Find the total emf and armature current., Electrical Engineering

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.



                  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%                                             

Posted Date: 7/23/2012 1:33:04 AM | Location : United States

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