SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 154 of 200

Marks: +1, -0

A consumer has annual consumption of 7,00,800 units. If his maximum demand is 200 kW. The load factor will be :

Solution:

Percent load factor

= {\Average \Consumption }/{ \Maximum \demand}×100

Now Average Consumption

= {\Daily \consumption \of \energy}/ {24 \hrs}

∴ Average Consumption =

{700800 ×1000}/{24 ×365}

= 80,000 W(∵ 1 unit = 1 kwhr)
Hence percent load factor =

80000/{200× 1000}×100=40%

∴ Option (c) is correct.

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