Correct Answer - Option 4 : 3903.47

__Concept:__

In a DC generator:

Demagnetizing ampere turn \(\left( A{{T}_{d}} \right)=ZI\frac{{{\theta }_{m}}}{360{}^\circ }\)

Cross-magnetizing ampere turn \(\left( A{{T}_{c}} \right)=ZI\left[ \frac{1}{2P}-\frac{{{\theta }_{m}}}{360{}^\circ } \right]\)

Where,

Z = Total number of conductors

I = Armature current (Ia) per conductor = Ia/A

A = Number of parallel paths

θm = Mechanical brush shift angle

P = Number of poles

__Calculation:__

P = 4, Z = 730, Ia = 110 A, θm = 10°

A = 2 (for wave winding)

\(\therefore I=\frac{110}{2}=55~A\)

Cross-magnetizing ampere turn \(\left( A{{T}_{c}} \right)=730\times 55\left[ \frac{1}{2\times 4}-\frac{10}{360} \right]\)

= 3903.47