Interpret negative loop current and source absorption

Establish the model and evidence

Choose a loop traversal and an assumed positive current direction before writing equations. Traversing an ideal source from negative to positive terminal gives a potential rise; reversing the traversal gives a drop. Across a resistor, moving with the assumed current gives −IR. Consistent signs let the calculation reveal the actual direction rather than requiring a guess to be correct.

The supplied series loop contains a 6 V rise, a 10 Ω resistor, an opposing source of selectable emf and a second resistor. Its equation is 6−I(10)−emf₂−IR₂=0. If the opposing source exceeds 6 V, I comes out negative. That means actual current flows opposite the assumed direction; it does not mean the resistor has negative resistance or produces energy.

The node-potential graph follows the chosen traversal, not necessarily the current. When current is negative, traversing a resistor in that chosen direction raises potential. Returning to the starting node must recover the initial potential. The loop rule is energy conservation per charge, while the junction rule separately requires no net charge accumulation. They are compatible constraints, not interchangeable formulas.

Source power also needs a sign convention. Here positive delivered power means energy leaves a source into the loop; a negative value means the ideal source absorbs energy. Resistor dissipation is I²R and remains nonnegative. This abstract absorption is not permission to charge an arbitrary battery: real charging compatibility and safety are outside the model. The lesson never connects unequal ideal voltage sources directly in parallel.

Worked evidence

For opposing emf 4 V and R₂=20 Ω, I=(6−4)/(10+20)=0.06667 A. The 6 V source delivers 0.40 W, the opposing source absorbs 0.26667 W and the resistors dissipate 0.13333 W. The signed source powers sum to the resistor loss.

Predict and investigate

If opposing emf rises to 8 V, which signs change and which dissipation remains nonnegative?

Capture opposing emfs 4, 6 and 8 V at R₂=20 Ω. Reconstruct every node step, signed current and source power. Explain the zero-current middle case without assuming every element has zero potential difference.

Transfer to changed conditions

A series loop has opposing ideal sources 5 V and 11 V and resistors 10 Ω and 20 Ω. With positive traversal through the 5 V rise, calculate current and delivered source powers.

Compare the worked transfer

I=(5−11)/30=−0.20 A. The 5 V source delivers −1.0 W (absorbs), the 11 V source delivers +2.2 W, and resistor dissipation is 1.2 W. The energy ledger closes.

Explain the boundary

Why does reversing the chosen traversal not change the physical current or total resistor heating?

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. Original cases and computed/synthetic evidence, not actual laboratory measurements. These activities are virtual only: do not reproduce wiring, short-circuit, battery charging, heating or charged-capacitor experiments. Any real practical work requires qualified supervision and an independently approved safety procedure.