Suppose a single-phase transformer is subjected to open-circuit and short-circuit tests. Additionally, the core loss (given in some arbitrary Watts) is at unity power factor. It just so happens that for this transformer,
Copper losses (I^2 * R) = calculated short-circuit power
Core losses = calculated open-circuit power
Strictly qualitatively speaking, can you imagine why/what constitutes both losses being equal to their respective short-circuit and open-circuit power ratings?
P.S. This is stated in the solution for Problem #4.29, p. 143, of the Kaplan Power Afternoon Sample Test
I owe you guys big time for all your efforts in helping me out...
Copper losses (I^2 * R) = calculated short-circuit power
Core losses = calculated open-circuit power
Strictly qualitatively speaking, can you imagine why/what constitutes both losses being equal to their respective short-circuit and open-circuit power ratings?
P.S. This is stated in the solution for Problem #4.29, p. 143, of the Kaplan Power Afternoon Sample Test
I owe you guys big time for all your efforts in helping me out...