What if we use multiple cables in parallel? Then the load current will be divided amongst them and cable size can also be reduced. I dont know which will cost less.
Thank you cranky108. If we consider load current as 1 pu resistive and fault current as 5 pu inductive and add them vectorially, the result would be sqrt(26) which is very close to 5. So even if load current is 20% we can ignore it. Beautiful explanation.
Thanks
Why load current is usually neglected in fault studies? Fault current is usually 5-6 times the load current. How we can neglect load current which is as high as 20%?
regards
Sunil
