Question

A 200 V separately excited DC motor has an armature resistance of 0.05 draws an armature current of 20 A, running at 1,200 rpm. Calculate the approximate torque developed for given armature current.
1. π/200 Nm
2. 10/4 Nm
3. 100/π Nm
4. π/100 Nm

Answer 1

Correct Answer - Option 3 : 100/π Nm

Concept:

In a separately excited DC motor, the back emf is given by

E_b = V_t - I_aR_a

Where V_t is the terminal voltage

I_a is the armature current

R_a is the armature resistance

The power developed by the motor (P) = E_bI_a

The torque developed by motor,

Where N is the speed

Calculation:

Given that, terminal voltage (V_t) = 200 V

Armature current (I_a) = 20 A

Armature resistance (R_a) = 0.05 Ω

Back emf (E_b) = 200 – 20 × 0.05 = 199 V

The power developed = 199 × 20 = 3980

The torque developed, \(T = \frac{{30}}{{1200 \times \pi }} \times 3980\)

\( = \frac{{99.5}}{\pi } \approx \frac{{100}}{\pi }Nm\)