Question

A 12-pole, 165 A, dc shunt generator has 662 lap connected armature conductors. Field winding takes current of 5 A. At full load, brushes are shifted from GNA by $4^{\circ }$ mechanical. Then the cross magnetizing ampere turns per pole are____ AT/pole.

• A. 286.56

Answer 1

The correct option is A 286.56
$\text{At full load,}$



$\text{Armature current,}{I}_a=165+5=170A$
${\theta }_e=\frac{P}{2}.{\theta }_m=\frac{12}{2}\times 4={24}^{\circ }$

$\text{Cross magnetizing AT/Pole}$
$=\left(\frac{{180}^{\circ }-2{\theta }_e}{{180}^{\circ }}\right)\times \frac{Z/2}{P}\times \frac{I_a}{A}$ $\text{(For lap connected windings A = P = 12)}$

$=(\frac{180-2\times 24}{{180}^{\circ }}\times \frac{662/2}{12}\times \frac{170}{12})$

$=286.56AT/Pole$