Question

A 50 kVA, 6600/230V, single phase transformer has h.v & l.v winding resistances of 7 Ω and
0.008 Ω respectively. With l.v winding open, a current of 0.3A at a p.f of 0.3 lagging is recorded
on h.v side with the application of full rated voltage. Calculate efficiency at full load and 0.8 p.f
lagging. Determine also the load current at which maximum efficiency occurs.

Answer 1

Answer:

very big and vegshekmLahdnzk

Answer 2

Answer:

    The maximum efficiency is 6.61 Ampere.

Explanation:

During an open circuit, the power drawn is approximately equal to iron loss (if rated voltage is being applied).

       So Iron Loss Pc is  $Pc=6600*0.3*0.3=594watt$.

Equivalent resistance at hv side$=7+(0.008)∗(6600230)2=13.58$ ΩAt .        maximum efficiency, Cu loss =Iron LossSo    

           Culoss$=I^{2} *13.58$

                       $=594$

                I=$\frac{\sqrt{594} }{13.58}$

                  $=6.61$Ampere