Why is there current flow in the sub-circuits?
Step 1) Use Th&eacute;venin's Theorem to mathematically derive the Th&eacute;venin equivalent circuit for subcircuit V1, V2, Ri_1, and Ri_2. (Hint: The Th&eacute;venin equivalent should be 6&nbsp;VDC in series with 250&nbsp;&mu;&Omega;).

Step 2) Use Th&eacute;venin's Theorem to mathematically derive the Th&eacute;venin equivalent circuit for subcircuit V3, V4, Ri_3, and Ri_4. (Hint: The Th&eacute;venin equivalent should be 6&nbsp;VDC in series with 250&nbsp;&mu;&Omega;).

Step 3) Redraw the circuit, replacing the two subcircuits with their Th&eacute;venin equivalent circuits. From the Load resistor's perspective, this simplified circuit is equivalent to the original circuit.

Step 4) Use Kirchhoff's Voltage Law (KVL) to calculate the current in the simplified and equivalent series circuit. (Hint: Your calculated current should be zero amps.)

So the Load resistor has zero amps flowing through it, and each subcircuit (in the original circuit) is a 6&nbsp;kA current loop.