Question

A DC shunt motor of 200 V, 10.5 A, 2000 rpm has an armature resistance of 0.5 Ω and field resistance of 400 Ω. It drives a load whose torque is constant at rated motor torque. What is the value of armature current if the source voltage drops to 175 V?
1. 9.7 A
2. 12.4 A
3. 11.4 A
4. 10.7 A

Answer 1

Correct Answer - Option 3 : 11.4 A

Given V₁ = 200 V

Armature resistance (R_a) = 0.5 Ω

Field winding resistance (R_sh) = 400 Ω

Field current = 200/400 = 0.5 A

Load current = 10.5 A

Armature current = 10.5 – 0.5 = 10 A

Given that load torque is constant.

\({I_{a2}} = \frac{{200}}{{175}} \times 10 = 11.42\;A\)