Determine the maximum primary fuse for installation in a 250-kva, 3-phase, 13.2- kv to 480/277-volt transformer with an impedance of 3.25% installed in an unsupervised location.

Determine the maximum primary fuse for installation in a 250-kva, 3-phase, 13.2- kv to 480/277-volt transformer with an impedance of 3.25% installed in an unsupervised location.

• A. 35 A
• B. 25 A
• C. 50 A
• D. 75 A

Answer: (A)

The key idea is protection coordination on the transformer’s primary: use the transformer's rated current and its impedance to pick a fuse size that carries normal load and startup inrush but will still clear a fault without letting dangerous damage occur. First, find the primary full-load current: I_primary,FL = S / (√3 × V_primary) = 250 kVA / (1.732 × 13.2 kV) ≈ 10.9 A. The transformer impedance is 3.25%. That means the available fault current on the primary is quite high, roughly I_sc ≈ I_primary,FL / Z% ≈ 10.9 A / 0.0325 ≈ 336 A. So the protective device must be able to interrupt faults well before this level, yet not trip during normal energization inrush. For an unsupervised location, a practical sizing approach is to select a primary fuse rating around a few times the full-load current to tolerate startup inrush while still providing protection. A factor around 3× I_fl gives about 32–33 A, so the next standard size is 35 A. Among the choices, 35 A provides the needed balance: it carries the normal 11 A load, tolerates typical startup surge, and will clear faults, whereas a smaller option could nuisance-trip, and larger options would reduce protection.

The key idea is protection coordination on the transformer’s primary: use the transformer's rated current and its impedance to pick a fuse size that carries normal load and startup inrush but will still clear a fault without letting dangerous damage occur.

First, find the primary full-load current:

I_primary,FL = S / (√3 × V_primary) = 250 kVA / (1.732 × 13.2 kV) ≈ 10.9 A.

The transformer impedance is 3.25%. That means the available fault current on the primary is quite high, roughly I_sc ≈ I_primary,FL / Z% ≈ 10.9 A / 0.0325 ≈ 336 A. So the protective device must be able to interrupt faults well before this level, yet not trip during normal energization inrush.

For an unsupervised location, a practical sizing approach is to select a primary fuse rating around a few times the full-load current to tolerate startup inrush while still providing protection. A factor around 3× I_fl gives about 32–33 A, so the next standard size is 35 A. Among the choices, 35 A provides the needed balance: it carries the normal 11 A load, tolerates typical startup surge, and will clear faults, whereas a smaller option could nuisance-trip, and larger options would reduce protection.