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A 4-pole generator has a wave wound armature with 730 conductors and it delivers 110 A on full load. If the brush lead is 10 degrees mechanical, calculate the cross-magnetising ampere turn per pole.
1. 2000.33
2. 5801.45
3. 1200.12
4. 3903.47

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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, θ= 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

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