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