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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%                                             

Electrical & Electronics, Engineering

  • Category:- Electrical & Electronics
  • Reference No.:- M9504820

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