190103-2035 EST
julian padilla:
Your question as posted is not real clear to me.
First, assume all the loads are resistive.
Then the question is what wires are to be fused?
If you assume each load is to be fused individually, then clearly the current per load is 49,000 / 480 = 102.1 A.
If you want to fuse a leg from the delta transformer, which is what I think your question is, then you need vector addition of the two currents connected. The easy intuitive way to attack the problem is to convert the balanced delta load to a balanced wye load. No math needed. Just redrawing the load. Since the load is balanced and the source is balanced this means the center point of the star resistor network is at the same potential as the neutral point of a wye source used to replace the delta source. Thus, we now have a wye load connected to a wye source. All we need to know is the wye line to neutral voltage.
The wye voltage you probably know off the top of your hat, 480 / 1.732 = 277.1 V. Thus, line current from the transformer for a resistive load is 49,000 / 277.1 = 176.8 A. This is because the line to neutral voltage is in phase with the load current. If the power factor is not 1, then you need to do its correction.
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