SSC JE Electrical Engineering 2014 Set 1 Paper

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Question : 177 of 200

Marks: +1, -0

The maximum demand of a consumer is 2 kW and his daily energy consumption is 24 units. His load factor is _____ %.

24
41.6
50
80

Solution:

Load factor =

{ \Unit \generated}/{\Maximum \Demand ×\Hour}

% Load factor =

{24×1000}/{2×1000×24}×100

= 50%

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