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Why must all the current arriving at a junction equal all the current leaving it?

Topic 11.7 Kirchhoff's Junction Rule: apply conservation of charge to the currents at a junction in a circuit.

A focused answer to AP Physics 2 Topic 11.7, covering Kirchhoff's junction rule as conservation of charge, how current splits and recombines at junctions, writing junction equations, and combining the junction and loop rules to solve multi-loop circuits, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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Quick answer

Kirchhoff's junction rule states that the total current into a junction equals the total current out: $\sum I_{in} = \sum I_{out}$ . It expresses conservation of charge, because charge cannot accumulate at a point in a steady circuit, so it must flow out as fast as it flows in. At a junction where a wire splits, the incoming current divides among the branches and the branch currents add back to the original when they recombine. Writing a junction equation gives one relation between the branch currents; combined with the loop rule (conservation of energy), the junction rule lets you solve multi-loop circuits by setting up and solving simultaneous equations. The junction rule is also why parallel branch currents add up to the total: it is the charge-conservation half of circuit analysis, paired with the energy-conservation loop rule.

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What this topic is asking
What the junction rule says
Current splitting and recombining
Combining with the loop rule
Try this

What this topic is asking

The College Board (Topic 11.7) wants you to apply Kirchhoff's junction rule: the total current entering a junction equals the total current leaving it, a statement of conservation of charge. You must write junction equations and combine them with the loop rule.

What the junction rule says

A junction is any point where three or more wires meet. The rule is simply charge conservation: charge is neither created nor destroyed, and in a steady (unchanging) circuit it cannot pile up anywhere, so the rate of charge arriving must equal the rate leaving. Picture water pipes meeting at a junction: the water flowing in must equal the water flowing out, because none is stored.

Current splitting and recombining

The junction rule is what makes parallel circuits work: the source current splits at the entrance node, each branch carries a share, and the shares recombine at the exit node. This is exactly why the branch currents in a parallel combination add up to the total current drawn from the source. The branch with the smallest resistance takes the largest share, since each branch has the same voltage across it.

Combining with the loop rule

The junction and loop rules together solve any circuit. For a multi-loop network: assign an unknown current to each branch, write junction equations ( $\sum I_{in} = \sum I_{out}$ ) at the nodes, write loop equations ( $\sum \Delta V = 0$ ) for the independent loops, then solve the simultaneous equations for the currents. The two rules are complementary statements of the two great conservation laws: the junction rule conserves charge, the loop rule conserves energy. The strategic role of this topic is to complete the circuit-analysis toolkit. Series and parallel reduction (Topic 11.5) handles simple networks, but circuits with multiple sources or bridge configurations need the full Kirchhoff treatment, and the junction rule supplies the charge-conservation equations that, with the loop rule, pin down every current. This is the most powerful and general method in the unit.

Currents at a junction in a parallel circuit

A $12$ V source drives a junction that splits into two parallel resistors, $6.0$ ohms and $4.0$ ohms, which rejoin. Find each branch current and verify the junction rule.

step 1 Branch currents from Ohm's law

Each branch has the full $12$ V across it (parallel). $I_1 = \dfrac{12}{6.0} = 2.0$ A; $I_2 = \dfrac{12}{4.0} = 3.0$ A.

step 2 Total current into the junction

The source current is $I = I_1 + I_2 = 2.0 + 3.0 = 5.0$ A.

step 3 Apply the junction rule

At the splitting node, $I_{in} = 5.0$ A and $I_{out} = I_1 + I_2 = 5.0$ A, so $\sum I_{in} = \sum I_{out}$ , as required.

step 4 Interpret

The lower-resistance $4.0$ -ohm branch carries the larger current ( $3.0$ A), and the branch currents add back to the $5.0$ A drawn from the source, confirming charge conservation.

Try this

Q1. State the conservation law that Kirchhoff's junction rule expresses. [1 point]

Cue. Conservation of charge.

Q2. A $7.0$ A current enters a junction and splits into branches of $4.0$ A and an unknown current. Find the unknown. [1 point]

Cue. $7.0 = 4.0 + I_2$ , so $I_2 = 3.0$ A.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)5 marksSection II (short FRQ). At a junction, a current of

5.0

A flows in, and two wires carry current out: one carries

2.0

A. (a) State Kirchhoff's junction rule and the conservation law it expresses. (b) Calculate the current in the second outgoing wire. (c) Explain why current cannot accumulate at a junction in a steady circuit.

Show worked answer →

A 5-point FRQ on the junction rule.

(a) Statement (2 points): the total current flowing into a junction equals the total current flowing out. It expresses conservation of charge.
(b) Current (2 points): $I_{in} = I_{out}$ , so $5.0 = 2.0 + I_2$ , giving $I_2 = 3.0$ A.
(c) No accumulation (1 point): charge is conserved and cannot pile up at a point in a steady (constant) circuit, so whatever flows in must flow out at the same rate.

Markers reward the in-equals-out statement with charge conservation, the arithmetic for the second current, and the no-accumulation explanation.

AP 2023 (style)1 marksSection I (multiple choice). A current of

4.0

A splits at a junction into two branches; one branch carries

1.5

A. What does the other branch carry? (A)

1.5

A (B)

2.5

A (C)

4.0

A (D)

5.5

A. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the junction rule. The answer is (B).

By conservation of charge, the currents leaving a junction sum to the current entering: $4.0 = 1.5 + I_2$ , so $I_2 = 2.5$ A. The trap is (D): the branch currents add to the incoming current, they are not added to it.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)