Question

A compensated generator has 12000 armature ampere turns per pole. The ratio of pole arc/pole pitch = 0.7, length of interpole airgap = 1.25 cm and flux density in the interpole airgap $=0.3Wb/m^2.$ The ampere turns per pole for interpole winding_____ AT.

• A. 6585

Answer 1

The correct option is A 6585
$\text{Ampere turns for interpole}=\frac{B}{{\mu }_0}{l}_g+\text{total AT (1-0.7)}$

$=\frac{0.3}{4\pi \times {10}^{-7}}\times 1.25\times {10}^{-2}+12000\times 0.3$

$=6584.15AT\approx 6585AT$