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.

Same downstream nodes, different source dissipation

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Read the explanation

The default circuit has R one of four ohms feeding two equal twelve ohm downstream branches. One branch is R two, twelve ohms. The other is R three plus R four: four plus eight, also twelve ohms. Their parallel equivalent is six ohms. Together with R one, the total is ten ohms. Twelve volts divided by ten gives a source current of one point two amperes. A fifty percent reduction lowers R one from four to two ohms. The downstream equivalent remains six, so the total is eight ohms. At twelve volts the source current becomes one point five amperes. The downstream node voltages become nine and six volts. The original nodal solver obtains these same values by solving its two by two conductance matrix. Keep R one at four ohms and raise the source to fifteen volts. Dividing by the unchanged ten ohm network gives the same one point five amperes, with the same downstream node voltages. The lower resistance case takes eighteen watts from its twelve volt source. The higher voltage case takes twenty two point five watts. The difference is four point five watts of additional dissipation in R one. The power ledger conserves energy; lowering resistance does not create energy.