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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 11.4

Electrical Energy Transfer and Power

Track how sources supply energy and how circuit elements convert it into thermal, light, sound, chemical, or mechanical forms. Derive electrical power from energy per charge, choose the correct resistor power equation, compare series and parallel loads, and connect watts, joules, kilowatt-hours, ratings, and efficiency.

Learning Goals

  • Derive and apply \(P=I\Delta V\) as an electrical energy-transfer rate.
  • Use \(P=I^2R\) and \(P=(\Delta V)^2/R\) appropriately for resistive elements.
  • Distinguish energy from power and convert between joules and kilowatt-hours.
  • Construct an energy ledger for sources, loads, internal resistance, and useful output.
  • Compare power in series and parallel networks by identifying the fixed quantity.
  • Interpret device power ratings, efficiency, and thermal-safety limits.
\(E\)transferred energyjoule, \(\mathrm J\)
\(P\)energy-transfer ratewatt, \(\mathrm{J/s}\)
\(I\)charge-flow rateampere, \(\mathrm{C/s}\)
\(\Delta V\)energy per chargevolt, \(\mathrm{J/C}\)

1. Energy and Power Are Different

Energy is an amount transferred or stored. Power is the rate of that transfer:

\(P_{\mathrm{avg}}=\frac{\Delta E}{\Delta t},\quad P=\frac{dE}{dt}\)

One watt is one joule per second. A high-power device can transfer a modest energy if used briefly; a low-power device can transfer a large energy if used long enough.

2. Voltage Is Energy per Charge

When charge \(dq\) moves through a potential difference \(\Delta V\), the magnitude of electrical energy transferred is

\(dE=|\Delta V|\,dq\)

Whether the circuit element supplies or absorbs that energy depends on polarity and current direction.

3. Deriving Electrical Power

Divide the energy transferred during a short interval by the time:

\(P=\frac{dE}{dt}=|\Delta V|\frac{dq}{dt}\)

Since \(I=dq/dt\), the magnitude of the electrical power is

\(P=I|\Delta V|\)

Unit analysis confirms the result:

\(\mathrm A\cdot\mathrm V=\frac{\mathrm C}{\mathrm s}\frac{\mathrm J}{\mathrm C}=\frac{\mathrm J}{\mathrm s}=\mathrm W\)

This relationship applies to sources, resistors, motors, lamps, capacitors, and other elements when \(I\) and the voltage across the same element are used.

Worked Example 1 · Power and Energy

Keep Rate and Duration Separate

A \(12\,\mathrm V\) device carries \(2.0\,\mathrm A\) for \(5.0\,\mathrm{min}=300\,\mathrm s\).

\(P=I\Delta V=(2.0)(12)=24\,\mathrm W\)
\(E=P\Delta t=(24)(300)=7.2\times10^3\,\mathrm J\)

The \(24\,\mathrm W\) value is a rate; \(7.2\,\mathrm{kJ}\) is the energy transferred during the interval.

4. Sign: Is an Element Supplying or Absorbing?

If conventional current enters an element at its higher-potential terminal, the element absorbs electrical power. If current leaves its higher-potential terminal, it supplies power to the rest of the circuit.

  • A discharging battery usually supplies electrical power.
  • A resistor or operating lamp absorbs electrical power and transfers it mainly to thermal or radiant energy.
  • A charging battery absorbs electrical power and stores part of it chemically.

A negative result from a consistent signed calculation means the element is supplying energy rather than violating energy conservation.

Power is redistributed, not consumedThe input rate from a source equals the total rate of useful output, thermal transfer, and any other energy-storage or transfer channels.

5. Three Power Equations for a Resistive Element

Start with \(P=I\Delta V\). For an ohmic resistor at the operating temperature, \(\Delta V=IR\):

\(P=I(IR)=I^2R\)

Or substitute \(I=\Delta V/R\):

\(P=\frac{(\Delta V)^2}{R}\)

Therefore,

\(P=I\Delta V=I^2R=\frac{(\Delta V)^2}{R}\)

The first form is the general electrical power magnitude. The resistance forms combine it with a resistive \(V\)-\(I\) relation and must use the voltage, current, and resistance of the same element at the same operating state.

Worked Example 2 · Fixed Voltage

Use \(P=V^2/R\)

A \(6.0\,\Omega\) resistor is connected across \(12\,\mathrm V\).

\(I=\frac{12}{6.0}=2.0\,\mathrm A\)
\(P=\frac{12^2}{6.0}=24\,\mathrm W\)
Worked Example 3 · Fixed Current

Use \(P=I^2R\)

A \(3.0\,\Omega\) resistor carries \(4.0\,\mathrm A\).

\(P=(4.0)^2(3.0)=48\,\mathrm W\)
\(\Delta V=IR=(4.0)(3.0)=12\,\mathrm V\)

6. The Constraint Determines the Power Trend

  • At fixed current, \(P=I^2R\), so a larger resistance receives more power.
  • At fixed voltage, \(P=V^2/R\), so a smaller resistance receives more power.

These are not contradictory statements. They describe different experimental constraints. Before comparing resistors, decide whether they share current, share voltage, or are part of a changing entire network.

Never discuss \(P\) versus \(R\) without the constraintSeries elements share current, while parallel elements share voltage. That difference reverses the resistance–power comparison.
Worked Example 4 · Series Power

Same Current, Different Voltage Drops

A \(4.0\,\Omega\) and an \(8.0\,\Omega\) resistor are in series across an ideal \(12\,\mathrm V\) source.

\(I=\frac{12}{4.0+8.0}=1.0\,\mathrm A\)
\(P_4=I^2R_4=4.0\,\mathrm W,\quad P_8=I^2R_8=8.0\,\mathrm W\)
\(P_{\mathrm{total}}=12\,\mathrm W=\mathcal E I\)

The \(8.0\,\Omega\) resistor receives more power because both resistors carry the same current.

Worked Example 5 · Parallel Power

Same Voltage, Different Branch Currents

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

\(P_6=\frac{12^2}{6.0}=24\,\mathrm W,\quad P_3=\frac{12^2}{3.0}=48\,\mathrm W\)
\(P_{\mathrm{total}}=72\,\mathrm W=(12\,\mathrm V)(6.0\,\mathrm A)\)

The smaller parallel resistance receives more power because each branch has the same voltage.

7. Conservation Check for an Entire Circuit

In steady operation, total power supplied equals total power absorbed:

\(\sum P_{\mathrm{supplied}}=\sum P_{\mathrm{absorbed}}\)

For one ideal source and resistors,

\(\mathcal E I=\sum_j I_j\Delta V_j=\sum_jP_j\)

If the two totals disagree, recheck branch currents, element voltages, polarity, units, or an omitted energy-storage element.

8. Energy for Constant and Changing Power

For constant power,

\(E=P\Delta t\)

If power changes with time, energy is the signed area under a power–time graph:

\(\Delta E=\int_{t_i}^{t_f}P(t)\,dt\)

Positive absorbed power increases energy stored or transferred into an element; negative absorbed power means the element supplies energy.

9. Kilowatt-Hours Are Energy

\(1\,\mathrm{kW\,h}=(1000\,\mathrm W)(3600\,\mathrm s)=3.60\times10^6\,\mathrm J\)

A kilowatt-hour is not a power unit. It is convenient for measuring the energy transferred by household-scale devices over hours. Operating cost is energy in kWh multiplied by the price per kWh.

Worked Example 6 · Energy and Cost

Operate a Heater for 2.5 Hours

A \(1.5\,\mathrm{kW}\) heater runs for \(2.5\,\mathrm h\). Use a hypothetical energy price of \(0.18\) currency units per kWh.

\(E=(1.5)(2.5)=3.75\,\mathrm{kW\,h}=1.35\times10^7\,\mathrm J\)
\(\text{cost}=(3.75)(0.18)=0.675\approx0.68\text{ currency units}\)

The price is an explicit example assumption rather than a claim about a current local utility rate.

10. A Real Source Has an Internal Power Channel

For a source with emf \(\mathcal E\), internal resistance \(r\), and delivered current \(I\),

\(P_{\mathrm{source}}=\mathcal E I\)
\(P_{\mathrm{internal}}=I^2r,\quad P_{\mathrm{load}}=IV_{\mathrm{term}}\)
\(\mathcal E I=IV_{\mathrm{term}}+I^2r\)

Internal heating is part of the energy ledger. It explains why terminal voltage and useful output can fall under a heavy load.

Worked Example 7 · Source, Load, and Internal Loss

Close the Power Ledger

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

\(I=\frac{12}{5.5+0.50}=2.0\,\mathrm A\)
\(P_{\mathrm{source}}=(12)(2.0)=24\,\mathrm W\)
\(P_L=I^2R_L=22\,\mathrm W,\quad P_{\mathrm{internal}}=I^2r=2.0\,\mathrm W\)

The balance is \(24=22+2\). The fraction delivered to the load is \(22/24\approx91.7\%\).

11. Efficiency

Efficiency compares desired output to total input:

\(\eta=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}\)

Efficiency is often reported as a percentage and cannot exceed \(100\%\) for an ordinary energy-conversion device. “Loss” means energy transferred into less useful channels, not energy destroyed.

Worked Example 8 · Motor Efficiency

Separate Electrical Input from Mechanical Output

A motor operates at \(120\,\mathrm V\) and \(3.0\,\mathrm A\), producing \(270\,\mathrm W\) of mechanical power.

\(P_{\mathrm{in}}=IV=(3.0)(120)=360\,\mathrm W\)
\(\eta=\frac{270}{360}=0.75=75\%\)
\(P_{\mathrm{other}}=360-270=90\,\mathrm W\)

The remaining \(90\,\mathrm W\) is transferred mainly through heating, sound, and friction in this simplified ledger.

12. Interpreting a Power Rating

A rating states an intended operating condition. A \(60\,\mathrm W\), \(120\,\mathrm V\) resistive rating corresponds to \(I=P/V=0.50\,\mathrm A\) and \(R=V^2/P=240\,\Omega\) at that state.

It does not guarantee \(60\,\mathrm W\) at every applied voltage or temperature.

13. Thermal Limits and Protection

Wire and contact heating scale approximately as \(I^2R\). A fuse or circuit breaker limits current to reduce excessive heating. Replacing protection with a larger rating or bypassing it can allow conductors to exceed their safe thermal design.

14. Common Network Comparisons

  • Adding an identical resistor in series with a fixed-voltage source increases total resistance and decreases total source power.
  • Adding an identical parallel branch to an ideal fixed-voltage source decreases equivalent resistance and increases total source power.
  • Within one series circuit, the larger resistor receives more power because current is common.
  • Within one parallel circuit, the smaller resistor receives more power because voltage is common.

Statements about brightness or heating require a model connecting the observed effect to power and should account for temperature-dependent resistance when relevant.

15. Reliable Power Workflow

  1. Draw the element and label current direction and terminal voltage.
  2. Decide whether it supplies or absorbs energy.
  3. Use \(P=I\Delta V\) with values from the same element.
  4. Use \(I^2R\) only when current and resistance are known; use \(V^2/R\) only when element voltage and resistance are known.
  5. For energy, multiply by time or integrate changing power.
  6. Close the source–load power ledger and check units.

16. Common Traps

  • Confusing watts with joules or kilowatt-hours with kilowatts.
  • Using total source voltage with a single series resistor’s resistance.
  • Using total current with one parallel branch’s voltage incorrectly.
  • Claiming larger resistance always means larger power.
  • Applying \(V^2/R\) without checking which voltage is fixed.
  • Assuming current or charge is consumed by a load.
  • Omitting internal source heating from the energy ledger.
  • Treating a power rating as constant under every condition.
  • Calling inefficiently transferred energy “destroyed.”
Mastery Check

1. A device carries \(0.50\,\mathrm A\) at \(8.0\,\mathrm V\). Find its electrical power.

Show reasoning and answer

\(P=I\Delta V=(0.50)(8.0)=4.0\,\mathrm W\).

2. A \(10\,\Omega\) resistor carries \(2.0\,\mathrm A\). Find power and voltage.

Show reasoning and answer

\(P=I^2R=(2.0)^2(10)=40\,\mathrm W\), and \(\Delta V=IR=20\,\mathrm V\).

3. Two series resistors are \(3\,\Omega\) and \(9\,\Omega\). Which receives more power?

Show reasoning and answer

The same current passes through both, so \(P=I^2R\). The \(9\,\Omega\) resistor receives three times as much power.

4. The same resistors are connected in parallel. Which receives more power?

Show reasoning and answer

They share voltage, so \(P=V^2/R\). The \(3\,\Omega\) resistor receives three times as much power.

5. Convert \(0.80\,\mathrm{kW\,h}\) to joules.

Show reasoning and answer

\(E=(0.80)(3.60\times10^6)=2.88\times10^6\,\mathrm J\).

6. A source supplies \(50\,\mathrm W\), a load provides \(35\,\mathrm W\) useful output, and internal heating is \(5\,\mathrm W\). What other power channel is required?

Show reasoning and answer

Conservation requires \(50=35+5+P_{\mathrm{other}}\), so \(P_{\mathrm{other}}=10\,\mathrm W\).

Investigation: Build a Circuit Power Ledger

Use the open PhET Circuit Construction Kit: DC simulation. Build one battery and resistor loop. Measure current through the resistor and voltage across it, then calculate power with \(IV\), \(I^2R\), and \(V^2/R\). The three results should agree within displayed precision.

Add an identical resistor first in series and then in parallel. Predict each resistor’s power and total source power before measuring. Explain the results by identifying the quantity shared by the resistors. Use only simulation or instructor-approved low-voltage equipment; never probe mains circuits.

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