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      R_a = 0.06?, R_sh = 25?, R_sc = 0.04?

                                   I_I = 200 × 55/110 = 100 Amp

 (i) For long shunt

                                   I_f = 110/25 = 4.4 Amp

                                   I_a = I_c + I_f = 100 + 4.4 = 104.4 Amp

                                   E_a = V + I_a (R_a + R_sc)

                                        = 110 + 104.4 (0.06 + 0.04) = 120.4 volt

 (ii) For short shunt

                                   V_a = V + I_LR_sc = 110 + 100 × 0.04

                                    V_a = 114 Volt

                                    I_f = V/R_f

                                    I_f = 114/25

                                    I_f = 4.56 Amp

                                    I_a = I_L + I_f  = 100 + 4.56 = 104.56 Amp

                                    E_a = V_a + I_a + R_a

                                         = 114 + 104.56 × 0.06 = 120.03 volt

(iii) Now with diverter

                                    I_dS_e = 104.4 × 0.1/0.14 = 74.57 Amp

                                    I_Sc = I_a = 104.4 Amp

                       Series field A_r reduce to