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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

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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

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