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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 11.6

Loop Equations from Energy Conservation

Turn energy conservation into equations for circuit loops. Track potential rises and drops with a consistent traversal convention, solve networks with multiple sources or multiple loops, interpret negative currents correctly, and verify every solution with potential and power ledgers.

Learning Goals

  • Derive the loop rule \(\sum_{\mathrm{loop}}\Delta V=0\) from conservation of energy.
  • Assign signs to source and resistor terms from the chosen traversal direction.
  • Write and solve loop equations with one or more sources.
  • Interpret a negative current as direction information rather than a failed solution.
  • Form independent equations for a two-loop network with a shared element.
  • Check loop solutions using node potentials and power conservation.
\(\sum\Delta V\)net loop voltagezero for a closed loop
\(\mathcal E\)source riseenergy supplied per charge
\(IR\)resistor changedrop along current
\(I_k\)assumed currentsign reports actual direction

1. Voltage Is an Energy Ledger per Charge

Moving charge \(q\) through a potential change \(\Delta V\) changes its electric potential energy by

\(\Delta U=q\Delta V\)

A source raises electric potential by transferring energy into the charges. A resistive load lowers potential while transferring that energy into thermal or other forms.

2. A Closed Trip Returns to the Same State

After following one complete loop back to the starting node, a charge has the same electric potential energy it had initially:

\(\Delta U_{\mathrm{loop}}=q\sum_{\mathrm{loop}}\Delta V=0\)

For nonzero test charge, divide by \(q\).

3. Kirchhoff’s Loop Rule

\(\boxed{\sum_{\mathrm{loop}}\Delta V=0}\)

The algebraic sum includes every source rise, source drop, resistor drop, resistor rise, and any other element voltage encountered during the selected closed traversal.

The rule does not say that every point has zero potential or that each element has zero voltage. It says that all changes around a completed loop cancel.

4. Sign Guide: Ask Whether Potential Rises or Falls

Element crossingPotential changeReasoning
Source from \(-\) to \(+\)\(+\mathcal E\)Move to the higher-potential terminal
Source from \(+\) to \(-\)\(-\mathcal E\)Move to the lower-potential terminal
Resistor along its assumed current\(-IR\)Potential falls in the current direction
Resistor opposite its assumed current\(+IR\)Traversal reverses the resistor drop
Ideal wire\(0\)All points on one ideal node share potential

This “rise or fall?” test is more reliable than memorizing a diagram orientation, because batteries and resistors can be drawn in any direction.

5. Current Direction and Loop Direction Are Independent Choices

Assign an arrow to each unknown current. Then choose either clockwise or counterclockwise traversal for each loop. The current arrow controls resistor polarity; the traversal arrow controls which side of the element is crossed first.

Reversing a loop traversal multiplies every term in that loop equation by \(-1\), producing an equivalent equation. Reversing an assumed current direction changes the sign of that current’s solution but not the physical circuit.

Worked Example 1 · One Source, Two Loads

Write the Equation Before Combining Resistance

An ideal \(12\,\mathrm V\) source drives \(4.0\,\Omega\) and \(8.0\,\Omega\) resistors in one loop. Assume clockwise current and traverse clockwise from the negative source terminal.

\(+12-I(4.0)-I(8.0)=0\)
\(I=\frac{12}{12}=1.0\,\mathrm A\)

The familiar series-resistance result is therefore a special case of the loop rule.

A potential walk must closeThe source rise \(+\mathcal E\) is exactly balanced by resistor drops. Ideal-wire sections are horizontal because their modeled potential change is zero.
Worked Example 2 · Node-by-Node Potential Walk

Verify the One-Loop Solution

Set the negative source terminal to \(0\,\mathrm V\). Cross the \(12\,\mathrm V\) source, then the \(4.0\,\Omega\) and \(8.0\,\Omega\) resistors in the \(1.0\,\mathrm A\) current direction.

\(0\xrightarrow{+\mathcal E}12\,\mathrm V\xrightarrow{-IR_1}8\,\mathrm V\xrightarrow{-IR_2}0\,\mathrm V\)

Returning to \(0\,\mathrm V\) confirms the loop equation and identifies the intermediate node potential.

6. Sources That Aid

Series sources aid when the chosen loop traversal crosses both from negative to positive. Their emfs add:

\(\mathcal E_{\mathrm{net}}=\mathcal E_1+\mathcal E_2\)

7. Sources That Oppose

If traversal crosses one source from negative to positive and the other from positive to negative, their contributions oppose:

\(\mathcal E_{\mathrm{net}}=\mathcal E_{\mathrm{larger}}-\mathcal E_{\mathrm{smaller}}\)
Worked Example 3 · Aiding Sources

Add Source Rises Algebraically

A \(9.0\,\mathrm V\) and \(3.0\,\mathrm V\) source aid each other in a loop containing \(2.0\,\Omega\) and \(4.0\,\Omega\).

\(9.0+3.0-2.0I-4.0I=0\)
\(I=2.0\,\mathrm A\)
Worked Example 4 · Opposing Sources

The Stronger Source Sets the Direction

A \(12\,\mathrm V\) source opposes a \(5.0\,\mathrm V\) source in a loop with \(7.0\,\Omega\) total resistance.

\(12-5.0-7.0I=0\)
\(I=1.0\,\mathrm A\)

The current direction is the one driven by the \(12\,\mathrm V\) source.

8. A Negative Current Is Useful Information

You do not need to know the actual current direction before solving. If the calculated current is negative, its magnitude is correct and the physical current is opposite the assumed arrow.

Do not erase the negative sign halfway through the algebra. Finish the equations first, then redraw or state the actual direction.

Worked Example 5 · Reversed Assumption

Let Algebra Correct the Arrow

Assume clockwise current in a loop where a \(4.0\,\mathrm V\) source aids clockwise motion, a \(10\,\mathrm V\) source opposes it, and total resistance is \(3.0\,\Omega\).

\(4.0-10-3.0I=0\quad\Rightarrow\quad I=-2.0\,\mathrm A\)

The actual current is \(2.0\,\mathrm A\) counterclockwise. No new calculation is required.

9. Loop Equations Can Solve for More Than Current

Unknowns may include a source emf, resistance, current, or element voltage. Write the full loop relation symbolically before inserting values, then solve for the requested quantity.

Worked Example 6 · Unknown Source

Find the Emf Needed for a Target Current

A source drives \(0.50\,\mathrm A\) through series resistors \(4.0\,\Omega\) and \(6.0\,\Omega\).

\(\mathcal E-(0.50)(4.0)-(0.50)(6.0)=0\)
\(\mathcal E=5.0\,\mathrm V\)

The source must supply \(5.0\,\mathrm J\) of electric potential energy per coulomb.

10. Include Internal Resistance in the Loop

A real source modeled by emf \(\mathcal E\) and internal resistance \(r\) contributes a source rise and an internal resistor drop:

\(\mathcal E-Ir-IR_L=0\)
\(I=\frac{\mathcal E}{r+R_L},\quad V_{\mathrm{term}}=\mathcal E-Ir\)

Leaving out \(r\) would incorrectly set terminal voltage equal to emf under every load.

Worked Example 7 · Real Source

Use One Loop to Find Current and Terminal Voltage

A source has \(\mathcal E=12\,\mathrm V\), \(r=0.50\,\Omega\), and \(R_L=5.5\,\Omega\).

\(12-0.50I-5.5I=0\quad\Rightarrow\quad I=2.0\,\mathrm A\)
\(V_{\mathrm{term}}=12-(2.0)(0.50)=11\,\mathrm V\)

The load drop is \(IR_L=(2.0)(5.5)=11\,\mathrm V\), agreeing with terminal voltage.

11. Multiple Loops Need Independent Equations

A loop equation is independent only if it adds information not already contained in the others. In a two-loop network, the outer perimeter equation may be the sum or difference of the two inner-loop equations and therefore provide no new constraint.

Choose the smallest convenient loops first. Every unknown current must appear in enough independent equations to be determined.

A shared resistor belongs to both loop equationsIf clockwise mesh currents oppose in the shared branch, its downward current is \(I_1-I_2\). Each loop uses the same physical branch current with the appropriate traversal sign.

12. Mesh-Current Method for Two Loops

Assign clockwise mesh currents \(I_1\) and \(I_2\). If their directions oppose in a shared resistor \(R_s\), the physical shared-branch current in the \(I_1\) direction is

\(I_s=I_1-I_2\)

The left loop uses the shared drop \(R_s(I_1-I_2)\). Traversing the right loop clockwise uses \(R_s(I_2-I_1)\). These are the same element viewed in opposite directions.

Worked Example 8 · Two Coupled Loops

Solve a Shared-Resistor Network

The left loop has a \(10\,\mathrm V\) source and \(2.0\,\Omega\) outer resistor. The right loop has a \(6.0\,\mathrm V\) source and \(2.0\,\Omega\) outer resistor. The loops share \(4.0\,\Omega\). Both source orientations produce a rise in their clockwise mesh traversals.

\(10-2I_1-4(I_1-I_2)=0\quad\Rightarrow\quad6I_1-4I_2=10\)
\(6-2I_2-4(I_2-I_1)=0\quad\Rightarrow\quad-4I_1+6I_2=6\)
\(I_1=4.2\,\mathrm A,\quad I_2=3.8\,\mathrm A,\quad I_s=I_1-I_2=0.40\,\mathrm A\)

The shared current follows the \(I_1\) direction because \(I_1>I_2\).

13. Power Provides an Independent Check

For the two-loop example, source power is

\(P_{\mathrm{sources}}=(10)(4.2)+(6.0)(3.8)=64.8\,\mathrm W\)

Resistor power is

\(P_R=2(4.2)^2+2(3.8)^2+4(0.40)^2=64.8\,\mathrm W\)

Agreement supports the current directions and shared-resistor term. A mismatch indicates a sign, branch-current, or arithmetic error.

14. Experimental Loop Residual

Measured source and resistor voltages will rarely sum to exactly zero because meters, sources, wires, and components are not ideal and measurements have uncertainty. Define a loop residual

\(\delta V_{\mathrm{loop}}=\sum_k\Delta V_k\)

Compare \(|\delta V_{\mathrm{loop}}|\) with measurement resolution and expected uncertainty before deciding whether the model fails.

15. Reliable Loop-Equation Workflow

  1. Redraw and label every node, source polarity, resistor, and unknown current.
  2. Choose current arrows without worrying whether they are correct.
  3. Select independent loops and traversal directions.
  4. Walk one loop at a time, recording each potential rise or drop.
  5. Use the actual branch current for shared elements.
  6. Solve the simultaneous equations without changing conventions midway.
  7. Interpret negative currents and verify with potential and power ledgers.

16. Common Traps

  • Adding voltage magnitudes instead of signed changes.
  • Assigning a resistor sign from its drawing orientation rather than current and traversal.
  • Changing a loop direction halfway around the loop.
  • Treating every source as a positive term regardless of polarity.
  • Declaring a negative current unphysical.
  • Using a mesh current instead of the current difference in a shared resistor.
  • Writing several loop equations that are algebraically redundant.
  • Forgetting source internal resistance when it is specified.
  • Failing to return to the starting potential in a node-voltage check.
Mastery Check

1. Why must the algebraic potential change around a closed loop be zero?

Show reasoning and answer

The final point is the starting node, so a charge returns to the same electric potential energy. Thus \(q\sum\Delta V=0\), giving \(\sum\Delta V=0\).

2. You traverse a resistor opposite its assumed current. What term enters the loop equation?

Show reasoning and answer

Potential rises by \(+IR\) because the traversal is opposite the modeled resistor drop.

3. Reversing the traversal direction of an entire loop changes what?

Show reasoning and answer

Every term changes sign, multiplying the whole equation by \(-1\). The physical solution is unchanged.

4. A loop equation gives \(I=-0.70\,\mathrm A\). Interpret the result.

Show reasoning and answer

The actual current magnitude is \(0.70\,\mathrm A\), and its direction is opposite the initially assumed arrow.

5. Clockwise mesh currents \(I_1=2.5\,\mathrm A\) and \(I_2=1.0\,\mathrm A\) oppose in a shared resistor. Find the shared current in the \(I_1\) direction.

Show reasoning and answer

\(I_s=I_1-I_2=1.5\,\mathrm A\) in the \(I_1\) direction.

6. A \(9.0\,\mathrm V\) source drives current through \(2.0\,\Omega\) and \(7.0\,\Omega\) in one loop. Find \(I\).

Show reasoning and answer

\(9.0-2.0I-7.0I=0\), so \(I=1.0\,\mathrm A\).

Investigation: Measure a Closed Potential Walk

Use the open PhET Circuit Construction Kit: DC simulation. Build one source and two series resistors. Choose the negative source terminal as \(0\,\mathrm V\), measure successive node potentials around the loop, and record each signed change. Test whether the sum returns to zero.

Then build two loops with a shared resistor. Predict mesh-current directions, write the two loop equations, and compare calculated currents with non-contact ammeter readings. Reverse one assumed arrow on paper and show that only the sign interpretation changes. 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.