AP Physics C: E&M · Topic 11.4

Topic 11.4: Electric Power

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

Electric power is the rate at which a circuit element transfers, converts or dissipates energy. The AP Physics C E&M sheet prints one form, P = I times ΔV. The squared forms P = I^2 R and P = ΔV^2/R carry the CED's derived label and are not printed, so you produce them from Ohm's law yourself.

AP Physics: Unit 11 (topics 11.4 Electric Power). Topic 11.4 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, 11.4.A, describe the transfer of energy into, out of, or within an electric circuit, in terms of power, with two essential knowledge statements, 11.4.A.1 and 11.4.A.2. Statement 11.4.A.1 prints P = I times the potential difference as a relevant equation and P = I squared R = the potential difference squared over R as derived equations; only the first is on the equation sheet. Statement 11.4.A.2 says the brightness of a lightbulb increases with power, so power can be used to qualitatively predict brightness. The boundary statement reads in full: AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy. Suggested skills 1.A, 2.B, 2.C and 3.B; the AP Physics 2 version of this topic lists 1.C, 2.A, 2.D and 3.C, so the two courses share no suggested skill here, which happens on no other Unit 11 topic. Sample multiple-choice Question 2 in the course description is aligned to skill 2.B, learning objective 11.4.A and essential knowledge 11.4.A.1; its credited answer is a resistivity of 6.8 times ten to the minus eight ohm metres. The required content is identical to AP Physics 2 Topic 11.4, but the C: E&M sheet reprints the Mechanics table, which prints instantaneous power as dW/dt where the AP Physics 2 sheet prints force times velocity.

What Topic 11.4 requires

Topic 11.4 of AP Physics C: Electricity and Magnetism Unit 11 has one learning objective, two essential-knowledge statements, one relevant equation, one derived equation printed in two forms, and one boundary statement. Only Topic 11.7 is as short, at two statements, and it prints no boundary statement.

11.4.A, describe the transfer of energy into, out of, or within an electric circuit, in terms of power.

11.4.A.1 The rate at which energy is transferred, converted, or dissipated by a circuit element depends on the current in the element and the electric potential difference across it. Relevant equation $P = I \Delta V$ . Derived equation $P = I^2 R = \dfrac{\Delta V^2}{R}$ .
11.4.A.2 The brightness of a lightbulb increases with power, so power can be used to qualitatively predict the brightness of lightbulbs in a circuit.

Boundary statement, quoted whole: "AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy."

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; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Read 11.4.A.1 as a sentence before you read it as a formula. Power depends on two things, the current in the element and the potential difference across it. Getting those two prepositions right is most of the topic: current goes through, potential difference goes across, and mixing them up is the standard way to compute a wrong power in a compound circuit.

One printed equation, two derived ones, and the CED says which

The CED's Required Equations page draws a line that matters here. Not all equations in the course framework appear on the equation sheet; many are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam, and those are labelled Derived Equations.

Statement 11.4.A.1 is unusual in printing both kinds side by side:

Form	CED label	On the C: E&M sheet
$P = I \Delta V$	Relevant equation	yes
$P = I^2 R$	Derived equation	no
$P = \dfrac{\Delta V^2}{R}$	Derived equation	no

We checked the Table of Information appendix directly rather than working from memory: the only power entry in the whole Electricity and Magnetism table is $P = I \Delta V$ . The reprinted Mechanics table adds $P_{\text{avg}} = W/\Delta t = \Delta E/\Delta t$ and $P_{\text{inst}} = dW/dt$ , and neither squared form appears anywhere on either.

So you produce them, in one step each, by substituting Ohm's law from 11.3.B.1:

P = I\,\Delta V = I (IR) = I^2 R, \qquad P = I\,\Delta V = \left(\frac{\Delta V}{R}\right)\Delta V = \frac{\Delta V^2}{R}

Two things follow from the derived label. First, the exam is entitled to hand you a problem that cannot be solved without one of them, and it does: the CED's own sample multiple-choice question aligned to 11.4.A gives a power and a current and asks for a resistivity, which needs $R = P/I^2$ before $R = \rho\ell/A$ can be used. Second, an unprinted equation is one you have to be able to rebuild under pressure, so practise the substitution rather than the memorisation.

A third derived form worth having, because energy questions need it: $P = \mathcal{E} I$ for the power delivered by a source of emf $\mathcal{E}$ . It is the same equation with the source's potential difference in place of an element's.

Choosing between the two derived forms decides the answer

$P = I^2 R$ and $P = \Delta V^2 / R$ are algebraically identical for a single resistor. They give opposite rankings when you compare two resistors, and picking the wrong one is the most reliable way to get a brightness question backwards.

The rule is about which quantity is shared:

In series, the current is shared. Statement 11.5.A.1.i says the current in each element in series must be the same. So $I$ is the constant, $P = I^2 R$ is the useful form, and the larger resistance dissipates the most power.
In parallel, the potential difference is shared. Statement 11.5.A.1.ii says across each path the potential difference is the same. So $\Delta V$ is the constant, $P = \Delta V^2 / R$ is the useful form, and the larger resistance dissipates the least power.

The consequence, which the CED's essential questions gesture at and which shows up constantly: two bulbs swap their brightness ranking when you rewire them from series to parallel. Nothing about the bulbs changed. What changed is which quantity the circuit forces them to share.

A worked instinct that saves time in multiple choice: identify the shared quantity first, then pick the form whose other variable is the one you know. If you cannot see which is shared, you are not ready to use a squared form, and $P = I\Delta V$ with the actual current and the actual potential difference for that element is always safe.

One more check that catches sign and bookkeeping errors in compound circuits. Add up the power dissipated by every resistor and compare it with the power delivered by every source, $\sum \mathcal{E} I$ . They must be equal, because that equality is conservation of energy, which is the same principle 11.6.A.2 says Kirchhoff's loop rule is a consequence of. If your audit does not close, one of your currents is wrong.

What the C sheet gives you that the Physics 2 sheet does not

The essential knowledge in this topic is word for word the same in AP Physics C and AP Physics 2. Both frameworks state 11.4.A.1 and 11.4.A.2 in identical language, both label the squared forms as derived, and both print only $P = I \Delta V$ in their circuit tables. That is worth saying plainly instead of manufacturing a difference.

The suggested skills, though, share nothing. AP Physics C lists 1.A, 2.B, 2.C and 3.B for Topic 11.4; AP Physics 2 lists 1.C, 2.A, 2.D and 3.C. All four differ, which does not happen on any other topic in Unit 11. Read the C set and it points at calculation and comparison from a schematic; read the Physics 2 set and it points at sketching graphs, deriving symbolically and justifying a claim. Same content, opposite emphases.

The difference in what you are given is one line on the reprinted Mechanics table, and it is a genuine one because the C: E&M sheet reprints the C: Mechanics table in full while the Physics 2 sheet has its own.

Sheet	Instantaneous power printed as
AP Physics 2	$P_{\text{inst}} = F_{\parallel} v = F v \cos\theta$
AP Physics C: E&M	$P_{\text{inst}} = \dfrac{dW}{dt}$

The Physics 2 sheet's instantaneous power is a mechanical one, force times velocity. The C sheet's is the general definition, the derivative of energy transfer with respect to time, and inverting it gives the relation a Physics C energy question actually needs:

\Delta E = \int P(t)\, dt

That is the area under a power-versus-time graph. It is the only correct way to get the energy delivered when the power is not constant, and the power is not constant in every situation this unit cares about most: a charging capacitor, a discharging capacitor, an inductor after a switch closes, or any circuit with a time-varying current.

Both sheets print $P_{\text{avg}} = W/\Delta t = \Delta E/\Delta t$ , which is the constant-power version and the one that fails on those problems. A useful test of whether you need the integral: ask whether the current changes during the interval. If it does, the average form is only an approximation and the CED expects the calculus.

This is also where 11.4 quietly powers Topic 11.8. Integrating $I^2R$ over a whole capacitor discharge gives a total dissipated energy that must equal the energy the capacitor started with, $U_C = \tfrac{1}{2} Q \Delta V$ , which is printed on the sheet. That agreement is a complete, self-checking energy audit.

The boundary statement, exception clause included

Topic 11.4 prints one of Unit 11's three boundary statements, and it is the one most often quoted with its second half cut off:

"AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy."

Truncating it at "mechanical and electrical energy" reverses what it says. The boundary limits what you are asked to analyze, not what you are expected to know. Read whole, it says three things:

Quantitative energy accounting on the exam stays inside mechanical and electrical energy. You will not be asked to compute a temperature rise, a specific heat, or a heat flow.
You are still expected to be aware that electrical energy can be dissipated as thermal energy. A question can ask you to say so.
Because dissipation is real but not quantified thermally, the energy that leaves a resistor is accounted for as electrical energy transferred out of the circuit, which is what $P = I^2R$ measures.

Statement 11.3.B.1.iii is the matching content: resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment. Together with 11.3.A.2.ii, that resistivity typically increases with temperature, it is the chain behind the unit's essential question about why warming bulbs take several minutes to shine brightly. You can be asked to describe that chain; you will not be asked to put a number on the temperature.

The CED's own Topic 11.4 sample activity sits exactly on this boundary. It asks students to justify with evidence why a small 1 ohm resistor can only handle a small amount of power, such as a quarter of a watt, and why a large 1 kilohm resistor can handle a large power, such as 30 watts, and why computer processors have heat sinks attached to them. That is a qualitative thermal argument built on a quantitative electrical one, which is the level the boundary sets.

Brightness is a power question (11.4.A.2)

Statement 11.4.A.2 is short and it is the licence for a whole question type: the brightness of a lightbulb increases with power, so power can be used to qualitatively predict the brightness of lightbulbs in a circuit.

Read the qualifier. Brightness tracks power monotonically, so more power means brighter, but the CED does not give you a formula converting watts to lumens and never asks for one. Every brightness question is therefore a ranking question, and the answer is always "brighter", "dimmer", "unchanged" or "goes out", justified by a power comparison.

The standard moves:

Bulb goes out means zero power, which needs either zero current, from an open branch, or zero potential difference, from a short across it. Statement 11.2.A.2.iii defines the short case.
Bulb gets brighter means more power. In a series chain that means more current; in a parallel branch it means more potential difference across that branch.
Adding a parallel branch lowers the equivalent resistance by 11.5.A.2.iii, which raises the total current, which usually brightens whatever is in series with the whole combination and can dim what is in the branches.

The CED's Instructional Approaches case study runs this exact reasoning on a circuit of two identical bulbs and a switch connected to a battery, with Bulb 2 in parallel with the switch. Its skill 2.A question asks for the power delivered to Bulb 1 with the switch open and the credited expression is $\mathcal{E}^2/(4R)$ ; its skill 2.B question puts $R = 2.0\ \Omega$ and $\mathcal{E} = 9.0$ V into that expression for most nearly 10 W; and its skill 2.C question asks what closing the switch does, with the credited answer that Bulb 1 gets brighter and Bulb 2 goes out. One stimulus, three skills, all resting on 11.4.A.2. Topic 11.2 works that circuit in full.

One caution the CED is careful about and students often are not: brightness compares bulbs, and 11.4.A.2 is stated for lightbulbs specifically. The exam's reference information declares lightbulbs ohmic unless otherwise stated, so unless a question says the filament heats, treat each bulb's resistance as fixed and let the circuit decide the power.

Why the bill is in kilowatt-hours

One of Unit 11's four essential questions is: why does the electric company charge by kilowatt-hour instead of electrons used? It is a good question because it separates the two quantities students most often merge.

Power is a rate; energy is what gets paid for. A kilowatt-hour is a unit of energy, not power: one kilowatt sustained for one hour, which is $(1000\ \mathrm{W})(3600\ \mathrm{s}) = 3.6 \times 10^{6}$ J. Charging by the kilowatt-hour is charging by the joule in a convenient size.

Electrons are not consumed. Kirchhoff's junction rule is a consequence of conservation of electric charge, per 11.7.A.1, so every electron that enters a house leaves it. Charge is conserved and energy is transferred, so the transferred energy is the only thing there is to sell.

For the exam, that translates into a habit. When a question asks for energy rather than power, you need a time:

\Delta E = \int P(t)\, dt \quad \text{in general}, \qquad \Delta E = P \Delta t \quad \text{when } P \text{ is constant}

Both relations are on the C: E&M sheet via the reprinted Mechanics table, as $P_{\text{inst}} = dW/dt$ and $P_{\text{avg}} = \Delta E / \Delta t$ . Choosing between them is choosing whether the current is steady, and in a resistor fed by an ideal battery through a switch that has been closed a long time, it is.

A units check that catches most errors here: watts are joules per second, so watts times seconds are joules, and amps times volts are also joules per second because an amp is a coulomb per second and a volt is a joule per coulomb. If your answer to an energy question comes out in watts, you have skipped the time.

How Topic 11.4 is tested, and who each page is for

Topic 11.4 has one of the clearest assessment fingerprints in the unit, because the CED's own sample set includes a question written against it.

Sample multiple-choice Question 2 aligns to skill 2.B, learning objective 11.4.A and essential knowledge 11.4.A.1. It gives a cylindrical wire segment of length 0.25 m and diameter $3.0 \times 10^{-3}$ m that dissipates energy at $6.0 \times 10^{-4}$ W when the current in it is 0.50 A, and asks for the resistivity. The credited answer is $6.8 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ . Notice what the question requires: $P = I^2 R$ , which is not on the sheet, followed by $R = \rho\ell/A$ , which is. A student who only knows the printed form is stuck at the first step.

That is the pattern. Topic 11.4 questions are short, they are usually skill 2.B or 2.C, and they nearly always pair a power relation with something from Topic 11.3 or Topic 11.5.

Skill 1.A is also listed here, which is easy to overlook on a topic with no obvious diagram. It shows up as being asked to sketch or tabulate power against a varying quantity, or to mark on a schematic where energy is being dissipated.

The [AP Physics 2 Topic 11.4 page](/ap-physics-2/unit-11-electric-circuits/11-4-electric-power) is for AP Physics 2 students; this page is for AP Physics C students. The physics content of the two topics is the same, and that page covers it for the algebra-based exam. What differs on the C side is the sheet, where $P_{\text{inst}} = dW/dt$ replaces the mechanical instantaneous-power line and makes $\Delta E = \int P\, dt$ available, and the free-response style, where Science Practice 2 carries 40 to 45% of the section and symbolic answers are the norm.

The work and power calculator handles the arithmetic when you want to check a number quickly, and the work against power comparison covers the rate-against-total distinction from the mechanics side.

The CED's own sample question: resistivity from a dissipated power

A cylindrical wire segment of length $0.25$ m and diameter $3.0 \times 10^{-3}$ m dissipates energy at a rate of $6.0 \times 10^{-4}$ W when the current in the wire segment is $0.50$ A. Find the resistivity of the wire. This is sample multiple-choice Question 2 from the AP Physics C: Electricity and Magnetism course description, aligned to skill 2.B and essential knowledge 11.4.A.1.

Identify what connects the given quantities. You are handed a power and a current, and you want a resistivity. Power and current alone give resistance, and resistance plus geometry gives resistivity, so the route is $P \to R \to \rho$ .
First step, the derived form of 11.4.A.1: $P = I^2 R$ , so $R = \dfrac{P}{I^2} = \dfrac{6.0 \times 10^{-4}\ \mathrm{W}}{(0.50\ \mathrm{A})^2} = \dfrac{6.0 \times 10^{-4}}{0.25} = 2.4 \times 10^{-3}\ \Omega$ . This equation is not printed on the equation sheet, which is exactly why the CED labels it a derived equation.
Second step, the cross-sectional area. The diameter is $3.0 \times 10^{-3}$ m, so the radius is $1.5 \times 10^{-3}$ m and $A = \pi r^2 = \pi (2.25 \times 10^{-6}) = 7.069 \times 10^{-6}\ \mathrm{m^2}$ . Halving the diameter before squaring is the step most often skipped.
Third step, rearrange 11.3.A.2: $R = \dfrac{\rho \ell}{A}$ gives $\rho = \dfrac{RA}{\ell} = \dfrac{(2.4 \times 10^{-3})(7.069 \times 10^{-6})}{0.25}$ .
Numerator: $(2.4 \times 10^{-3})(7.069 \times 10^{-6}) = 1.696 \times 10^{-8}$ . Divide by $0.25$ : $\rho = 6.79 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ , which rounds to $6.8 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ .
Sanity check the magnitude. Copper's resistivity is about $1.7 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ , so a few times that is a plausible metal. The distractors in the original question are $9.0 \times 10^{-5}$ , $7.5 \times 10^{-3}$ and $840\ \Omega \cdot \mathrm{m}$ , none of which is a conductor, so an order-of-magnitude sense of metallic resistivity settles the question even faster than the algebra.
Units check: $\Omega \times \mathrm{m^2} / \mathrm{m} = \Omega \cdot \mathrm{m}$ .

$\rho = 6.8 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ . The question cannot be started without $P = I^2 R$ , which the CED labels a derived equation and does not print on the equation sheet.

Energy from a power that changes with time

A capacitor of $C = 1000\ \mu\mathrm{F}$ , initially charged to $\Delta V_0 = 20$ V, is discharged through a resistor of $R = 2.0\ \mathrm{k}\Omega$ . The current decays as $I(t) = I_0 e^{-t/\tau}$ with $\tau = RC$ . Find (a) the initial current and the initial power dissipated, (b) the total energy dissipated in the resistor over the whole discharge, and (c) the time at which half that energy has been dissipated. Then check (b) against the energy the capacitor started with.

(a) At $t = 0$ the full 20 V sits across the resistor, so $I_0 = \dfrac{\Delta V_0}{R} = \dfrac{20}{2000} = 1.0 \times 10^{-2}$ A. The derived form gives $P_0 = I_0^2 R = (1.0 \times 10^{-2})^2 (2000) = 0.20$ W.
The time constant, printed on the sheet as $\tau = R_{\text{eq}} C_{\text{eq}}$ , is $\tau = (2000)(1.0 \times 10^{-3}) = 2.0$ s.
(b) The power is not constant, so $P_{\text{avg}} = \Delta E / \Delta t$ will not do it. Use the other printed form, $P_{\text{inst}} = dW/dt$ , inverted: $\Delta E = \displaystyle\int_0^{\infty} P(t)\, dt$ with $P(t) = I^2 R = I_0^2 R\, e^{-2t/\tau}$ .
The sheet's calculus table prints $\int e^{ax} dx = \dfrac{1}{a}e^{ax}$ , so $\displaystyle\int_0^{\infty} e^{-2t/\tau} dt = \frac{\tau}{2}$ , giving $\Delta E = I_0^2 R \dfrac{\tau}{2} = P_0 \dfrac{\tau}{2} = (0.20)(1.0) = 0.20$ J.
Check it against the stored energy, using the printed $U_C = \tfrac{1}{2} Q \Delta V$ with $Q = C \Delta V_0 = (1.0 \times 10^{-3})(20) = 2.0 \times 10^{-2}$ C. Then $U_C = \tfrac{1}{2}(2.0 \times 10^{-2})(20) = 0.20$ J. The two agree exactly, which is the energy audit closing: every joule the capacitor held is dissipated in the resistor.
(c) Energy dissipated up to time $t$ is $\displaystyle\int_0^{t} P_0 e^{-2t'/\tau} dt' = P_0 \frac{\tau}{2}\left(1 - e^{-2t/\tau}\right) = \Delta E \left(1 - e^{-2t/\tau}\right)$ . Setting that equal to half of $\Delta E$ gives $e^{-2t/\tau} = 0.5$ , so $\dfrac{2t}{\tau} = \ln 2$ and $t = \dfrac{\tau \ln 2}{2} = \dfrac{(2.0)(0.693)}{2} = 0.69$ s.
Read the result. Half the energy is gone in about a third of a time constant, because power goes as the square of the current and so decays with a time constant of $\tau/2$ , not $\tau$ . That is a common trap: the energy half-life is not the charge half-life.

(a) $I_0 = 10$ mA and $P_0 = 0.20$ W. (b) $\Delta E = 0.20$ J in total, which equals the capacitor's initial stored energy $U_C = \tfrac{1}{2}Q\Delta V = 0.20$ J exactly. (c) $t = \tau \ln 2 / 2 = 0.69$ s, because the power decays with time constant $\tau/2$ rather than $\tau$ .

Three identical bulbs, and one gets unscrewed

Three identical ohmic bulbs, each of resistance $R = 6.0\ \Omega$ , are connected to an ideal battery of emf $\mathcal{E} = 12$ V. Bulb 1 is in series with a parallel combination of Bulbs 2 and 3. Find, symbolically and numerically, the power in each bulb. Then Bulb 3 is unscrewed. Say what happens to the brightness of Bulbs 1 and 2, and to the total power drawn from the battery.

Reduce the network. The parallel pair has $\dfrac{1}{R_p} = \dfrac{1}{R} + \dfrac{1}{R}$ , so $R_p = R/2$ , and in series with Bulb 1 that gives $R_{\text{eq}} = R + \dfrac{R}{2} = \dfrac{3R}{2} = 9.0\ \Omega$ .
Battery current: $I = \dfrac{\mathcal{E}}{R_{\text{eq}}} = \dfrac{2\mathcal{E}}{3R} = \dfrac{12}{9.0} = 1.33$ A. Bulb 1 carries all of it; the junction rule splits it evenly between the identical parallel bulbs, so each carries $I/2 = 0.667$ A.
Powers, using $P = I^2 R$ because the currents are what we know: $P_1 = \left(\dfrac{2\mathcal{E}}{3R}\right)^2 R = \dfrac{4\mathcal{E}^2}{9R} = \dfrac{4(144)}{9(6.0)} = 10.7$ W, and $P_2 = P_3 = \left(\dfrac{\mathcal{E}}{3R}\right)^2 R = \dfrac{\mathcal{E}^2}{9R} = \dfrac{144}{54} = 2.67$ W.
Audit the energy: $10.7 + 2.67 + 2.67 = 16.0$ W, and the battery delivers $\mathcal{E} I = (12)(1.33) = 16.0$ W. It closes, so the currents are right. Bulb 1 is four times as bright as each of the others, because it carries twice the current and power goes as $I^2$ at fixed $R$ .
Now unscrew Bulb 3. Its branch is an open circuit, so it carries no current, and Bulbs 1 and 2 are left in series: $R_{\text{eq}}' = 2R = 12\ \Omega$ and $I' = \dfrac{\mathcal{E}}{2R} = 1.0$ A.
New powers: $P_1' = P_2' = I'^2 R = (1.0)^2(6.0) = 6.0$ W each, or $\dfrac{\mathcal{E}^2}{4R}$ symbolically. Total is 12.0 W, which matches $\mathcal{E}^2/R_{\text{eq}}' = 144/12 = 12.0$ W.
Compare. Bulb 1 falls from 10.7 W to 6.0 W, so it gets dimmer. Bulb 2 rises from 2.67 W to 6.0 W, so it gets brighter. That reversal is the point: removing a bulb raised the total resistance, which lowered the current through Bulb 1, while Bulb 2 now receives the whole of that smaller current instead of half of a larger one.

Initially $P_1 = 4\mathcal{E}^2/(9R) = 10.7$ W and $P_2 = P_3 = \mathcal{E}^2/(9R) = 2.67$ W, totalling 16.0 W. After Bulb 3 is unscrewed, $P_1 = P_2 = \mathcal{E}^2/(4R) = 6.0$ W each, totalling 12.0 W. Bulb 1 gets dimmer, Bulb 2 gets brighter, and the battery delivers less power overall.

Frequently asked questions

Is P = I squared R on the AP Physics C E&M equation sheet?

No. The only power equation in the Electricity and Magnetism table of the AP Physics C: Electricity and Magnetism equation sheet is P equals I times the potential difference. The course description prints P = I squared R and P equals the potential difference squared over R at essential knowledge 11.4.A.1, but both carry the Derived equation label, which the framework's Required Equations page defines as a final result of a derivation expected of students on the exam rather than something provided on the sheet. You produce them in one step by substituting Ohm's law into the printed form. The reprinted Mechanics table adds average power as energy over time and instantaneous power as dW/dt, and neither squared form appears anywhere.

Which power formula should you use, I squared R or delta V squared over R?

Use the one whose variable is shared by the elements you are comparing. Elements in series all carry the same current, because essential knowledge 11.5.A.1.i requires it, so P = I squared R is the useful form and the largest resistance dissipates the most power. Elements in parallel all have the same potential difference across them, by essential knowledge 11.5.A.1.ii, so P equals the potential difference squared over R is the useful form and the largest resistance dissipates the least. That is why two bulbs swap their brightness ranking when you rewire them from series to parallel, even though nothing about the bulbs has changed.

What is the boundary statement for AP Physics C Topic 11.4?

It reads in full: AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy. Both halves matter. The first limits what you are asked to analyze quantitatively, so no temperature rises or specific heats. The second says you are still expected to know that dissipation into thermal energy happens, and a question may ask you to say so. Essential knowledge 11.3.B.1.iii carries the matching content, that resistors can convert electrical energy to thermal energy and change the temperature of the resistor and its environment.

How do you find the energy dissipated when the current changes with time?

Integrate the power. The AP Physics C: Electricity and Magnetism equation sheet prints instantaneous power as dW/dt on its reprinted Mechanics table, and inverting that means the energy transferred is the time integral of the power, which is the area under a power-versus-time graph. The average-power form, energy over time interval, is only correct when the power is constant, so it fails for a charging or discharging capacitor, an inductor after a switch closes, or any circuit with a time-varying current. A useful check on a capacitor discharge is that the total energy dissipated must equal the energy the capacitor started with, one half Q times the potential difference, which is also printed on the sheet.

Why does bulb brightness depend on power rather than current or voltage?

Because brightness measures the rate at which the filament emits energy, and power is exactly that rate. Essential knowledge 11.4.A.2 in the AP Physics C course description states that the brightness of a lightbulb increases with power, so power can be used to qualitatively predict the brightness of lightbulbs in a circuit. The word qualitatively is deliberate: the framework never supplies a relation between watts and light output, so every brightness question is a ranking rather than a calculation. A bulb goes out when its power is zero, which needs either zero current from an open branch or zero potential difference from a short across it.

Is AP Physics C Topic 11.4 different from AP Physics 2 Topic 11.4?

The required content is the same. Both frameworks state essential knowledge 11.4.A.1 and 11.4.A.2 in identical wording, both label the two squared power forms as derived equations, and both print only P equals I times the potential difference in their circuit tables. Two differences are real. The AP Physics C: Electricity and Magnetism sheet reprints the Mechanics table, which gives instantaneous power as dW/dt, where the AP Physics 2 sheet prints instantaneous power as force times velocity instead, so only the C course has energy as an integral of power on its sheet. And the C exam weights symbolic derivation more heavily, with Science Practice 2 carrying 40 to 45 percent of the free-response section.

What is a kilowatt-hour and why is electricity billed in them?

A kilowatt-hour is a unit of energy, not of power: one kilowatt sustained for one hour, which is 3.6 million joules. Electricity is billed in energy because energy is what the utility actually delivers. Charge is not consumed by a circuit, since Kirchhoff's junction rule is a consequence of conservation of electric charge, so every electron entering a building also leaves it. What changes is the energy those charges carry. This is one of the four essential questions the AP Physics C course description lists for Unit 11, and the exam version of it is that any question asking for energy rather than power needs a time interval, either multiplying by it or integrating over it.