Lower R, not raise V?

Test the claim on a real two-node network. Current rises when resistance falls only under a fixed-voltage boundary. The source still supplies every watt.

Source voltage V

R1 reduction %

R1 source series Ω

R2 node A load Ω

R3 A to B Ω

R4 node B load Ω

Two-node matrix ready.

Nodal solveG · [VA, VB]ᵀ = [VS/R1, 0]ᵀ

Power ledgerVS · IS = Σ I²R for passive resistors

NODE A

NODE B

R1 connects the ideal voltage source to node A. R2 loads A to ground; R3 connects A to B; R4 loads B to ground.

Boundary matters. An ideal current source behaves differently: lowering resistance would reduce required voltage, not increase imposed current.

Matched downstream flow

Same current. Different source cost.

Both interventions reach 1.50 A and the same node voltages, but they do not ask the source for the same power.

Lowering resistance changes the circuit's demand under fixed voltage. It removes dissipation from R1; it does not create energy.

Three conductance-matrix solves

Awaiting solve

Observed consequence

Lower R and higher V can produce the same downstream node state while changing source-side dissipation.

Conservation proof

Source power equals summed resistor dissipation in each solved state.