Control power transformer - Secondary's view of primary's short circuit
Control power transformer - Secondary's view of primary's short circuit
(OP)
Hello,
Suppose that a 480/120V, 350VA transformer is fused on the secondary, and that there are 10,000A fault current available on the primary. Is it correct that the secondary fuse should be rated to interrupt the available 10,000A? If so, has anyone witnessed such an event in practice? Thanks for your thoughts.
Suppose that a 480/120V, 350VA transformer is fused on the secondary, and that there are 10,000A fault current available on the primary. Is it correct that the secondary fuse should be rated to interrupt the available 10,000A? If so, has anyone witnessed such an event in practice? Thanks for your thoughts.






RE: Control power transformer - Secondary's view of primary's short circuit
RE: Control power transformer - Secondary's view of primary's short circuit
Bill
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"Why not the best?"
Jimmy Carter
RE: Control power transformer - Secondary's view of primary's short circuit
RE: Control power transformer - Secondary's view of primary's short circuit
On another note, if you are bound to NEC rules, you should put a fuse on the primary side to protect the transformer in case there was an internal short.
RE: Control power transformer - Secondary's view of primary's short circuit
RE: Control power transformer - Secondary's view of primary's short circuit
RE: Control power transformer - Secondary's view of primary's short circuit
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If we learn from our mistakes I'm getting a great education!
RE: Control power transformer - Secondary's view of primary's short circuit
Using secondary as base,
I_base = 150VA / 120V = 1.25 A
Z_base = 120V / 1.25A = 96 Ohm
I_fault = 10000A / 1.25A = 8000 A pu
Z_fault = 1.0V / 8000 Ohm = 1.25e-4 Ohm
Then fault current seen by secondary fuse is I = 1.0 / (0.035 + 1.25e-4) = 28.47A ?
RE: Control power transformer - Secondary's view of primary's short circuit
Longer answer... ignore the system impedance and assume only the transformer impedance, giving you:
If(pu) = 1 / 0.035 = 28.57 pu
If(A) = 28.57 x (150VA/120V) = 35.71 A
You could add in the system impedance, but you'll probably find the change is negligible.