appunni
Electrical
- Feb 11, 2003
- 52
Sir,
In solidly earthed transformer, the fault current varies with the fault position in the winding in a complex manner. The voltage is proportional to turns and impedance is proportional to square of turns. Then, how does the fault current reduces at about the middle portion of the winding and how does the fault current reduces immediately at points very near to neutral? Can anybody give a detailed explanation?
In solidly earthed transformer, the fault current varies with the fault position in the winding in a complex manner. The voltage is proportional to turns and impedance is proportional to square of turns. Then, how does the fault current reduces at about the middle portion of the winding and how does the fault current reduces immediately at points very near to neutral? Can anybody give a detailed explanation?