SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 173 of 200

Marks: +1, -0

A DC generator delivers a 30-A current to a load at 120 V by consuming 4500 J/s of mechanical energy. Determine the efficiency of the generator.

Solution:

Current delivered to load, I = 30 A
Voltage at the load, V = 120 V
Mechanical power input,

\P_\m

= 4500 J/s = 4500 W
Electrical power output,

\P_\e

= V × I = 120 × 30 = 3600 W
From concept,

\η={3600}/{4500}=0.8

=80 \%

(In percentage)

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