Ans will be D. half load cu loss=f.l cu loss×(1/2)^2 so, h.l cu loss= 4000×1/4=1000w Btw options given r wrng... one more 0 will be add on three options

To find the demagnetizing and cross magnetizing ampere turns per pole of a 6 pole wave wound DC generator, we need to consider the armature reaction.

Armature reaction refers to the effect of armature current on the magnetic field produced by the field winding. It causes a shift in the magnetic field, leading to a change in the flux distribution and effective field strength.

Armature Reaction Ampere Turns:
The armature reaction ampere turns can be divided into two components: demagnetizing ampere turns and cross-magnetizing ampere turns.

1. Demagnetizing Ampere Turns:
Demagnetizing ampere turns are the component of armature reaction that reduces the effective field strength. It opposes the main field and tends to demagnetize the poles. The formula to calculate demagnetizing ampere turns is given by:

Demagnetizing Ampere Turns = (Z/2) * Ia * cos(δ)

Where,
Z = Number of armature conductors
Ia = Armature current
δ = Angle of brush shift from the Geometrical Neutral Axis (G.N.A.)

In this case, Z = 460, Ia = 180 A, and δ = 6 degrees.

Demagnetizing Ampere Turns = (460/2) * 180 * cos(6)
= 230 * 180 * 0.9945
≈ 82272

2. Cross-Magnetizing Ampere Turns:
Cross-magnetizing ampere turns are the component of armature reaction that creates a magnetic field perpendicular to the main field. It causes a distortion in the flux distribution. The formula to calculate cross-magnetizing ampere turns is given by:

Cross-Magnetizing Ampere Turns = (Z/2) * Ia * sin(δ)

Using the same values as above:

Cross-Magnetizing Ampere Turns = (460/2) * 180 * sin(6)
= 230 * 180 * 0.1045
≈ 4296

Therefore, the demagnetizing and cross-magnetizing ampere turns per pole are approximately 82272 and 4296, respectively. Hence, the correct answer is option 'D' which states (230, 3220).