Mistake Master
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CED objectives

Kirchhoff's Junction Rule

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletJunction Lab · three branches off one junction, a switch in each, and the even split that never happens printed underneath

The junction rule states that the current arriving at a node equals the current leaving it, which fixes the total and nothing else. Each branch's share follows from the potential difference the parallel branches have in common and that branch's own resistance, $I_k = \Delta V/R_k$, so the current divides in inverse proportion to the resistances and the larger share goes to the smaller one. Removing a branch raises the equivalent resistance, lowers the total current and redistributes every potential difference in the circuit, so a mixed circuit has to be solved again from the top.

Two errors dominate. Halving the current at every branch point regardless of what the branches contain, which conserves charge and still gets both branch currents wrong except when the branches happen to be equal. And assuming the rest of a circuit keeps its old currents and brightnesses when one bulb burns out, when in a mixed circuit a single removal routinely dims some bulbs while brightening others.

conservation fixes the total; the branches fix the shares 6 A in 2 Ω 4 Ω 4 A 2 A smaller R, bigger share 3 A and 3 A conserves charge perfectly and is still wrong the branches share ΔV, so I = ΔV/R divides AGAINST the resistances. An even split needs equal branches
Both candidate splits add to six. Only one of them also satisfies the shared potential difference across the two branches.
one removal, opposite effects on two bulbs bulb 1 bulb 2 bulb 3 before bulb 1 DIMS BRIGHTENS burned out after R(eq) rises → total current falls → bulb 1's drop falls → MORE ΔV left for the parallel section
Nothing was touched in bulb 2's branch, and its brightness still changed. That is why the circuit has to be solved twice.

The work

3 ways in · any order
Lesson
Kirchhoff's Junction Rule

Separates the total the junction rule fixes from the shares the branches decide, divides current in inverse proportion to resistance, and solves a mixed circuit twice when a branch is removed.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: splitting a current evenly at every junction, and assuming the untouched parts of a circuit keep their old brightness when one branch is removed. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions