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
Ia 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%
⇒ Ia2 = 150% Ia1 = 1.5 Ia1
As T ∝ ϕ Ia
\(\frac{{{T_2}}}{{{T_1}}} = \frac{{{\phi _2}}}{{{\phi _1}}} \times \frac{{{I_{a2}}}}{{{I_{a1}}}}\)
\(\frac{{{T_2}}}{{150}} = \frac{{0.9{\phi _2}}}{{{\phi _1}}} \times \frac{{1.5{I_{a2}}}}{{{I_{a1}}}} = 1.35\)
T2 = 202.5 N-m
Therefore the new value of torque is 202.5 N-m