SSC JE Electrical Engineering 2015 Paper

Question : 140 of 199

Marks: +1, -0

A short shunt compound generator supplies a loud current of 100 A at 250 V. The generator has the following winding resistances : shunt field = 130Ω, Armature = 0.1Ω and series field = 0. 1Ω. Find the Emf generated if the brush drop is IV per brush.

272.2 volt
262.2 volt
272.0 volt
262.0 volt

Validate

Solution:

E = V + I_1 R_{\se} + I _aR_a + V_B

I _L=( {V+I_LR_{\se} }/R_{\SH})

={ 250+ 100 ×.1}/ 130= 2A

I_ a = I_L + I_F = 100 + 2 = 102 A

Because armature current is 55 A,so we use wave winding and wave winding has two brush so brush voltage drop

(V_B) 2× 1 =2v

E = 250 + 100 × .01 + 102 × .1 + 2

= 262.2

volt

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