SSC JE Electrical Engineering 2011 Paper

Question : 121 of 150

Marks: +1, -0

A dc shunt generator is delivering 500 A at 220 V. The shunt field current is 10 A. The armature resistance is 0.01 Ω. The stray power is 5000 W. The efficiency of the generator is

91.09%
95.82%
95.64%
91.82%

Validate

Solution:

Out Put

= V I_L

= 220 × 500
= 110 KW

I _a = I_1 + I_\f

= 500 +10 = 510 A
Armature power loss

I^2_a R=( 510) ^2 ×0.01 = 2601 W

R_\f=v/I_\f=220/10= 22l

field power loss

I^2_\f R_\f= (100) ^2 22 = 2200

Stray loss = 500
loss = stray loss +

I^2_a R+I^2_\f R_\f

2.601+ 2.2 5 = 9.801 W

\efficiency = \output/{\output +\losses}

=110/{110+9.801}=91.82%

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