Find the total emf and armature current., Electrical Engineering

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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 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

= 74.57/104.4 × 100 = 71.4%

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