What is an infinite bus

[@eepower]

This is by far the best best explanation of what infinite bus. Thanks for make this great video.

[@educationpower1823]

Nice job Tom Appreciate it Robert from Omaha

[@theworldine]

1:55 utility Z p.u. calculation results was 0.0104 but at 2:55 why it became 0.0083 p.u.?

[@uttamkumarbanerjee3666]

THANKS A LOT . I like your leaching style

[@namsogen4729]

Can someone explain why Utility Z p.u. suddenly become 0.0083 p.u.? Thanks

[@TDUNPLUGGED]

I just put together a short video to answer your question as well as another on a different topic. here is the video. I would suggest you look at the derrivation of the equation to convert the available fault current into an %impedance and you'll see how I came up with that value of impedance. thanks for the question!!! kzbin.info/www/bejne/maW8l3yIo86nack

[@MrDands]

@@TDUNPLUGGED Tom, thanks for the link to the other video, but the Utility ISC here is a different value and no X/R was given in your example here. Could you walk us through how you got the 0.0083 p.u.? Thanks

[@deanmacy8291]

@@MrDands I think it was an accident and 25,000A was used for the utility Isc. X/R ratio doesn't matter when dealing with impedances in pu analysis... unless of course you are trying to find the real power/voltage drop.

[@deanmacy8291]

@@MrDands Unless he forgot to put the X/R ratio of the Utility which could be 1.32 and he is only using the reactive part of the impedance (which makes sense cause phase angle of 90 deg) which would give an impedance value of 0.0083 ang 90 deg. But that wouldn't make sense either cause why would you not want to use the real part of the impedance when calculating Isc? Especially when the real part of the transformer is being used in the calculation. Never mind now I'm just confusing myself lol.

[@teddyacero]

A lot in so little, thanks.

[@nadersad]

Could you speak slower!! its really hard for me!