PPT Slide
Section 12.9: The Nonideal Transformer and its Equivalent Circuits cont.
The full-load power losses are 2.18 KW.
When half-loaded, the load draws half its rated apparent power at rated voltage so the magnitude of the winding currents is halved
So the total half-load copper power losses are:
(63.65A/2)2´0.2W + (625A/2)2´0.002W = 398 W.
The core loss remains 589 W. (since the supply voltage doesn’t change)
So the total half-load power losses are: (398 + 589)W. = 987 W.
Calculate the previous example’s all-day efficiency if it’s load schedule is as follows: Full-load at 0.8 lagging power factor for 12 hours, no-load for 4 hours and half-load at unity power factor for 8 hours