How can you accurately calculate fault current on a finite bus by hand?

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Calculating fault current on a finite bus differs from an infinite bus due to the limitations in available current and system impedance. In the example of a 225KVA transformer with 5% impedance, the primary current is affected by the load on the secondary side, which influences the fault current calculation. The principle of power conservation (power in equals power out) is crucial, as it dictates that the source must supply the necessary current based on the secondary load. The discussion highlights confusion over the application of Ohm's law in this context, particularly regarding resistance and the implications of finite versus infinite current sources. Understanding these concepts is essential for accurate fault current calculations in practical scenarios.
danielandham
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so calculating FC on an infinite bus is easy but how do you calculate it by hand for a finite bus?

here is the concept i am having an issue understanding:

let say i have a transformer that is 225KVA with 5%z and 3 phase 480 to 208.

with infinite bus i have 12491A available on the secondary of this transformer IF 12291A is flowing on the secondary then based on the voltage ratio and OHMs law 5412.8 A is flowing on the primary.

If this is true then what does it matter if i have infinite amps available on the primary or i have 5412.9 amps available 12491A should still flow in my mind.

I have run this in SKM and i see that it does matter but i just don't understand why.
 
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Let's talk in an ideal world. If you have no load, no current will flow. As you add a load to the secondary, current will flow based on the secondary voltage. Since power in == power out, then equal power will flow in the primary and you can calculate the primary current based on the primary voltage. So the key is power in == power out. Your source merely needs to be able to supply the required current.

I'm not sure where ohms law comes in since you aren't talking resistance yet. I don't quite understand your infinite vs finite either. Am I missing some basic assumption?
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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