Allday 24 0/P Allday = Allday 24 0/P+ (Pivan) 24 + (Pau) 201 24 A10 KVA transformer During the day it is loaded as follows: For 6 hours For 8 hours For 10 hours full load at 0.9 p.f lag halve load at 0.8 p.f lag no load If the iron loss equal 60 watt and copper loss equal 100 watt, determine its all day efficiency. ANS. 97.4%

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I need the solution to this problem by an expert using the above-mentioned law.
Allday 24 0/P
Allday = Allday 24 0/P+ (Pivan) 24 + (Pau) 201
24
A10 KVA transformer During the day it is loaded as follows:
For 6 hours
For 8 hours
For 10 hours
full load at 0.9 p.f lag
halve load at 0.8 p.f lag
no load
If the iron loss equal 60 watt and copper loss equal 100 watt, determine its all day
efficiency.
ANS. 97.4%
Transcribed Image Text:Allday 24 0/P Allday = Allday 24 0/P+ (Pivan) 24 + (Pau) 201 24 A10 KVA transformer During the day it is loaded as follows: For 6 hours For 8 hours For 10 hours full load at 0.9 p.f lag halve load at 0.8 p.f lag no load If the iron loss equal 60 watt and copper loss equal 100 watt, determine its all day efficiency. ANS. 97.4%
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