A 4-pole shunt generator with lap-connected armature having field and armature
resistance of 50 ohm and 0.1 ohm respectively supplies sixty 100 V, 40W lamps.
Calculate the total armature current.
Answers
Answer:
Explanation:
Field copper loss. In the case of shunt generators, it is practically constant and I 2 R (or VIsh). In the case of series generator, it is = Ise2Rse where Rse is resistance of the series field winding.
This loss is about 20 to 30% of F.L. losses.
(iii) The loss due to brush contact resistance. It is usually included in the armature copper loss.
(b) Magnetic Losses (also known as iron or core losses),
(i) hysteresis loss, Wh µ B max f and (ii) eddy current loss, We µ Bmax f These losses are practically constant for shunt and compound-wound generators, because in their case, field current is approximately constant.
Both these losses total up to about 20 to 30% of F.L. losses.
(c) Mechanical Losses. These consist of :
(i) friction loss at bearings and commutator.
(ii) air-friction or windage loss of rotating armature.
These are about 10 to 20% of F.L. Losses.
Answer:
Load current = 60 × 40 100 = 24A
Shunt field current= 100/50 = 2A
Armature current = 24+2 = 26A
... Current in each conductor 26 4 = = 6.5A