SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 194 of 200

Marks: +1, -0

The maximum demand of a consumer is 2 kW and his daily energy consumption is 20 units. His load factor is :

Solution:

Load factor =

{\Average \demand}/{\Maximum \demand}

∴ Average demand =

{\Daily \consumed \energy}/{24 \hour}

{20 ×1000}/24

(∵ 1 unit = 1 kwhr)
∴ %Load factor =

[{20×1000}/{24×2×1000}×100]%=41.6%

Go to Question:

Prev Question

Next Question