# Available short circuit current



## RubyTuesday (Oct 19, 2014)

Oh I think I got it. I=P/E.:jester:


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## PlugsAndLights (Jan 19, 2016)

Here's a more long winded way to look at it: 
First consider the transformer to be 100% efficient. -> Pin=Pout
So whatever power is available to the primary is also available 
at the secondary. Now, as you said, apply I=P/E and you have 
a higher available fault current when the voltage is lower. 
Of course, there are other factors. 
Good Luck,
P&L


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