# 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 I_{a} = 20 A

Speed N = 1500 rpm

Armature current I_{a} = 20 A

Torque T =?

From the voltage equation, the back EMF of DC motor is

**E _{b} = V − I_{a}R_{a}**

= 250 − 20 × 0.5

**E _{b} = 240**

The torque of DC motor is

**T = P ⁄ ω**

Where

P = Output power of separately excited motor and it is given as **P = E _{b}I_{a}**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 =** **E _{b}I_{a}**

^{ }

**× 60 ⁄ 2πN**

= 240 × 10 × 60 **⁄ **2π × 1500

**= 15.28 N-m**