6 resistors and a 3-phase power source?

Question

Two leads coming off of each 136 ohm resistor. A three phase power source with three leads. I need 10KW.

How do I wire up a delta? What voltage?

How do I wire up a WYE? What voltage?

Answer 1

Wire per first answer.

For 10kW, 3333W in each of 3 resistors, you need V^2/R = 3333; V = (136X3333)^0.5 = 673V across each resistor. For delta load, 3-phase L-L voltage is 673V. For wye, load L-N V = 673V; source L-L voltage = 673X3^0.5 = 1166V.

Oops
I see you said 6 resistors not 3. Wire pairs of resistors in series or parallel then treat each pair as one 272 ohm or 68 ohm resistor. You have 4 choices of voltage: parallel D or Y, series D or Y; 476, 824, 952, 1648V as calculated above.

Answer 2

In a DELTA you connect one end of each resistor to an end of the other so they end up forming a triangle. You connect one wire to each corner of the triangle.

In a WYE you connect all three of one end together. Then you connect one wire to each of the three other ends of the resistors.

Either way power is then available across each resistor.

You did not provide enough info about the incoming power to answer voltage questions. Are these three resistors actually heating elements whose power consumption at the unnamed voltage is to total 10KW?