Question

A DC Motor develops a torque of 150 N-m. A 10 percent reduction in the field flux causes a 50 percent increase in armature current. The new value of torque is:
1. 102.5 N-m
2. 202.5 N-m
3. 172.5 N-m
4. 232.5 N-m

Answer 1

Correct Answer - Option 2 : 202.5 N-m

Concept:

\(T = \frac{ZPϕ(I_a)}{A60}\)

Where, 

T is the torque of the DC motor

Z is the number of conductors

A is the number of parallel paths

P is the number of poles of the motor

ϕ is the field flux of the motor

I_a is armature current

For any DC machine,   \((\frac{ZP}{A60})\) remains constant

∴ T ∝ ϕ Ia

Calculation:

Given,

Torque T₁ = 150 N-m

There is 10% reduction in field flux 

⇒ ϕ₂ = 90% ϕ₁ = 0.9 ϕ₁

Also armature current increased by 50%

⇒ I_a2 = 150% I_a1 = 1.5 I_a1

As T ∝ ϕ I_a

\(\frac{{{T_2}}}{{{T_1}}} = \frac{{{\phi _2}}}{{{\phi _1}}} \times \frac{{{I_{a2}}}}{{{I_{a1}}}}\)

\(\frac{{{T_2}}}{{150}} = \frac{{0.9{\phi _2}}}{{{\phi _1}}} \times \frac{{1.5{I_{a2}}}}{{{I_{a1}}}} = 1.35\)

T₂ = 202.5 N-m

Therefore the new value of torque is 202.5 N-m