AP Course

AP Physics 2

Study AP Physics 2 Units 9–15, with the Fall 2026 radioactive-decay clarification applied and official exam updates linked separately.

Study Units 9–15 through original models, derivations, experiments, quizzes, and practice sets.

Lessons
Particle Motion Behind Temperature and PressureConnecting Gas Variables with the Ideal-Gas ModelHeat Flow and the Approach to Thermal BalanceEnergy Accounting for Thermodynamic ProcessesMaterial Response to Heating and Heat ConductionEntropy, Probability, and the Direction of Thermal ChangeCharge Interactions and Coulomb-Force ModelsTracking Charge During Contact and InductionMapping Electric Fields from Source ChargesEnergy of Configurations of ChargesElectric Potential and Equipotential ReasoningCapacitance and Energy Stored in Electric FieldsConserving Energy in Charged-Particle MotionCharge Flow and Conventional CurrentModeling Sources, Wires, and Loads in Simple NetworksHow Geometry and Material Set ResistanceElectrical Energy Transfer and PowerReducing Series-Parallel DC NetworksLoop Equations from Energy ConservationJunction Equations from Charge ConservationTransient Charging and Discharging in RC NetworksSources, Direction, and Strength of Magnetic FieldsMagnetic Forces and Curved Paths of ChargesForces Between Fields and Current-Carrying ConductorsChanging Magnetic Flux and Induced EMFReflection and Absorption at Material BoundariesLocating Images with Mirror Ray ModelsRefraction, Index, and Total Internal ReflectionLocating Images with Thin-Lens ModelsPulses, Wave Types, and Propagation SpeedPeriodic-Wave Measures and GraphsBoundary Changes and PolarizationElectromagnetic-Wave BehaviorFrequency Shifts from Relative MotionSuperposition, Interference, and Standing WavesDiffraction from Openings and EdgesTwo-Slit and Grating Pattern GeometryPhase Change and Thin-Film ColorQuantum Models and Dual Wave-Particle EvidenceEnergy Levels in a Hydrogen-Like AtomConnecting Spectral Lines to Energy TransitionsThermal Spectra and Blackbody CurvesPhoton Thresholds in the Photoelectric EffectPhoton-Electron Scattering and MomentumMass-Energy Accounting in Fission and FusionRandom Nuclear Decay, Activity, and Half-Life
Quizzes
Practice Problems Formula notes, diagrams, and practice sets will be added here.

AP Physics 2 · Unit 11 · Topic 11.7

Junction Equations from Charge Conservation

Convert conservation of electric charge into equations for branching circuits. Distinguish nodes from branches, assign current directions consistently, solve current-divider and multi-junction networks, interpret negative answers, and use a charge ledger to test whether measured currents are physically consistent.

Learning Goals

  • Derive the junction rule from charge conservation and \(I=dQ/dt\).
  • Write the same junction equation using either an in/out ledger or one signed-current convention.
  • Find unknown currents at a junction and interpret negative solutions correctly.
  • Relate junction equations to current division in parallel resistive branches.
  • Select independent junction equations in a network with several nodes.
  • Diagnose missing branches, charge accumulation, or measurement inconsistency from a nonzero current residual.
Nodeconnected conductor regionone modeled potential
Branchpath between nodesone branch current
\(I=dQ/dt\)charge-flow rate\(1\,\mathrm A=1\,\mathrm{C/s}\)
\(\sum I=0\)signed current ledgersteady ideal node

1. What Counts as a Junction?

A node is a continuous conducting region modeled at one potential. A junction is a node where three or more branches meet. A bend in one wire is not a junction, and wires connected without an intervening element belong to the same node.

2. Current Is a Rate, Not an Amount

Current reports how quickly signed charge crosses a chosen section:

\(I=\frac{dQ}{dt}\)

A \(2.0\,\mathrm A\) branch transports \(2.0\,\mathrm C\) of conventional charge per second through that section.

3. Derivation from a Charge Ledger

Draw a small boundary around a junction. During a short interval \(\Delta t\), incoming currents add charge to the region and outgoing currents remove charge:

\(\Delta Q_{\text{node}}=\left(\sum I_{\text{in}}-\sum I_{\text{out}}\right)\Delta t\)

Divide by \(\Delta t\) and take the instantaneous limit:

\(\frac{dQ_{\text{node}}}{dt}=\sum I_{\text{in}}-\sum I_{\text{out}}\)

In the steady-state lumped-circuit model, charge does not continually accumulate at an ideal node, so \(dQ_{\text{node}}/dt=0\). Therefore

\(\boxed{\sum I_{\text{in}}=\sum I_{\text{out}}}\)

This is Kirchhoff’s junction rule: a circuit form of conservation of electric charge.

A junction is a charge-flow checkpointFor the directions shown, \(I_1+I_2=I_3\). Reversing any assumed arrow changes the sign assigned to that branch, not the conservation principle.

4. Two Equivalent Sign Conventions

ConventionEquation formHow to use it
Separate ledgers\(\sum I_{\mathrm{in}}=\sum I_{\mathrm{out}}\)Place current magnitudes on the side indicated by their arrows
Signed algebraic sum\(\sum_k s_kI_k=0\)For example, incoming \(+\), outgoing \(-\)

Either choice is correct. State the convention once and keep it throughout the equation. Do not attach both a signed variable and a second direction sign to the same current.

Worked Example 1 · One Unknown Branch

Balance Incoming and Outgoing Currents

Currents \(2.4\,\mathrm A\) and \(1.1\,\mathrm A\) enter a junction. A current of \(0.80\,\mathrm A\) and unknown \(I_x\) leave.

\(2.4+1.1=0.80+I_x\)
\(I_x=2.7\,\mathrm A\)

Unit check: each term is a charge-flow rate, so only currents may be added in this equation.

5. Arrow Directions May Be Chosen Before Solving

If an unknown branch direction is unclear, choose an arrow and write the junction equation consistently. A positive answer confirms the assumed direction. A negative answer means the physical conventional current points opposite the arrow.

The negative sign is not evidence that charge was destroyed or that the algebra failed; it is direction information encoded by the variable definition.

Worked Example 2 · Negative Current

Let Conservation Correct the Assumed Arrow

At a junction, \(5.0\,\mathrm A\) enters and \(1.5\,\mathrm A\) leaves. Assume another \(I_x\) enters and a \(2.5\,\mathrm A\) branch leaves.

\(5.0+I_x=1.5+2.5\)
\(I_x=-1.0\,\mathrm A\)

The actual current is \(1.0\,\mathrm A\) leaving along the \(I_x\) branch. The algebra has preserved charge conservation and corrected the assumed arrow.

6. A Current Residual Measures the Imbalance

Define incoming current as positive and outgoing current as negative:

\(I_{\text{res}}=\sum I_{\text{in}}-\sum I_{\text{out}}=\frac{dQ_{\text{node}}}{dt}\)

For a steady ideal node, \(I_{\text{res}}=0\). A nonzero measured residual can indicate an omitted branch, a reversed meter sign, component or meter uncertainty, or time-dependent charge storage in the chosen region.

Worked Example 3 · Diagnose an Incomplete Ledger

Find the Missing Charge-Flow Rate

A student records \(3.0\,\mathrm A\) entering a steady junction and \(1.2\,\mathrm A\) plus \(0.80\,\mathrm A\) leaving.

\(I_{\text{res}}=3.0-1.2-0.80=1.0\,\mathrm A\)

If the node truly accumulated charge at this rate, it would gain \(1.0\,\mathrm C\) every second. That is not a steady-state description, so the data imply a missing \(1.0\,\mathrm A\) outgoing branch or an equivalent measurement/sign error.

7. The Same Series Current Follows from the Junction Rule

At a two-terminal connection between series elements, the current entering the small connecting region must equal the current leaving it. Therefore all elements in a single unbranched path carry the same current in steady state.

\(I_1=I_2=I_{\text{series}}\)

The charge carriers are not consumed by a resistor. The resistor transfers energy, while charge continues through the branch.

Current divides and recombinesAt the first node \(I_T=I_1+I_2\); at the second, \(I_1+I_2=I_T\). These are the same conservation statement, so only one is independent.

8. Derive the Two-Branch Current Divider

Parallel resistors \(R_1\) and \(R_2\) share potential difference \(V\), so \(I_1=V/R_1\), \(I_2=V/R_2\), and the junction rule gives

\(I_T=I_1+I_2=V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)

Eliminating \(V\) produces

\(\boxed{I_1=I_T\frac{R_2}{R_1+R_2}},\qquad \boxed{I_2=I_T\frac{R_1}{R_1+R_2}}\)

The smaller-resistance branch receives the larger current. The “other resistance” appears in each numerator because these formulas are obtained by normalizing branch conductance.

Worked Example 4 · Current Division

Use Shared Voltage and Verify at the Junction

A \(6.0\,\Omega\) resistor and a \(3.0\,\Omega\) resistor are connected in parallel across \(12\,\mathrm V\).

\(I_1=\frac{12}{6.0}=2.0\,\mathrm A,\qquad I_2=\frac{12}{3.0}=4.0\,\mathrm A\)
\(I_T=I_1+I_2=6.0\,\mathrm A\)

The \(3.0\,\Omega\) path carries twice the current of the \(6.0\,\Omega\) path. Their currents recombine to the original \(6.0\,\mathrm A\).

9. Combine Element Relations with the Junction Rule

A junction equation alone relates currents but does not usually determine every current. Add element equations such as \(V=IR\), capacitor relations, and loop equations until the number of independent equations matches the number of unknowns.

Keep the roles distinct: junction equations enforce charge conservation; loop equations enforce energy conservation; element equations describe device behavior.

Worked Example 5 · Series Feed into Parallel Loads

Use Charge and Energy Conservation Together

A \(12\,\mathrm V\) ideal source drives a \(2.0\,\Omega\) resistor in series with parallel \(6.0\,\Omega\) and \(3.0\,\Omega\) branches.

\(R_p=\left(\frac{1}{6.0}+\frac{1}{3.0}\right)^{-1}=2.0\,\Omega\)
\(I_T=\frac{12}{2.0+2.0}=3.0\,\mathrm A\)

The series resistor drops \((3.0)(2.0)=6.0\,\mathrm V\), leaving \(6.0\,\mathrm V\) across both branches:

\(I_{6\Omega}=1.0\,\mathrm A,\qquad I_{3\Omega}=2.0\,\mathrm A\)
\(\underbrace{3.0}_{\text{enters}}=\underbrace{1.0+2.0}_{\text{leaves}}\)

The junction check closes the charge ledger, while the loop calculation supplies the branch voltage.

10. Not Every Junction Equation Is Independent

Suppose one parallel network has a splitting node A and recombination node B:

\(\text{A: }I_T-I_1-I_2=0\)
\(\text{B: }I_1+I_2-I_T=0\)

The second equation is exactly \(-1\) times the first, so it adds no new information. For a connected network with \(N\) essential nodes, at most \(N-1\) node charge-balance equations are independent.

Worked Example 6 · Recognize Redundancy

Do Not Count the Same Conservation Statement Twice

A total current \(I_T\) divides into \(I_1\) and \(I_2\), then recombines. Writing both \(I_T=I_1+I_2\) and \(I_1+I_2=I_T\) does not provide two equations. If all three currents are unknown, an element or loop relationship is still needed.

11. Multi-Junction Networks: Work One Node at a Time

Label each branch current once and keep that label along the entire branch. At every selected node, copy the arrow direction from the master diagram rather than deciding direction again. This prevents one physical current from receiving contradictory signs in different equations.

Worked Example 7 · Three-Junction Ledger

Propagate Known Currents Through a Network

At node A, \(I_0=8.0\,\mathrm A\) enters and divides into \(I_1=3.0\,\mathrm A\) and \(I_2\). At node B, \(I_1\) and an additional \(I_3=1.5\,\mathrm A\) enter and form \(I_4\). At node C, \(I_2\) and \(I_4\) combine into \(I_5\).

\(\text{A: }I_2=8.0-3.0=5.0\,\mathrm A\)
\(\text{B: }I_4=3.0+1.5=4.5\,\mathrm A\)
\(\text{C: }I_5=5.0+4.5=9.5\,\mathrm A\)

The final current exceeds \(I_0\) because a separate \(1.5\,\mathrm A\) branch entered the chosen network boundary at B. For the whole boundary, \(8.0+1.5=9.5\,\mathrm A\).

12. Choose the System Boundary Explicitly

A current can be internal to one selected circuit boundary but external to a smaller node boundary. When checking a whole subnetwork, internal branch currents cancel because each one leaves one internal node and enters another. Only currents crossing the outer boundary remain.

\(\sum I_{\text{boundary,in}}-\sum I_{\text{boundary,out}}=\frac{dQ_{\text{inside}}}{dt}\)

This boundary view is a powerful check for complex diagrams.

Worked Example 8 · Measurement with Uncertainty

Decide Whether a Residual Is Meaningful

At a steady node, measured incoming currents are \(2.00\pm0.03\,\mathrm A\) and \(0.80\pm0.02\,\mathrm A\). The outgoing current is \(2.76\pm0.04\,\mathrm A\).

\(I_{\text{res}}=2.00+0.80-2.76=0.04\,\mathrm A\)

A conservative absolute uncertainty estimate is \(0.03+0.02+0.04=0.09\,\mathrm A\). Because the \(0.04\,\mathrm A\) residual is smaller, these measurements are consistent with zero imbalance at this precision.

13. Steady Nodes and Charge-Storing Elements

The familiar form \(\sum I_{\mathrm{in}}=\sum I_{\mathrm{out}}\) assumes no continuing net charge accumulation inside the chosen node region. During a capacitor’s charging process, currents can change with time and charge intentionally accumulates on the plates. The full charge ledger remains valid:

\(\sum I_{\text{in}}-\sum I_{\text{out}}=\frac{dQ_{\text{stored}}}{dt}\)

For an ideal zero-size circuit node, no separate storage is assigned to the node; capacitor storage belongs to the capacitor element and is handled through its branch current.

14. Reliable Junction-Equation Workflow

  1. Mark continuous nodes and distinct branches.
  2. Assign one arrow and one variable to each unknown branch current.
  3. Select a sign convention and state it.
  4. Draw a boundary around one junction.
  5. List every branch crossing that boundary exactly once.
  6. Write the symbolic equation before inserting values.
  7. Use only independent junction equations.
  8. Solve, interpret negative signs, and check a larger boundary.

15. Common Traps

  • Treating current as something consumed by a resistor.
  • Calling every wire bend or two-terminal connection a junction.
  • Adding voltages or resistances in a junction current equation.
  • Counting one branch twice because it bends near the node.
  • Changing a current arrow between neighboring nodes.
  • Using electron-flow direction when arrows represent conventional current.
  • Discarding a negative current instead of reversing its interpretation.
  • Counting redundant node equations as independent information.
  • Assuming equal branch currents merely because branches are parallel.
  • Claiming charge accumulation before comparing residual with uncertainty.
Mastery Check

1. Currents \(4.0\,\mathrm A\) and \(1.5\,\mathrm A\) enter a node. A \(2.0\,\mathrm A\) current and \(I_x\) leave. Find \(I_x\).

Show reasoning and answer

Charge conservation gives \(4.0+1.5=2.0+I_x\), so \(I_x=3.5\,\mathrm A\) leaving.

2. An unknown current is drawn leaving a junction, but the solution is \(-0.60\,\mathrm A\). What is the physical interpretation?

Show reasoning and answer

The current magnitude is \(0.60\,\mathrm A\), and conventional current actually enters the junction along that branch.

3. Why do the two ends of a simple parallel section not provide two independent junction equations?

Show reasoning and answer

The split equation \(I_T=I_1+I_2\) and recombination equation \(I_1+I_2=I_T\) are algebraic multiples of the same charge-conservation statement.

4. A \(9.0\,\mathrm A\) total current divides between \(3.0\,\Omega\) and \(6.0\,\Omega\) parallel branches. Find both branch currents.

Show reasoning and answer

The \(3.0\,\Omega\) branch carries \(I_1=9.0(6.0)/(3.0+6.0)=6.0\,\mathrm A\). The \(6.0\,\Omega\) branch carries \(3.0\,\mathrm A\). Their sum is \(9.0\,\mathrm A\).

5. Measurements give \(1.20\,\mathrm A\) and \(0.50\,\mathrm A\) entering, while \(1.55\,\mathrm A\) leaves. Find the signed residual using incoming positive.

Show reasoning and answer

\(I_{\mathrm{res}}=1.20+0.50-1.55=+0.15\,\mathrm A\). For a claimed steady node, investigate an omitted outgoing branch, sign error, or measurement uncertainty.

6. In \(0.20\,\mathrm s\), \(0.70\,\mathrm C\) enters a small region and \(0.62\,\mathrm C\) leaves. Find the average charge-accumulation rate.

Show reasoning and answer

\(\Delta Q_{\mathrm{node}}=0.08\,\mathrm C\), so \(\Delta Q/\Delta t=0.08/0.20=0.40\,\mathrm A\). If the region is supposed to be a steady ideal node, the ledger is incomplete or inconsistent.

Investigation: Audit Current at a Branching Node

Use the open PhET Circuit Construction Kit: DC simulation. Build a source feeding two unequal parallel resistors. Place non-contact ammeters immediately before the split, in each branch, and after recombination. Predict every current from the resistances before reading the meters.

Test whether \(I_T=I_1+I_2\) at both junctions. Change only one branch resistance and record which currents change. Then reverse one arrow in your paper model without changing the simulated circuit; demonstrate that the signed equation changes while the physical meter readings do not. Use simulation or instructor-approved low-voltage equipment only.

Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.