Q. In a 110V compound generator, the armature, shunt and series winding resistance are 0.06?, 25? and 0.04? respectively. The load consists of 200 lamps each rated 55W, 110V connected on parallel. Find the total emf and armature current, when the machines is conducted for
(i)Long shunt
(ii)Short shunt
Ignore the armature and brush drop.
Ans.
Given Ra = 0.06?, Rsh = 25?, Rsc = 0.04?
II = 200 × 55/110 = 100 Amp
(i) For long shunt
If = 110/25 = 4.4 Amp
Ia = Ic + If = 100 + 4.4 = 104.4 Amp
Ea = V + Ia (Ra + Rsc)
= 110 + 104.4 (0.06 + 0.04) = 120.4 volt
(ii) For short shunt
Va = V + ILRsc = 110 + 100 × 0.04
Va = 114 Volt
If = V/Rf
If = 114/25
If = 4.56 Amp
Ia = IL + If = 100 + 4.56 = 104.56 Amp
Ea = Va + Ia + Ra
= 114 + 104.56 × 0.06 = 120.03 volt
(iii) Now with diverter
IdSe = 104.4 × 0.1/0.14 = 74.57 Amp
ISc = Ia = 104.4 Amp
Series field Ar reduce to
= 74.57/104.4 × 100 = 71.4%