Question

Topic : single phase transformer , Short-circuit & Open-Circuit , Efficiency of a transformer My question is : Is it logical to say that the Copper loss or Short-circuit watt meter reading is a FULL LOAD reading @ 3-kVA? And if the load is rated for 2-kVA do you have to take 2 parts of the 3-KVA's watt meter reading for the short circuit test ?

Answer 1

i'm horny

Answer 2

when you are calculating for the current at load side you will take 2/3 of full load current
\[I_2=\frac{ rated\ KVA }{ V_2 }\times \frac{2}{3}\]\[I_2^{'}=\frac{ I_2 }{ a }\]