Note the HT side VOLTAGE,Current,P.F at the same time calculate the L.T side Voltage,current ,P.F.Convert the L.T side V,I,P.F to H.T side by using transformation ratio and calculate the Power in KW for both H.T side and L.T with respect to H.T Then Loss=H.T side power in K.W-L.T side power in K.W with respect to H.T
To calculate the no load current from transformer & core loss is also calculated.
the efficiency is maximum in a transformer when no load loss is equal to load loss.
You need to specify the phase. I assume it is 3 phase system. Then the HT current is 30.3 amps
Sir I am Imran khan in nepal my problem is 6000KVA Step down Transformer is not charging. Transformer Pri- 33000V /105A Sec- 6900V / 502Atransformer all test (open & short circuit ) good . but problem is HT(IR)= value this time IR value 50M ohmBefore 3days HT(IR) = value is 7000M ohm suddenly start Transformer result is Transformer not charging & HT(IR) value is down 50M ohm.
there are several losses in a transformer that prevent it from attaining 100% efficiency. One is core loss, which can be divided into Hysteresis losses, Eddy currents and Magnetostriction loses. see for more details http://en.wikipedia.org/wiki/Transformer#Energy_losses
Voltage transformer or potential Transformers are used in conjunction with Current transformers for metering and protection purpose in HT and LT power system.
why mentioning loss angle 0.002 in transformer
The maximum efficiency condition in distribution transformer is said to be occurred when iron loss = copper loss
Frist putt of the main incomer HT side then Rack out the breaker and put the LOTO after that Discharge the cable Both side (HT/LT).
112.5
Copper loss varies with the load.
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