# 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

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