current, the torque developed for an armature current of 10 A will be
A separately excited DC motor has an armature resistance of 0.5 W. It runs from a 200 V DC supply drawing an armature current of 20 A at 1500 rpm. For the same field current, the torque developed for an armature current of 10 A will be
Right Answer is:
15.28 N-m
SOLUTION
Given parameters Ra = 0.5 ohm
Supply Voltage V = 250 V
Armature current Ia = 20 A
Speed N = 1500 rpm
Armature current Ia = 20 A
Torque T =?
From the voltage equation, the back EMF of DC motor is
Eb = V − IaRa
= 250 − 20 × 0.5
Eb = 240
The torque of DC motor is
T = P ⁄ ω
Where
P = Output power of separately excited motor and it is given as P = EbIa
ω = Angular speed in rad/sec. and it is given as ω = 2πN ⁄ 60
Hence for the armature current of 10 A the torque developed is
T = EbIa × 60 ⁄ 2πN
= 240 × 10 × 60 ⁄ 2π × 1500
= 15.28 N-m