You're right to be questioning the different scenarios. Here's the breakdown of the correct approach and why

Understanding the Impedance:
The key is the per-unit impedance (%Z) given is referred to the transformer's kVA rating and voltage rating. This means the 2.2% impedance is a percentage of the base impedance calculated from the transformer's nameplate ratings (37.5 kVA and 240V). Impedance is a property of the transformer as a whole, not specific to a particular voltage tap. It represents the internal opposition to current flow within the transformer windings. While the actual impedance in ohms is different on the primary and secondary sides due to the turns ratio, the per-unit impedance remains the same when referred to the appropriate base kVA and voltage.

Correct Approach


Choose a Base Voltage

Since you're interested in the fault current on the secondary side, choose either 240V or 120V as your base voltage. It's often easiest to work with the highest voltage available on the secondary, which is 240V in this case.2.

Calculate the Base Impedance (if needed)

This is rarely needed because the %Z is already provided. However, for understanding: Using 240V as the base voltage: Base Impedance (Z_base) = (V_base)² / VA_base = (240V)² / 37500VA = 1.536 ohms The 2.2% impedance means the transformer's impedance (referred to the 240V secondary) is 2.2% of this base impedance.
3.

Calculate the Actual Impedance

If using 240V as the base: Z_actual = %Z Z_base = 0.022 1.536 ohms = 0.033792 ohms4.

Calculate the Fault Current

Fault Current (I_fault) = V_base / Z_actual = 240V / 0.033792 ohms = 7102.27 A

Why the Other Scenarios are Incorrect



Scenario 1B and Scenario 2B (Using 120V)

These are incorrect because you are fundamentally using the wrong transformer impedance %Z in the calculation. The 2.2% given in the transformer nameplate is based on the highest available voltage on the secondary of the transformer, 240V in your case.

Scenario 2A (Using 4.4% Z)

You're attempting to adjust the impedance based on the fact you're tapping off half the voltage. While the voltage is halved, the relationship between voltage, current, and impedance is more complex in a transformer. It's much easier and more accurate to work with the base voltage and given %Z. In addition, you still need to consider the connection that the voltage is supplied in, which is not directly apparent with the information that you have given.

General Issue

The biggest mistake is mixing and matching %Z and voltage levels without properly considering the base values they are referenced to.

Therefore, your Scenario 1A is the correct approach.Important Considerations Beyond the Calculation



X/R Ratio

This calculation assumes a purely resistive impedance. In reality, transformers have both resistance (R) and reactance (X). The X/R ratio (reactance/resistance) influences the asymmetry of the fault current waveform and the peak fault current. For a more accurate assessment, particularly for breaker interrupting ratings, you need to consider the X/R ratio (or estimate it – a typical value is around 10 for a small distribution transformer). This is very important to prevent equipment failure.

Source Impedance

This calculation assumes an infinite source (zero impedance) on the primary side. The utility's impedance will limit the fault current. You need to obtain the utility's available fault current at the transformer's primary voltage to determine the source impedance.

Upstream Protection

The utility's overcurrent protection (fuses or breakers) will likely interrupt the fault before the full calculated fault current is reached.

Downstream Impedance

This calculation assumes a bolted fault right at the transformer secondary terminals. The impedance of conductors (wire, busbar) between the transformer and the point of fault will reduce the actual fault current.

Available Fault Current vs. Interrupting Rating

The calculated fault current must be compared to the interrupting rating of any downstream protective devices (circuit breakers, fuses) to ensure they can safely interrupt the fault.

Cable Short Circuit Current Withstand

Cable need to be properly specified so that if a fault occurs, that the cable is able to withstand the short circuit current.

In Summary:

Your initial Scenario 1A provides a good starting point for estimating the maximum fault current. However, for a complete and accurate assessment for electrical safety and equipment selection, you must* consider the factors listed above (X/R ratio, source impedance, downstream impedance, etc.). Consult with a qualified electrical engineer for a proper short circuit study. Flag for review