AP Physics C: E&M · Topic 11.7

Topic 11.7: Kirchhoff's Junction Rule

Unit 11: Electric Circuits15-25% of the multiple-choice section

Kirchhoff's junction rule says the charge entering a junction per unit time equals the charge leaving per unit time, so the currents in equal the currents out. It follows from conservation of electric charge: a junction cannot store charge. It is not printed on the AP Physics C E&M sheet.

AP Physics: Unit 11 (topics 11.7 Kirchhoff's Junction Rule). Topic 11.7 of the current AP Physics C: Electricity and Magnetism course and exam description, in Unit 11 Electric Circuits (15 to 25% of the multiple-choice section, about 12 to 24 class periods). One learning objective, 11.7.A, describe a circuit or elements of a circuit by applying Kirchhoff's junction rule, with just two essential knowledge statements and no sub-statements: 11.7.A.1, that the rule is a consequence of the conservation of electric charge, and 11.7.A.2, that the total amount of charge entering a junction per unit time must equal the total amount exiting per unit time, with the relevant equation that the sum of the inward currents equals the sum of the outward currents. The topic prints no boundary statement. The junction rule is not printed on the equation sheet, and neither is the loop rule. Suggested skills 1.B, 2.B, 3.A and 3.B; the AP Physics 2 version of this topic has the same two statements but lists 1.A, 2.B, 2.C and 3.C instead, so the C course attaches experimental design and quantitative data plotting to it. Sample multiple-choice Question 10 in the course description is aligned to skill 3.B, learning objective 11.7.A and essential knowledge 11.7.A.2, with credited answer D. Learning objective 11.7.A is also one of the five objectives aligned to sample free-response Question 4.

What Topic 11.7 requires

Topic 11.7 of AP Physics C: Electricity and Magnetism Unit 11 has one learning objective and two essential-knowledge statements, with no sub-statements and no boundary statement. Only Topic 11.4 is as short, and it prints a boundary statement and three equation forms where 11.7 prints one equation and no boundary at all.

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. Relevant equation $\sum I_{\text{in}} = \sum I_{\text{out}}$ .

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Read 11.7.A.2 carefully, because the CED states the rule in terms of charge per unit time rather than current. That is the same thing, since 11.1.A.1 defines current as $I = dq/dt$ , but the phrasing is doing work: it makes the connection to conservation of charge explicit, and it is what allows a capacitor branch to enter a junction equation as a derivative rather than as a labelled current.

Two statements and one equation is the whole of the required content. Everything else this topic asks of you is application, which is why three of its four suggested skills are about doing something with it rather than knowing it.

The same two statements as AP Physics 2, with a different skill set

The required content of Topic 11.7 is identical in AP Physics C and AP Physics 2. Both frameworks contain exactly 11.7.A.1 and 11.7.A.2, in the same words, with the same relevant equation and no boundary statement in either. There is no calculus in the statements themselves, and pretending otherwise would be inventing a difference.

The suggested skills are not the same, and this is one of the few topics in Unit 11 where they diverge:

	AP Physics C: E&M	AP Physics 2
Science Practice 1	1.B quantitative graphs, including plotting data	1.A diagrams and schematics
Science Practice 2	2.B calculation	2.B calculation
Science Practice 2		2.C comparison between scenarios
Science Practice 3	3.A create experimental procedures
Science Practice 3	3.B apply a law to make a claim	3.C justify with evidence

The C course attaches experimental design and quantitative data plotting to the junction rule, where the algebra-based course attaches schematic drawing and scenario comparison. Skill 3.A appears on only three of Unit 11's eight topics, 11.3, 11.7 and 11.8, and the unit's exam-preparation note explains why: the third free-response question is the Experimental Design and Analysis question, in which students will be required to justify their selection of the kind of data needed and then design a plan to collect these data, and will also be required to derive relevant equations, linearize and analyze data.

So the C version of this topic is likelier to arrive as "design a way to verify that the currents in these branches sum correctly, and plot the data so a linear fit tests it" than as "find the missing current".

The AP Physics 2 Topic 11.7 page covers the rule itself for the algebra-based exam and is written for those students. This page is for AP Physics C students, and past the shared statements it covers the two things the C course does differently: putting $dq/dt$ into a junction equation, and expressing the rule in terms of current density.

Why it is true: charge cannot pile up at a point

Statement 11.7.A.1 gives the whole justification in one line: Kirchhoff's junction rule is a consequence of the conservation of electric charge. Statement 11.7.A.2 then supplies the mechanism, which is that a junction is a point, and a point has no volume in which to store charge.

Spell it out and it becomes a one-sentence derivation you can write in a free-response answer. Charge is conserved, so the charge in any region changes only by charge crossing its boundary. A junction is a region of negligible size that holds no net charge, so the rate at which charge arrives must equal the rate at which it leaves, and rates of charge transfer are currents. Hence

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

The conservation law being invoked is a different one from the loop rule's, and the pairing is worth keeping straight because it is the cleanest way to remember which rule is which:

Rule	Conservation law	Statement	Sums to
Loop	energy (11.6.A.2)	around a closed loop	zero potential difference
Junction	electric charge (11.7.A.1)	at a point	equal current in and out

One qualification that AP Physics C makes visible and AP Physics 2 does not. A capacitor plate genuinely does store charge, so it is not a junction. When a branch containing a capacitor meets other branches, the junction rule still applies at the node where the wires meet, but the current in the capacitor branch is the rate at which charge accumulates on the plate, which is $dq/dt$ rather than a steady labelled current. That is not an exception to charge conservation; it is charge conservation applied to a region that is storing charge.

Statement 11.8.A.2 makes the same point from the capacitor side: as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.

Writing it down without sign errors

There are two equivalent ways to write the rule, and picking one and staying with it prevents most errors.

Form 1, the CED's: $\sum I_{\text{in}} = \sum I_{\text{out}}$ . List the currents you have assumed to arrive on one side and the ones you have assumed to leave on the other. No signs at all.

Form 2, the signed version: $\sum I = 0$ with currents into the junction counted positive and currents out counted negative. This is more compact for a network and it is what most textbook systems use, but it needs a declared sign convention.

Both require the same first step, and it is the one people skip: assign and draw a direction for every branch current before writing anything. Guess freely. A negative answer means the guess was backwards, the magnitude is still right, and no marks are lost. Topic 11.1 explains why guessing is safe: statement 11.1.A.4 says current is a scalar with a direction that does not obey vector addition, so there is nothing geometric to get wrong, only a sign.

That last point deserves emphasis because it is where a Physics C student can overthink. You never resolve currents into components at a junction. Three wires meeting at 120 degrees still obey $\sum I_{\text{in}} = \sum I_{\text{out}}$ with no cosines anywhere. Statement 11.1.A.4 is explicit: current does not obey the laws of vector addition and has no vector components. The angles in the drawing are decoration.

Two more practical points:

A junction needs three or more wires. Two wires meeting is not a junction; it is a bend, and the current is the same on both sides trivially. Statement 11.5.A.1.i covers that case as a series connection.
The number of useful junction equations is one fewer than the number of junctions. In a two-node network the equation at the top node and the equation at the bottom node are the same equation, so only one of them adds information.

Where the parallel rule comes from

The junction rule is not an extra fact bolted onto circuit analysis. Combined with the loop rule it generates the parallel-resistance formula, and deriving that in three lines is a skill 2.A exercise worth practising.

Take resistors $R_1$ and $R_2$ between the same two nodes, with a total current $I$ arriving at the upper node.

Junction rule at the upper node: $I = I_1 + I_2$ .
Parallel connection, by 11.5.A.1.ii, means both branches have the same potential difference $\Delta V$ across them, so Ohm's law gives $I_1 = \Delta V / R_1$ and $I_2 = \Delta V / R_2$ .
Substitute: $I = \Delta V\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$ , and since $I = \Delta V / R_{\text{eq}}$ by definition of the equivalent resistance,

\frac{1}{R_{\text{eq,p}}} = \sum_i \frac{1}{R_i}

That is statement 11.5.A.2.ii, and the reciprocals appear because the currents add while the potential difference is shared. The series rule comes out of the loop rule by the mirror argument: the potential differences add while the current is shared, so the resistances add directly.

The same logic explains 11.5.A.2.iii, that adding a parallel branch lowers the equivalent resistance: another branch is another path for charge to leave the junction, so more total current flows for the same potential difference, and more current at the same potential difference means less resistance.

It also explains the capacitor rules, which are swapped relative to the resistor ones. For capacitors in parallel, the charges add at the same potential difference, so $C_{\text{eq,p}} = \sum_i C_i$ . For capacitors in series, conservation of charge forces the same magnitude of charge on each plate, which is 11.8.A.2, so the potential differences add and the reciprocals of the capacitances add. Rather than memorising which rule flips, remember which quantity the connection forces to be shared, and let the algebra follow.

dq/dt as a branch current, and the rubric point it earns

This is the one place where AP Physics C uses the junction rule differently from the algebra-based course, and the CED's own scoring guidelines say so explicitly.

In a circuit where one branch contains a capacitor, the current in that branch is not a constant to be labelled and solved for. It is the rate at which charge accumulates on the plate, so by 11.1.A.1 it is $dq/dt$ . Write the junction rule and that derivative enters the equation directly:

I_{\text{total}} = I_{\text{resistor}} + \frac{dq}{dt}

Substituting Ohm's law for the resistor branch and $q = C\,\Delta V$ for the capacitor turns an ordinary junction equation into a differential equation. That is one of the two routes to statement 11.8.B.1, and Topic 11.8 develops it fully.

The CED's sample free-response Question 4 is built on exactly this, and its scoring guidelines award three separate points in part B:

one point for a multi-step derivation starting with a correct application of either Kirchhoff's junction rule or loop rule, with $I_{\text{const}} = I_R + I_C$ given as an acceptable junction-rule opening;
one point for substituting correct expressions for the current in the resistor, $\Delta V / R$ , and the current in the capacitor branch, $dq/dt$ , into an equation that expresses Kirchhoff's junction rule;
one point for correctly substituting the current in the capacitor branch as $\dfrac{d(C\,\Delta V)}{dt}$ .

That third point is the subtle one. If the question asks for a differential equation in the potential difference rather than in the charge, you replace $q$ by $C\Delta V$ before differentiating, and because $C$ is constant it comes out front: $\dfrac{dq}{dt} = C\dfrac{d(\Delta V)}{dt}$ .

Learning objective 11.7.A is one of the five objectives that question is aligned to, alongside 10.3.A, 11.1.A, 11.3.B and 11.8.B. So on the C exam, the junction rule's most likely appearance in a free-response question is as the opening line of an RC derivation.

The junction rule in terms of current density

AP Physics C carries a quantity AP Physics 2 never meets, and it gives the junction rule a second reading that is worth having.

Statement 11.1.A.2 defines current density and prints $I = \int \vec{J} \cdot d\vec{A}$ on the equation sheet. Draw a closed surface around a junction. The total current out of that surface is the flux of $\vec{J}$ through it, and conservation of charge with no charge accumulating inside says that flux is zero. Break the surface into the pieces each wire passes through and you have $\sum I_{\text{in}} = \sum I_{\text{out}}$ again. The junction rule is the closed-surface statement of charge conservation, in the same way Gauss's law is the closed-surface statement about charge and flux.

The practical consequence catches people out: the junction rule conserves current, not current density. If a wire splits into branches whose total cross-sectional area differs from the original, the current density changes even though the current is exactly accounted for. So does the drift speed, since 11.1.A.2.i gives $\vec{J} = nq\vec{v}_d$ , and so does the internal field, since 11.1.A.2.iii gives $\vec{E} = \rho\vec{J}$ .

A related check that is easy to state and easy to forget: the current is the same at every cross-section of an unbranched wire, whatever its shape. A wire that narrows carries the same current through the narrow part as the wide part, because there is no junction and nowhere for charge to accumulate. What rises in the narrow section is the current density, and with it the field and the rate of energy dissipation per unit volume. That is the physics behind a fuse, and behind the CED's Topic 11.4 activity about why a small resistor can only handle a small power.

None of this is separately stated in Topic 11.7; it follows from putting 11.7.A.1 next to the Topic 11.1 statements. That kind of joining up is what skill 3.B, applying an appropriate law or model to make a claim, is asking for.

What is printed, and how Topic 11.7 is tested

Checked against the Table of Information appendix rather than recalled:

Equation	CED statement	On the sheet
$\sum I_{\text{in}} = \sum I_{\text{out}}$	11.7.A.2, relevant	no

The junction rule is not printed on the AP Physics C: Electricity and Magnetism equation sheet, and neither is the loop rule. Two of Unit 11's eight topics are named after these rules and neither appears anywhere in the Table of Information. The CED introduces both as consequences of conservation laws, so the intended route is to write them from the principle rather than to look them up. Also absent from the sheet, and often looked for: the squared power forms, and the terminal potential difference of a battery with internal resistance.

Topic 11.7 gets a disproportionate share of the CED's own sample assessment for a topic with two statements.

Sample multiple-choice Question 10 is aligned to skill 3.B, learning objective 11.7.A and essential knowledge 11.7.A.2, and its credited answer is D. It shows a junction of three wires and four candidate diagrams with currents of 2 A, 3 A and 5 A, asking which could indicate the directions and magnitudes of the currents. The whole question is whether the arrows are consistent with charge conservation.

Sample free-response Question 4 lists 11.7.A among its five learning objectives, and as described above, the junction rule appears there as the opening move of a differential-equation derivation.

The skill list tells you the third shape to expect. 1.B and 3.A point at the Experimental Design and Analysis question: designing a measurement that tests whether branch currents sum, choosing what data to collect, and plotting it so that a straight line is the prediction. 2.B is the routine calculation.

The CED's Topic 11.6 sample activity covers both rules together and is the best single practice task for either: solve a typical multi-loop circuit problem with batteries and resistors, then construct a representation for each possible loop that visually shows Kirchhoff's loop rule and a representation for each junction that visually shows Kirchhoff's junction rule.

Which junction diagrams are possible, and which are not

A junction in a circuit consists of three wires connected together, carrying currents of magnitude $2$ A, $3$ A and $5$ A. (a) Which arrangements of arrow directions are consistent with Kirchhoff's junction rule? (b) A student proposes replacing the 5 A wire with a 4 A wire, keeping the other two at 2 A and 3 A. Show that no arrangement of directions can then satisfy the rule. (c) Generalise: what condition must three magnitudes satisfy at a three-wire junction? This is the setting of the CED's sample multiple-choice Question 10, aligned to skill 3.B and essential knowledge 11.7.A.2.

(a) State the rule from 11.7.A.2: the total charge entering per unit time equals the total charge exiting per unit time, so $\sum I_{\text{in}} = \sum I_{\text{out}}$ . With three wires, every arrangement has either one wire in and two out, or two in and one out. A three-in or three-out arrangement would pile charge up at the junction or create it there, which conservation of charge forbids by 11.7.A.1.
Case one wire in, two out: the single inward current must equal the sum of the two outward ones. The only way to make that work from $\{2, 3, 5\}$ is $5$ in with $2$ and $3$ out, since $5 = 2 + 3$ .
Case two wires in, one out: $2$ and $3$ in with $5$ out, again because $2 + 3 = 5$ . Any diagram showing $2$ and $5$ arriving with $3$ leaving fails, because $7 \neq 3$ .
So exactly two arrangements work and they are reverses of each other. The 5 A wire is always the one on its own, and the 2 A and 3 A wires are always on the same side.
(b) With $\{2, 3, 4\}$ , test all three ways of choosing the odd wire out: $2 + 3 = 5 \neq 4$ ; $2 + 4 = 6 \neq 3$ ; $3 + 4 = 7 \neq 2$ . None works, so no assignment of directions satisfies the junction rule and the proposed set is physically impossible at a three-wire junction.
(c) The condition is that one magnitude must equal the sum of the other two. Equivalently, with a signed convention where inward currents count positive, some choice of signs must make $\pm I_1 \pm I_2 \pm I_3 = 0$ .
Note what plays no part in this. The angles between the wires are irrelevant, because 11.1.A.4 says current is a scalar with a direction that does not obey the laws of vector addition and has no vector components. Three wires meeting at 120 degrees give exactly the same equation as three wires meeting at right angles.

(a) Two arrangements, mirror images of each other: the 5 A wire alone on one side and the 2 A and 3 A wires together on the other, because $2 + 3 = 5$ . (b) Impossible, since none of $2+3$ , $2+4$ or $3+4$ equals the remaining magnitude. (c) At a three-wire junction one current magnitude must equal the sum of the other two, and the geometry of the wires is irrelevant.

A junction with a capacitor branch, at both ends of time

An ideal battery of emf $\mathcal{E} = 20$ V is in series with $R_1 = 5.0\ \mathrm{k}\Omega$ . That branch reaches a node where it splits into two parallel branches back to the battery: one containing $R_2 = 20\ \mathrm{k}\Omega$ , the other containing an initially uncharged capacitor $C = 10\ \mu\mathrm{F}$ . The switch closes at $t = 0$ . Write the junction equation, then evaluate every current immediately after closing and long afterwards.

Junction equation first, at the node where the branch splits. Two currents leave it: the resistor current $I_2$ and the capacitor branch current. By 11.1.A.1 the latter is the rate at which charge accumulates on the plate, so $I_1 = I_2 + \dfrac{dq}{dt}$ . That is 11.7.A.2 with a derivative in it, and writing it is the step the CED's sample Question 4 rubric awards a point for.
Substitute the element relations. Both parallel branches share a potential difference by 11.5.A.1.ii, call it $\Delta V_C$ . Then $I_2 = \dfrac{\Delta V_C}{R_2}$ from 11.3.B.1, and $q = C \Delta V_C$ from the printed $C = Q/\Delta V$ .
Immediately after closing, statement 11.8.B.3.i says an uncharged capacitor acts like a wire, so $\Delta V_C = 0$ and therefore $I_2 = 0$ . The whole battery emf sits across $R_1$ : $I_1 = \dfrac{20}{5.0 \times 10^{3}} = 4.0 \times 10^{-3}$ A, that is 4.0 mA.
Check the junction equation at that instant: $4.0 = 0 + \dfrac{dq}{dt}$ in milliamps, so all 4.0 mA is going onto the capacitor plate. The capacitor branch carries the largest current it will ever carry, and the resistor branch carries none.
Long afterwards, statement 11.8.B.3.iv says a charging capacitor reaches a maximum potential difference at which there is zero current in its branch, so $dq/dt = 0$ . The junction equation collapses to $I_1 = I_2$ , and the battery sees $R_1$ and $R_2$ in series: $I_1 = I_2 = \dfrac{20}{25 \times 10^{3}} = 8.0 \times 10^{-4}$ A, that is 0.80 mA.
The capacitor's final potential difference is the one across $R_2$ : $\Delta V_C = (8.0 \times 10^{-4})(20 \times 10^{3}) = 16$ V. Its final charge is $q = C\Delta V_C = (10 \times 10^{-6})(16) = 1.6 \times 10^{-4}$ C, or $160\ \mu\mathrm{C}$ .
Sanity check the whole loop at the end: $20 = I_1 R_1 + \Delta V_C = (8.0 \times 10^{-4})(5000) + 16 = 4.0 + 16 = 20$ V. The loop rule closes, so the two ends are consistent.
Read the pattern. The junction equation is the same equation at both instants; all that changes is which of its three terms is zero. That is why the two limiting cases can be answered without solving anything.

The junction equation is $I_1 = I_2 + dq/dt$ with $I_2 = \Delta V_C/R_2$ and $q = C\Delta V_C$ . Immediately after closing, $I_1 = 4.0$ mA, $I_2 = 0$ and $dq/dt = 4.0$ mA. Long afterwards, $I_1 = I_2 = 0.80$ mA and $dq/dt = 0$ , with $\Delta V_C = 16$ V and $q = 160\ \mu\mathrm{C}$ .

Current is conserved at a junction; current density is not

A wire of radius $R = 1.0$ mm carrying a current of $6.0$ A splits at a junction into three identical wires, each of radius $r = 0.50$ mm, made of the same metal. Find (a) the current in each branch, (b) the current density in the main wire and in each branch, (c) the ratio of the branch current density to the main one, symbolically, and (d) what happens to the drift speed and to the electric field inside the metal.

(a) Statement 11.7.A.2 gives $\sum I_{\text{in}} = \sum I_{\text{out}}$ , so $6.0 = 3I_{\text{branch}}$ by symmetry, and each branch carries $2.0$ A. Charge conservation fixes this exactly and says nothing at all about the wire radii.
(b) The current is uniform across each wire, so the flux integral of 11.1.A.2 collapses to $J = I/A$ . Main wire: $A = \pi R^2 = \pi(1.0 \times 10^{-3})^2 = 3.142 \times 10^{-6}\ \mathrm{m^2}$ , so $J_{\text{main}} = \dfrac{6.0}{3.142 \times 10^{-6}} = 1.91 \times 10^{6}\ \mathrm{A/m^2}$ .
Each branch: $A = \pi r^2 = \pi(0.50 \times 10^{-3})^2 = 7.854 \times 10^{-7}\ \mathrm{m^2}$ , so $J_{\text{branch}} = \dfrac{2.0}{7.854 \times 10^{-7}} = 2.55 \times 10^{6}\ \mathrm{A/m^2}$ .
(c) Symbolically, $\dfrac{J_{\text{branch}}}{J_{\text{main}}} = \dfrac{I/3}{\pi r^2} \cdot \dfrac{\pi R^2}{I} = \dfrac{R^2}{3r^2}$ . With $R = 2r$ that is $\dfrac{4r^2}{3r^2} = \dfrac{4}{3}$ , and $\dfrac{2.55}{1.91} = 1.33$ confirms it numerically.
The reason is geometric. Halving the radius quarters the area, so the three branches together present $3 \times \tfrac{1}{4} = \tfrac{3}{4}$ of the original cross-section. The same total current squeezed through three quarters of the area gives four thirds of the current density.
(d) Statement 11.1.A.2.i gives $\vec{J} = nq\vec{v}_d$ . The metal is the same, so $n$ and $q$ are unchanged and the drift speed scales with $J$ : it rises by a factor of $4/3$ in the branches. Statement 11.1.A.2.iii gives $\vec{E} = \rho\vec{J}$ , and $\rho$ is a material property unchanged by the split, so the internal field also rises by $4/3$ .
The reading to keep: the junction rule conserves current, not current density, not drift speed and not the internal field. All three of those depend on geometry, and only the current is fixed by charge conservation.

(a) $2.0$ A in each branch. (b) $J_{\text{main}} = 1.91 \times 10^{6}\ \mathrm{A/m^2}$ and $J_{\text{branch}} = 2.55 \times 10^{6}\ \mathrm{A/m^2}$ . (c) The ratio is $R^2/(3r^2) = 4/3$ , because the three branches present only three quarters of the original cross-section. (d) Drift speed and internal field both rise by the same factor of $4/3$ , since both are proportional to the current density.

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, so the currents into a junction sum to the currents out of it. The AP Physics C course description gives it at essential knowledge 11.7.A.2 and says at 11.7.A.1 that it is a consequence of the conservation of electric charge. The reason is that a junction is a point with no capacity to store charge, so whatever arrives must leave immediately. It applies at any node where three or more wires meet; two wires meeting is just a bend, and the current is trivially the same on both sides.

Is Kirchhoff's junction rule on the AP Physics C E&M equation sheet?

No. Neither Kirchhoff rule is printed anywhere in the Table of Information for AP Physics C: Electricity and Magnetism, even though two of Unit 11's eight topics are named after them. The course description introduces both as consequences of conservation laws, the junction rule from conservation of electric charge at essential knowledge 11.7.A.1 and the loop rule from conservation of energy at 11.6.A.2, so the expectation is that students write each from the principle rather than look it up. Two other Unit 11 results students often expect to find are also missing from the sheet: the squared forms of the power equation, and terminal potential difference for a battery with internal resistance.

How does a capacitor branch enter Kirchhoff's junction rule?

As dq/dt, the rate at which charge accumulates on the plate. Essential knowledge 11.1.A.1 in the AP Physics C course description defines current as the derivative of charge with respect to time, so a branch containing a capacitor contributes that derivative to the junction equation rather than a fixed labelled current. The scoring guidelines for the course description's own sample free-response Question 4 award a point for substituting the resistor current as the potential difference over R and the capacitor branch current as dq/dt into an equation expressing Kirchhoff's junction rule, and a further point for writing that capacitor current as the derivative of C times the potential difference. This is one of the two routes to the RC differential equation of essential knowledge 11.8.B.1.

Do you resolve currents into components at a junction?

No, never. Essential knowledge 11.1.A.4 in the AP Physics C course description states that although current is a scalar quantity it does have a direction, and because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components. So the angles between wires at a junction are irrelevant to the junction rule: three wires meeting at 120 degrees give exactly the same equation as three wires meeting at right angles. The only bookkeeping needed is a sign for each branch, decided by whether you assumed the current arrives at or leaves the junction, and a negative answer simply means the assumed direction was backwards.

How do the parallel resistance rules follow from the junction rule?

In three steps. The junction rule says the total current equals the sum of the branch currents. A parallel connection means every branch has the same potential difference across it, by essential knowledge 11.5.A.1.ii, so each branch current is that shared potential difference divided by its own resistance. Substituting gives the total current as the potential difference times the sum of the reciprocals of the resistances, and comparing that with the definition of equivalent resistance shows that the reciprocal of the equivalent resistance is the sum of the reciprocals. The series rule is the mirror image, derived from the loop rule instead: the potential differences add while the current is shared, so the resistances add directly.

Does the junction rule conserve current density too?

No, only current. Conservation of electric charge fixes the total current arriving and leaving a junction, but it says nothing about the cross-sectional areas of the wires. If a wire splits into branches whose total area differs from the original, the current density changes even though every ampere is accounted for. Because essential knowledge 11.1.A.2.i relates current density to drift velocity and 11.1.A.2.iii relates it to the electric field inside the conductor, the drift speed and the internal field change in the same proportion. The same point applies inside an unbranched wire that narrows: the current is identical at every cross-section, while the current density, the field and the rate of energy dissipation per unit volume all rise in the narrow part.

Is AP Physics C Topic 11.7 different from AP Physics 2 Topic 11.7?

The required content is identical. Both frameworks contain exactly two essential knowledge statements, 11.7.A.1 that the rule follows from conservation of electric charge and 11.7.A.2 stating the rule, with the same relevant equation and no boundary statement in either course. The suggested skills differ, and that is the real distinction. AP Physics C lists 1.B on quantitative graphs including plotting data, 2.B, 3.A on creating experimental procedures, and 3.B, where AP Physics 2 lists 1.A, 2.B, 2.C and 3.C. So the C version leans toward experimental design and data analysis. In practice the C course also uses the rule with dq/dt as a branch current to build differential equations, which AP Physics 2 does not do.