Question

A compensated DC machine has 19000 armature ampere turns per pole. The ratio of pole arc to pole pitch is 0.7. Interpolar air gap length and flux density are respectively 1 cm and 0.3 T. For rated armature current of 1000 A. The number of turns on each interpole and compensating winding conductors per pole would be respectively.

• A. 7 and 28
• B. 8 and 32
• C. 7 and 32
• D. 8 and 28

Answer 1

The correct option is D 8 and 28

$\text{Compensating winding AT/pole = armature AT/pole}\times \frac{Polearc}{Polepitch}$
$=19000\times 0.7=13300$

$\text{Turn/pole}$$=\frac{A{T}_{cw/pole}}{\text{Armature current}}=\frac{13300}{1000}=13.3=14$

$\text{No.of compensating conductor per pole,}$
$=14\times 2=28$

$\text{AT for airgap under interpole}$ $=\frac{B_g}{{\mu }_g}{l}_g=\frac{0.3}{4\pi \times {10}^{-7}}\times 1\times {10}^{-2}=2387.324AT$

$\text{Net AT for interpole}$ $=19000+2387.324-14000$

$\text{No.of turns in interpole}$ $=\frac{19000+2387.324-14000}{1000}=8$