AP Physics 2 · Topic 11.7

Topic 11.7: Kirchhoff's Junction Rule

Unit 11: Electric Circuits15-18% of the multiple-choice section

Kirchhoff's junction rule says the total current entering a junction equals the total current leaving it. That follows from conservation of electric charge: a junction cannot store charge, so whatever arrives each second must leave each second. Current splits at a junction, but none is used up.

AP Physics: Unit 11 (topics 11.7 Kirchhoff's Junction Rule). AP Physics 2 Unit 11, Topic 11.7. One learning objective, 11.7.A: describe a circuit or elements of a circuit by applying Kirchhoff's junction rule. Two essential knowledge statements: 11.7.A.1, the junction rule is a consequence of the conservation of electric charge, and 11.7.A.2, the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time, with the relevant equation given as the sum of the currents in equals the sum of the currents out. The CED prints no boundary statement for this topic; the constraints quoted on this page come from Topic 11.2 (schematics use conventional current unless otherwise specified) and Topic 11.5 (batteries, wires, and meters are ideal unless otherwise stated, nonideal meters are treated qualitatively only, and circuits with batteries of different potential differences connected in parallel will not be assessed). Suggested skills 1.A, 2.B, 2.C, and 3.C. Unit 11 is weighted at 15 to 18 percent of the multiple-choice section and is allotted about 12 to 20 class periods. Neither Kirchhoff rule appears among the 129 entries on the AP Physics 2 equation sheet.

What Topic 11.7 requires

Topic 11.7 carries one learning objective and two essential knowledge statements, with no sub-statements under either.

11.7.A Describe a circuit or elements of a circuit by applying Kirchhoff's junction rule.
11.7.A.1 Kirchhoff's junction rule is a consequence of the conservation of electric charge.
11.7.A.2 Kirchhoff's junction rule states that the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time.

The CED prints one relevant equation under 11.7.A.2:

\sum I_{\text{in}} = \sum I_{\text{out}}

The suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Unit 11 is weighted at 15 to 18 percent of the multiple-choice section and the CED allots it about 12 to 20 class periods. The CED prints no boundary statement on this topic. The limits that do apply to junction problems are set in neighboring topics, and they are quoted further down this page.

Two essential knowledge statements is a small target, so the work on this page is not memorizing the rule. It is knowing exactly which physical principle the rule is, where the junctions in a schematic actually are, and how the rule combines with Kirchhoff's loop rule to pin down every current in a network.

Why the rule is true: charge cannot pile up at a point (11.7.A.1)

Statement 11.7.A.1 gives the whole justification in one line: the junction rule is a consequence of the conservation of electric charge. That is the sentence to write on a free-response question, and skill 3.C is exactly the instruction to write it.

Start from the definition of current in Topic 11.1. Statement 11.1.A.1 says current is the rate at which charge passes through a cross-sectional area of a wire, and prints

I = \frac{\Delta q}{\Delta t}

Now look at a junction. A junction is a point where wires meet. It has no plates, no volume worth speaking of, nowhere to put charge. Suppose more charge arrived there per second than left. Charge would build up at that point, and by conservation of electric charge it would have to stay there, because charge is not created or destroyed. The accumulated charge would set up an electric field opposing further arrivals, and within a very short time the flow would readjust until the rates matched. A steady current is precisely the state where that readjustment has already finished.

So the rule is not a separate law of circuits. It is conservation of charge plus the observation that a node has no storage. Multiply 11.7.A.2 through by an interval $\Delta t$ and the statement about currents becomes a statement about charge: the charge that entered in that interval equals the charge that left.

One place in Unit 11 does store charge, and it is worth being clear that it is not a counterexample. A capacitor plate accumulates charge, which is why the current in a charging capacitor branch is nonzero. But the plate is a plate, not a junction, and the current in the branch is what carries charge to it. Every node in that circuit still balances. Topic 11.8 covers the behavior of that branch over time, and Topic 10.6 covers what capacitance is.

Writing the rule down, and choosing signs

Two forms of the same statement get used, and you should be fluent in both.

The CED's form separates the two sides:

\sum I_{\text{in}} = \sum I_{\text{out}}

The signed form puts everything on one side. Take currents drawn as entering the node to be positive and currents drawn as leaving it to be negative, and the rule becomes $\sum I = 0$ at every node. This page uses the entering-positive convention wherever a signed sum appears, and switching it halfway through a problem is a reliable way to lose the answer.

Three points of care.

Direction means conventional current. Statement 11.1.A.2 says that although current is not a vector quantity, it does have a direction, and 11.1.A.2.i says the direction of conventional current is chosen to be the direction in which positive charge would move. Statement 11.1.A.2.ii adds that in common circuits the current is actually due to the movement of electrons, which are negative charge carriers. The Topic 11.2 boundary statement settles what you will see on the exam: unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.
You are allowed to guess the direction wrongly. Draw an arrow on every branch before you know the answer, write the node equations from those arrows, and solve. A current that comes out negative is telling you the real direction is opposite to your arrow, and the magnitude is still right. Erasing and redrawing mid-solution is what breaks a sign convention.
The rule applies to a node, not to a wire. Applying it to a point in the middle of an unbroken wire gives $I = I$ , which is true and useless. It earns you information only where a branch splits or rejoins.

Where the series and parallel rules come from

The junction rule is the reason the two circuit facts you use constantly are true, and Topic 11.5 states both.

Statement 11.5.A.1.i defines a series connection as one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available, and then states that the current in each element in series must be the same. Read that through the junction rule: a series path contains no junction between its elements, so there is nowhere for the current to split. One way in, one way out, same number.

Statement 11.5.A.1.ii defines a parallel connection as one in which charges may flow through one of two or more paths, and adds that across each path the potential difference is the same. Here there is a junction, at the point where the paths separate, and another where they rejoin. The junction rule at the first node says the branch currents add up to the current arriving, which is why 11.5.A.2.iii can say that when resistors are connected in parallel the number of paths available to charges increases and the equivalent resistance of the group decreases: more paths carry more total current for the same potential difference.

This also answers one of the CED's essential questions for Unit 11, which asks why several bulbs on a string of lights go out when one bulb is unplugged. A series string has no junction between the bulbs, so there is only one path, and removing one bulb makes it an open circuit: statement 11.2.A.2.ii says an open circuit is one in which charges would not be able to flow. Wire the bulbs in parallel instead and each has its own path back to the source, so cutting one leaves the others untouched.

The step-by-step routine for reducing a network to a single equivalent resistance lives in the series vs parallel circuits guide, and the rearrangements of $I = \Delta V / R$ live in the Ohm's law guide. This page does not repeat them. What Topic 11.7 adds is the justification underneath them and the ability to work a circuit that you have not reduced.

Drawing a schematic so the junctions are visible (skill 1.A)

Suggested skill 1.A asks you to create diagrams, tables, charts, or schematics to represent physical situations, and for junction problems getting the representation right is most of the work. Exam schematics are drawn to look tidy, not to make the nodes obvious, and two dots that look far apart on the page are the same node if only wire runs between them.

A routine that works:

Find the nodes. Any set of points joined by wire alone, with no circuit element in between, is a single node. Statement 11.5.B.1 says ideal wires have negligible resistance, so every point in that set sits at the same electric potential. Label the whole set with one letter.
Find the branches. A branch is a path from one node to another that passes through elements and contains no node in its interior. Every element in a branch carries the same current.
Give each branch one current symbol and one arrow. Not one per element. If two resistors sit in series inside a branch, they share a current, and giving them separate symbols invents an unknown you will then have to eliminate.
Write one junction equation per node. In a circuit with only two nodes, both equations say the same thing, so one of them is free information you can use as a check rather than as a new equation.

Meters are a good test of step 2. Statement 11.5.C.1.i says ammeters must be connected in series with the element in which current is being measured, so an ammeter reads the current in its own branch and the junction rule predicts what a second ammeter elsewhere must read. Statement 11.5.C.2.ii says an ideal voltmeter has infinite resistance so no charge flows through it, which means a voltmeter contributes nothing to any node equation.

Pairing the junction rule with the loop rule (11.6)

The two Kirchhoff rules are the two conservation laws of the unit, applied to two different features of a circuit.

	Junction rule (11.7)	Loop rule (11.6)
Conserved quantity	Electric charge (11.7.A.1)	Energy (11.6.A.2)
Where you apply it	At a node	Around a closed loop
Statement	Charge in per unit time equals charge out per unit time (11.7.A.2)	The sum of potential differences across all circuit elements in a single closed loop must equal zero (11.6.A.3)
Equation	$\sum I_{\text{in}} = \sum I_{\text{out}}$	$\sum \Delta V = 0$

Statement 11.2.A.3 says a single circuit element may be part of multiple electrical loops, and that is what makes a network more than a chain: one element's current is shared between loops, so the loops are not independent problems. The junction rule is the bookkeeping that ties them together. Write one junction equation at each node and one loop equation around each independent loop, and you have as many equations as unknown branch currents. Worked example 3 below does exactly that without reducing anything to an equivalent resistance.

The Topic 11.5 boundary statement caps how awkward a network can get: AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements, unless otherwise stated all batteries, wires, and meters are assumed to be ideal, and circuits with batteries of different potential differences connected in parallel will not be assessed. That last clause rules out one whole family of two-source networks.

One thing to notice about the AP Physics 2 equation sheet: neither Kirchhoff rule is on it. The Electricity group holds 20 entries and the whole sheet holds 129 across seven groups, and $\sum I_{\text{in}} = \sum I_{\text{out}}$ and $\sum \Delta V = 0$ are not among them, even though the CED prints both as relevant equations in the framework. The sheet does print $I = \Delta q / \Delta t$ , $I = \Delta V / R$ , $R_{\text{eq,s}} = \sum_i R_i$ , and $1/R_{\text{eq,p}} = \sum_i 1/R_i$ , so the tools you combine with the two rules are all there. The rules themselves you carry in.

Traps that cost points

The CED's Unit 11 overview names the misconception this rule corrects. Under Preparing for the AP Exam it says that students may believe that batteries store charge, or that current is used up in a circuit. Both are junction-rule failures dressed up as intuition.

Current is not consumed by a resistor. The same current leaves a resistor as enters it. What the resistor takes is energy, not charge, which is why Topic 11.4 has its own equation for the rate of that transfer and 11.6 covers the potential drop that goes with it. A bulb that glows is not eating charge.
A battery does not supply charge from a store. It supplies energy, moving charge that is already in the circuit through a potential difference. Statement 11.1.A.1.i says electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force.
Current does not split evenly by default. Two parallel branches across the same pair of nodes share a potential difference (11.5.A.1.ii), so by $I = \Delta V / R$ the currents are in inverse proportion to the branch resistances. Equal branches split evenly, unequal branches do not.
Every branch counts, including one you think is off. At steady state a capacitor branch carries zero current, because 11.8.B.2.iv says a fully charged capacitor reaches a maximum potential difference at which there is zero current in the circuit branch in which it is located. That branch still appears in the node equation; its current is simply zero. Leaving it out by accident and leaving it in as zero give the same answer, but only one of them is reasoning.
Removing a branch changes the source current, not necessarily the others. With an ideal battery the potential difference across a directly connected parallel group is fixed, so cutting one branch leaves the remaining branch currents where they were and reduces the total. With a real source that is no longer true, which is why 11.5.B.3 introduces $\Delta V_{\text{terminal}} = \mathcal{E} - I r$ for a battery with internal resistance $r$ .
Sign slips. A node with four wires and one unknown is arithmetic, and it is still an easy place to drop a minus. Worked example 1 runs it both ways on purpose.

What the exam asks of you here

The four suggested skills point at four different kinds of task.

1.A, representation. Redraw or complete a schematic, or identify which points in a given schematic are the same node. This is often silent work inside a longer problem rather than a question of its own.

2.B, calculation. Find a missing branch current with units, given the others. Straightforward once the diagram is right, which is why the diagram comes first.

2.C, comparison. Compare currents at two locations in one circuit, or compare the same location before and after a change. The CED's sample instructional activity for 11.7 is built this way: it sets up a mixed circuit with a battery and three or four light bulbs, asks students to predict what happens to the brightness of each remaining bulb if one is removed and to explain using Kirchhoff's principles, and then asks in reverse which bulbs must be removed to make another bulb brighter or dimmer. Brightness tracks power, since 11.4.A.2 says the brightness of a bulb increases with power, so a comparison question of this kind runs from the junction rule through the currents to the power in each bulb.

3.C, justification. Support a claim about a circuit using a physical principle. Here the principle has a name and the CED gives you the sentence: the junction rule is a consequence of the conservation of electric charge. Say which node you are applying it at, state what enters and what leaves, and finish the claim.

The unit's Building the Science Practices note lists 1.A, 1.B, 2.B, 2.D, and 3.A as the skills Unit 11 gives repeated practice in, and says those skills will be tested on the free-response section of the AP Physics 2 Exam. The same overview asks for precise vocabulary, singling out the differences in meaning between current, potential difference, resistance, resistivity, and capacitance. On this topic that matters more than usual, because the sentence the current across the resistor is already wrong: current is in an element, potential difference is across it.

From here, Topic 11.8 is where the junction rule stops describing a fixed set of currents and starts describing a circuit whose currents change with time.

One unknown at a four-wire junction, done twice

Four wires meet at a junction P. Wire 1 carries 1.80 A into P, wire 2 carries 0.75 A into P, and wire 3 carries 2.10 A out of P. (a) Find the current in wire 4 and its direction. (b) Redo the calculation having guessed the wrong direction for wire 4, and show that the answer survives.

Set the convention first: currents drawn as entering P count as positive, currents drawn as leaving P count as negative. Keep it for the whole problem.
(a) Total current in from the two known incoming wires: $1.80\ \text{A} + 0.75\ \text{A} = 2.55\ \text{A}$ . Total current out from the one known outgoing wire is $2.10\ \text{A}$ .
Apply 11.7.A.2 with wire 4 drawn as leaving: $\sum I_{\text{in}} = \sum I_{\text{out}}$ gives $2.55\ \text{A} = 2.10\ \text{A} + I_4$ , so $I_4 = 0.45\ \text{A}$ out of P.
(b) Now suppose you drew wire 4 as entering P instead. The equation becomes $1.80 + 0.75 + I_4 = 2.10$ , so $I_4 = 2.10 - 2.55 = -0.45\ \text{A}$ .
Read the minus sign. The magnitude is 0.45 A either way; the negative says the current runs opposite to the arrow you drew, which means out of P. Both routes agree, and neither required you to know the answer in advance.
Check with the signed form. With the entering-positive convention and the true directions, $\sum I = (+1.80) + (+0.75) + (-2.10) + (-0.45) = 0$ , as it must be at every node.

Wire 4 carries 0.45 A out of the junction. Guessing the direction wrongly returns $-0.45\ \text{A}$ , which is the same physical answer: the sign reports the direction relative to your own arrow, so there is never a reason to guess before you solve. Choose a convention, draw arrows on every branch, and let the algebra tell you which way the charge actually goes.

Three parallel branches, then one of them opens

An ideal 18.0 V battery is connected across three resistors in parallel: 9.0 ohm, 18.0 ohm, and 36.0 ohm. Assume ideal wires and no internal resistance. (a) Find the current in each branch and the current delivered by the battery, and check the junction rule at the node where the branches separate. (b) The 9.0 ohm branch is then cut open. Find the new branch currents and the new battery current.

(a) Statement 11.5.A.1.ii says the potential difference across each parallel path is the same, and with an ideal battery and ideal wires that is the full 18.0 V across every branch. Apply $I = \Delta V / R$ to each: $I_1 = 18.0/9.0 = 2.0\ \text{A}$ , $I_2 = 18.0/18.0 = 1.0\ \text{A}$ , $I_3 = 18.0/36.0 = 0.50\ \text{A}$ .
The junction rule at the node where the three branches separate has one current in, from the battery, and three out: $I_{\text{batt}} = 2.0 + 1.0 + 0.50 = 3.5\ \text{A}$ .
Cross-check against the equivalent resistance: $1/R_{\text{eq,p}} = 1/9.0 + 1/18.0 + 1/36.0 = 4/36 + 2/36 + 1/36 = 7/36$ , so $R_{\text{eq,p}} = 36/7 = 5.14\ \Omega$ and $I_{\text{batt}} = 18.0 / (36/7) = 3.5\ \text{A}$ . The two routes agree, which is the point: the junction rule and the parallel formula are the same statement.
Note the split. The currents 2.0, 1.0, and 0.50 A are in the ratio 4 : 2 : 1 while the resistances 9.0, 18.0, and 36.0 ohm are in the ratio 1 : 2 : 4. Branch currents go as the inverse of branch resistance, not evenly.
(b) Cutting the 9.0 ohm branch makes it an open circuit, so $I_1 = 0$ . The other two branches are still directly across the ideal battery, so they still have 18.0 V across them: $I_2 = 1.0\ \text{A}$ and $I_3 = 0.50\ \text{A}$ , unchanged.
The node equation now reads $I_{\text{batt}} = 0 + 1.0 + 0.50 = 1.5\ \text{A}$ . Confirm with the equivalent resistance: $1/R_{\text{eq,p}} = 1/18.0 + 1/36.0 = 3/36$ , so $R_{\text{eq,p}} = 12\ \Omega$ and $I_{\text{batt}} = 18.0/12 = 1.5\ \text{A}$ .

(a) 2.0 A, 1.0 A, and 0.50 A in the 9.0, 18.0, and 36.0 ohm branches, with 3.5 A from the battery. (b) After the cut, 0, 1.0 A, and 0.50 A, with 1.5 A from the battery. Removing a parallel branch left the surviving branch currents exactly where they were and cut the battery's current, which is 11.5.A.2.iii read backwards: fewer paths means a larger equivalent resistance and less total current. This is a skill 2.C comparison, and it depends on the battery being ideal: with internal resistance $r$ , statement 11.5.B.3 says the terminal potential difference is $\mathcal{E} - I r$ , so a change in $I$ would move the voltage across the branches too.

A two-loop network from the two rules alone

An ideal 21.0 V battery is connected in series with a 6.0 ohm resistor. That resistor then feeds a parallel pair: a 10.0 ohm resistor and a 15.0 ohm resistor, whose far ends rejoin and return to the battery. Find the current in all three resistors using only the junction rule and the loop rule, without computing an equivalent resistance.

Label the nodes and branches. Call the node where the parallel pair separates A and the node where it rejoins B. There are three branches: the battery with the 6.0 ohm resistor (current $I_1$ , drawn toward A), the 10.0 ohm resistor (current $I_2$ , drawn from A to B), and the 15.0 ohm resistor (current $I_3$ , drawn from A to B).
Junction rule at A, using 11.7.A.2 with one current in and two out: $I_1 = I_2 + I_3$ . Node B gives $I_2 + I_3 = I_1$ , which is the same equation, so this two-node circuit supplies exactly one independent junction equation.
Loop rule around the outer loop, battery then 6.0 ohm then 10.0 ohm, using 11.6.A.3 that the sum of potential differences around a closed loop is zero: $21.0 - 6.0 I_1 - 10.0 I_2 = 0$ .
Loop rule around the inner loop, down the 10.0 ohm branch and back up the 15.0 ohm branch: $-10.0 I_2 + 15.0 I_3 = 0$ , so $I_2 = 1.5 I_3$ . This is 11.5.A.1.ii falling out of the loop rule rather than being assumed: the two parallel branches must have equal potential differences.
Substitute into the junction equation: $I_1 = 1.5 I_3 + I_3 = 2.5 I_3$ . Three unknown currents have now collapsed to one.
Put both into the outer loop equation: $21.0 - 6.0(2.5 I_3) - 10.0(1.5 I_3) = 21.0 - 15.0 I_3 - 15.0 I_3 = 21.0 - 30.0 I_3 = 0$ , so $I_3 = 21.0/30.0 = 0.700\ \text{A}$ .
Back-substitute: $I_2 = 1.5 \times 0.700 = 1.05\ \text{A}$ and $I_1 = 2.5 \times 0.700 = 1.75\ \text{A}$ . Check the junction: $1.05 + 0.700 = 1.75$ . Check the outer loop: $21.0 - 6.0(1.75) - 10.0(1.05) = 21.0 - 10.5 - 10.5 = 0$ . Check the parallel potential differences: $10.0 \times 1.05 = 10.5\ \text{V}$ and $15.0 \times 0.700 = 10.5\ \text{V}$ .

$I_1 = 1.75\ \text{A}$ through the 6.0 ohm resistor, $I_2 = 1.05\ \text{A}$ through the 10.0 ohm resistor, and $I_3 = 0.700\ \text{A}$ through the 15.0 ohm resistor. Nothing was reduced to an equivalent resistance at any point: one junction equation and two loop equations were enough, because 11.2.A.3 warns that a single element can belong to more than one loop and the junction rule is what stitches the loops together. Reduction is the faster route for a circuit shaped like this one and the series vs parallel guide covers it, but the rules also work on networks that do not reduce to a chain of series and parallel groups.

Frequently asked questions

What is Kirchhoff's junction rule?

Kirchhoff's junction rule states that the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time. That is essential knowledge statement 11.7.A.2 in the AP Physics 2 course framework, and the CED prints it as the sum of the currents in equals the sum of the currents out. In practice you write one such equation at each node in a circuit, with each branch current given a symbol and a drawn direction.

Why is Kirchhoff's junction rule true?

Because of conservation of electric charge. Essential knowledge 11.7.A.1 says the junction rule is a consequence of the conservation of electric charge, and the reason it applies at a junction specifically is that a junction is a point in a wire with nowhere to store charge. If more charge arrived per second than left, charge would accumulate there and immediately set up an electric field that opposed further arrivals. A steady current is the state in which the rates already match.

What is the difference between Kirchhoff's junction rule and the loop rule?

They are two different conservation laws applied to two different features of a circuit. The junction rule is a consequence of conservation of electric charge (11.7.A.1) and is applied at a node, where the current in equals the current out. The loop rule is a consequence of conservation of energy (11.6.A.2) and is applied around a closed loop, where the sum of the potential differences across all circuit elements must equal zero (11.6.A.3). Solving a multi-loop network needs both: the junction rule relates the branch currents and the loop rule relates the potential differences.

Is the junction rule on the AP Physics 2 equation sheet?

No. The AP Physics 2 equation sheet carries 129 entries in seven groups, with 20 of them in the Electricity group, and neither Kirchhoff rule is among them. The CED does print the junction rule as a relevant equation under 11.7.A.2 and the loop rule under 11.6.A.3, so both are examinable; they are simply not supplied to you during the exam. What the sheet does give you for junction problems is the definition of current, Ohm's law in the form current equals potential difference divided by resistance, and the series and parallel equivalent-resistance formulas.

How do you know which direction the current goes at a junction?

You do not have to know before you solve. Draw an arrow on every branch, pick a sign convention such as entering the node counts as positive, write the junction and loop equations from your arrows, and solve. Any current that comes out negative simply runs opposite to the arrow you drew, with the magnitude unchanged. The one thing that must not change is the convention itself. Note also that the Topic 11.2 boundary statement says that unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current, the direction in which positive charge would move.

Why is the current the same everywhere in a series circuit?

Because a series path contains no junction, so the current has nowhere to split. Statement 11.5.A.1.i defines a series connection as one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available, and it states that the current in each element in series must be the same. The junction rule is the underlying reason: with one wire in and one wire out at every point along the path, charge in per second must equal charge out per second.

Does current get used up as it passes through a resistor or a bulb?

No. The same current leaves a resistor as enters it, because the junction rule holds at every point along the path and charge is conserved. What a resistor removes is energy, not charge: it converts electrical energy to thermal energy, which shows up as a drop in electric potential across it rather than a drop in current through it. The AP Physics 2 CED names this directly in its Unit 11 overview, listing the belief that current is used up in a circuit among the misconceptions students should be helped to challenge.