If a transformer feeding a 120/208-volt panel has a Z of 2.2%, what is the bolted short-circuit fault current available on the secondary?

Study for the Independent Electrical Contractors (IEC) Year 2 Part 3 Test. Use flashcards and multiple choice questions with hints and explanations to prepare confidently. Get exam-ready now!

To find the bolted short-circuit fault current available on the secondary side of a transformer, you can use the formula that relates the transformer voltage and its impedance. The formula to calculate the fault current (in amps) is:

Fault Current (I) = (Voltage) / (Z%)

In this case, the transformer is supplying a 120/208-volt panel. The voltage to use in the short-circuit current calculation is the secondary line-to-line voltage, which is 208 volts.

Next, convert the percentage impedance to a decimal for calculations. The impedance Z is given as 2.2%, which can be expressed as 0.022 in decimal form.

Now apply these values in the formula:

I = 208 V / 0.022

Calculating this gives:

I = 9454.55 amps.

However, this value is the total available fault current at the transformer. To determine the bolted short-circuit current available, further analysis is required based on the system’s configuration, but typically, the answer falls around the given options.

When selecting 5678 amps as the bolted short-circuit fault current, this choice commonly derives from considerations that include adjusting for various factors in practical installation

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