To limit fault current to 5A in a 13,800-480Y/277V transformer, what size neutral resistor is needed?

To determine the appropriate size for a neutral resistor that limits fault current to 5A in a 13,800-480Y/277V transformer, we need to refer to the formula for calculating the resistor value based on the desired current and system voltage.

The formula for the neutral resistor is based on Ohm's Law, which states that voltage is equal to current multiplied by resistance (V = I * R). Rearranging this formula to find resistance, we get R = V / I.

In this case, the voltage involved is the phase-to-neutral voltage of the transformer, which can be calculated from the line-to-line voltage. For a 480Y system, the line-to-neutral voltage is:

[ V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{480V}{\sqrt{3}} \approx 277.1V ]

Using this voltage and the desired fault current of 5A, the resistance needed can then be calculated as follows:

[ R = \frac{V_{LN}}{I} = \frac{277.1V}{5A} \approx 55.42 \text{ ohms} ]

This calculation shows that a resistor