AP Physics 2 · Topic 11.4

Topic 11.4: Electric Power

Unit 11: Electric Circuits15-18% of the multiple-choice section

Electric power is the rate at which a circuit element transfers, converts, or dissipates energy, measured in watts. The AP Physics 2 sheet prints one form, P = IΔV. The CED labels P = I²R and P = (ΔV)²/R as derived, because each follows from substituting Ohm's law. Bulb brightness tracks power.

AP Physics: Unit 11 (topics 11.4 Electric Power). AP Physics 2 Unit 11, Topic 11.4. One learning objective, 11.4.A: describe the transfer of energy into, out of, or within an electric circuit, in terms of power. Essential knowledge 11.4.A.1 states that 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, giving P = I delta V as the relevant equation and P = I squared R = (delta V) squared over R as derived equations. Essential knowledge 11.4.A.2 states that the brightness of a bulb increases with power, so power can be used to qualitatively predict the brightness of bulbs in a circuit. The topic carries no boundary statement and no sub-statements. Suggested skills are 1.C, 2.A, 2.D and 3.C. Only P = I delta V is printed on the AP Physics 2 equation sheet, in the Electricity group. Unit 11 is weighted at 15 to 18 percent of the exam with a suggested 12 to 20 class periods.

What Topic 11.4 requires

Topic 11.4 carries one learning objective and two essential knowledge statements, with no sub-statements and 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. The CED gives $P = I \Delta V$ as the relevant equation, and then prints $P = I^2 R = \frac{(\Delta V)^2}{R}$ separately, labelled as derived equations.
11.4.A.2 The brightness of a bulb increases with power, so power can be used to qualitatively predict the brightness of bulbs in a circuit.

Read 11.4.A.1 again for its verbs. Transferred, converted, or dissipated: the CED is not restricting this to heat in a resistor. A battery transfers energy into the circuit, a resistor converts electrical energy to thermal energy, a bulb converts part of it to light, and $P = I \Delta V$ measures the rate for every one of them. That is why the objective says into, out of, or within.

The CED lists four suggested skills here: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Skill 2.A is not decoration here: the two derived power equations are the derivation the CED expects you to be able to produce on demand. Unit 11 is weighted at 15 to 18 percent of the exam, with a suggested 12 to 20 class periods.

One printed equation, two derived ones, and the CED marks which is which

P = I \Delta V

That is all the AP Physics 2 equation sheet prints for electric power. The Electricity group holds 20 equations, $P = I \Delta V$ is one of them, and neither $P = I^2 R$ nor $P = (\Delta V)^2 / R$ appears anywhere on the sheet. Counting the group is the only honest way to make a claim of that kind.

The CED is unusually explicit about the status of each form. Under essential knowledge 11.4.A.1 it prints $P = I \Delta V$ under the label relevant equation, then prints $P = I^2 R = \frac{(\Delta V)^2}{R}$ under the label derived equations. The sorting has been done for you: one is given, two are yours to produce.

Producing them takes one substitution each, using Ohm's law $I = \Delta V / R$ from Topic 11.3.

Substituting $\Delta V = IR$ into $P = I \Delta V$ :

P = I(IR) = I^2 R

Substituting $I = \Delta V / R$ into $P = I \Delta V$ :

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

Two consequences follow from these being derived rather than printed. First, they inherit Ohm's law's conditions. $P = I \Delta V$ holds for any circuit element, including a battery and a filament nowhere near ohmic, because it is a statement about energy and charge and nothing else. The two squared forms assume the element has a single well defined $R$ , which is precisely the ohmic assumption Topic 11.3 unpacks. Second, writing the substitution out is worth marks: suggested skill 2.A asks you to derive a symbolic expression by selecting and following a logical mathematical pathway, and these two lines are the smallest complete example of it in the course.

One notational hazard on this sheet: the letter $P$ is also pressure, appearing as $P = F_{\perp}/A$ in the Thermal Physics group and inside the ideal gas law. Context separates them, but check the units of whatever you substitute into. Power is in watts, pressure in pascals.

Where P equals I times delta V comes from

You can get the printed power equation out of two other printed equations in three lines, and doing so is the clearest way to see what it measures.

Start from the definition of average power, which the Mechanics and Fluids group of the AP Physics 2 sheet prints as

P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}

Power is an energy per unit time and it does not care what kind of energy. This is the same definition Topic 3.5 of AP Physics 1 is built on, reprinted unchanged on the Physics 2 sheet.

Now supply the energy. The Electricity group prints $\Delta U_E = q \Delta V$ : move a charge $q$ through a potential difference $\Delta V$ and its electric potential energy changes by $q \Delta V$ . That is Topic 10.4.

Divide by the elapsed time:

P = \frac{\Delta E}{\Delta t} = \frac{q \Delta V}{\Delta t} = \left(\frac{q}{\Delta t}\right)\Delta V

The bracket is the current, because the Electricity group also prints $I = \Delta q / \Delta t$ , which is Topic 11.1. So

P = I \Delta V

That derivation is why the Unit 11 opener describes the unit as revisiting the behavior of charged particles to deepen students' understanding of the law of conservation of energy and its application to electric circuits. Electric power is not a new principle. It is the energy accounting of the earlier units, applied to charge moving through a potential difference.

It also explains the two factors. Current counts how much charge passes per second, potential difference counts how much energy each coulomb gives up, and the product is joules per second: $1 \, \mathrm{A} \times 1 \, \mathrm{V} = 1 \, \mathrm{C/s} \times 1 \, \mathrm{J/C} = 1 \, \mathrm{J/s} = 1 \, \mathrm{W}$ .

Choosing between the two derived forms

All three expressions give the same number for the same element in the same circuit. They differ in what you must know, and more usefully in what they make obvious.

Form	Reach for it when you know	What it makes obvious
$P = I \Delta V$	the current and the potential difference	true for any element, ohmic or not
$P = I^2 R$	the current and the resistance	at fixed current, larger $R$ dissipates more
$P = \dfrac{(\Delta V)^2}{R}$	the potential difference and the resistance	at fixed potential difference, larger $R$ dissipates less

The last two rows point in opposite directions, and that is the point. "A bigger resistor dissipates more power" is not a fact. It is an answer that depends entirely on what the circuit is holding fixed.

In series, every element carries the same current, so current is the fixed quantity and $P = I^2 R$ is the form to reach for. The largest resistance dissipates the most power.
In parallel, every branch has the same potential difference across it, so potential difference is the fixed quantity and $P = (\Delta V)^2 / R$ is the form to reach for. Now the smallest resistance dissipates the most power.

Topic 11.5 is where those two connection types get defined and reduced, and the series and parallel guide walks the reduction procedure. What Topic 11.4 adds is the habit of asking, before you pick a form, which quantity the circuit is holding constant for the element you care about.

When you are unsure, there is a route that never misleads: find the current in the element and the potential difference across it, then use $P = I \Delta V$ . It needs no ohmic assumption and it never asks which quantity is fixed. The derived forms are shortcuts, and shortcuts are where the scaling errors live.

Power is a rate, so energy needs a time

Power is in watts and energy in joules, and questions move between them often enough that the conversion deserves its own line. Rearranging the definition of average power gives

\Delta E = P \Delta t

A resistor dissipating 3.0 W for 40 s converts 120 J of electrical energy to thermal energy. This is essential knowledge 11.3.B.1.iii from the previous topic getting a number attached at last: resistors can convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment. Topic 11.4 supplies the rate. How much the temperature actually rises is a specific-heat question, handled by $Q = mc\Delta T$ in the Thermal Physics group of the sheet and by Topic 9.5.

The kilowatt hour on an electricity bill is an energy, not a power, and it converts with the same equation: $1 \, \mathrm{kWh} = (1000 \, \mathrm{W})(3600 \, \mathrm{s}) = 3.6 \times 10^6 \, \mathrm{J}$ . Nothing on the AP sheet uses it, but a context-rich question may, and the conversion is one multiplication.

Two habits keep the units honest.

Watts are already per second. "The bulb uses 60 W per second" is like saying "60 miles per hour per hour". A bulb dissipates 60 W, and it uses 60 J each second.
Convert the time before multiplying. Five minutes is 300 s, so a 300 W element run for that long delivers $(300 \, \mathrm{W})(300 \, \mathrm{s}) = 9.0 \times 10^4 \, \mathrm{J}$ .

The energy comes from the battery. The rate at which a battery delivers it is $P = I \Delta V$ using the potential difference across its terminals, and in a single-loop circuit that total equals the sum of the powers dissipated by the elements. Real batteries also dissipate some of it internally, which is Topic 11.5 territory. The exam conventions treat batteries as ideal unless a question says otherwise.

Bulb brightness is a power question (11.4.A.2)

Essential knowledge 11.4.A.2 is a single sentence: the brightness of a bulb increases with power, so power can be used to qualitatively predict the brightness of bulbs in a circuit. That is why so many circuit questions are phrased in terms of bulbs going dim.

What it licenses is a ranking. Rank the powers and you have ranked the brightnesses, with no need to turn watts into anything optical. What it does not license is a number: the CED says qualitatively, so brightness has no units in this course, and three times the power does not mean three times as bright to the eye.

Unit 11's essential questions include how you could make a 120 watt bulb brighter than a 40 watt bulb, which sounds like a trick until you notice that a rating describes one operating point rather than the device. A bulb labelled 120 W is labelled at some rated potential difference, and by $P = (\Delta V)^2 / R$ a higher rating at the same rated voltage means a lower resistance. Wire both bulbs in series and the current through them is equal, so $P = I^2 R$ takes over and the 40 W bulb, with the larger resistance, is the brighter one. Wire them in parallel across the rated voltage and each runs at its rating, so the 120 W bulb is brighter. The third worked example below runs the numbers.

The other classic is a bulb dimming when something changes elsewhere in the circuit. The reasoning is always the same chain, and it is worth rehearsing as a chain rather than as a result:

What did the change do to the equivalent resistance of the circuit?
What did that do to the current from the battery?
What did that do to the current in, or the potential difference across, this particular bulb?
What does that do to $P$ for this bulb, and so to its brightness?

Suggested skill 3.C, justify or support a claim using evidence from physical principles or laws, is asking for that chain written down. An answer that jumps from step 1 to step 4 has skipped the part being scored.

One caveat, which the exam conventions handle for you: a real filament is not ohmic, so its resistance changes as its brightness changes. The convention list assumes lightbulbs are ohmic unless otherwise stated, and that is what keeps the ranking argument clean. Topic 11.3 covers what the assumption is worth and when a question takes it away.

Sketching the graphs (skill 1.C)

Suggested skill 1.C asks for qualitative sketches of graphs that represent features of a model or the behavior of a physical system. Topic 11.4 is a natural place to be asked for one, because power depends on three quantities and the shape of the graph changes with which of them you hold still.

Hold fixed	Vary	Shape of the power graph
$R$	$I$	$P = I^2 R$ , an upward parabola through the origin
$R$	$\Delta V$	$P = (\Delta V)^2 / R$ , an upward parabola through the origin
$I$	$R$	$P = I^2 R$ , a straight line through the origin of slope $I^2$
$\Delta V$	$R$	$P = (\Delta V)^2 / R$ , a falling curve that approaches zero without reaching it
$\Delta V$	$I$	$P = I \Delta V$ , a straight line through the origin of slope $\Delta V$

Every one passes through the origin except the fourth, which is undefined at $R = 0$ and falls away as $R$ grows. Sketching well means getting three things right: the intercept, the curvature, and whether the curve rises or falls. Axis labels and a correct shape earn the point; a plotted table of values is not what 1.C is asking for.

The two parabolas explain the sensitivity that shows up in factor-of-change questions. Because both squared forms carry an exponent of 2, a 10 percent rise in current at fixed resistance raises the power by 21 percent rather than 10, since $1.10^2 = 1.21$ . Halving the potential difference across a fixed resistor cuts the power to a quarter. Skill 2.D lives on these ratios, and the second worked example below drills them.

How Topic 11.4 is tested, and where it leads

Four suggested skills, four recognizable question shapes.

Derive a power expression (2.A). Start from $P = I \Delta V$ , substitute Ohm's law, and land on whichever form the question's variables call for. Show the substitution rather than quoting the result.
Predict a factor of change (2.D). The potential difference triples, or the resistance is halved, and you report the multiplier on the power. State which quantity you are holding fixed, because the answer turns on it.
Rank bulb brightnesses and justify the ranking (3.C with 11.4.A.2). Equivalent resistance, then current, then the individual element, then the power.
Sketch a power graph (1.C) for a described change, with the right shape and the right intercept.

A fifth pattern carries no skill code but appears constantly: energy over an interval, from $\Delta E = P \Delta t$ , sometimes followed by a temperature change through $Q = mc\Delta T$ .

The topic sits between others that supply its inputs and consume its outputs. Topic 11.3 provides the $R$ and the ohmic assumption the derived forms rest on, Topic 11.1 provides the current, Topic 11.5 supplies the equivalent resistances most brightness questions turn on, and Topic 11.6 states the energy conservation that makes the individual potential differences add up around a loop. Further back, the definition of power is the one from Topic 3.5 in AP Physics 1, unchanged. The Unit 11 overview puts all eight topics in order, and the AP Physics 2 formula sheet shows which of the equations on this page you will be handed on exam day.

Which pencil lead heats the water faster?

This finishes an argument started in Topic 11.3. Two graphite pencil leads, each 60.0 mm long, are used as heating elements in water. The 0.50 mm diameter lead has a resistance of $5.0 \, \Omega$ and the 0.70 mm lead has a resistance of $2.6 \, \Omega$ . Each is connected directly across its own ideal 1.5 V cell. Which lead delivers thermal energy to the water faster, and by what factor?

Each lead sits directly across a cell, so each has the full 1.5 V across it. The potential difference is the quantity being held fixed, so the useful derived form is $P = (\Delta V)^2 / R$ .
Thin lead: $P_{0.50} = \dfrac{(1.5 \, \mathrm{V})^2}{5.03 \, \Omega} = \dfrac{2.25}{5.03} = 0.45 \, \mathrm{W}$ , carrying the unrounded resistance from Topic 11.3.
Thick lead: $P_{0.70} = \dfrac{(1.5 \, \mathrm{V})^2}{2.57 \, \Omega} = \dfrac{2.25}{2.57} = 0.88 \, \mathrm{W}$ .
Check both against the printed equation rather than the derived one. The currents are $I_{0.50} = 1.5/5.03 = 0.30 \, \mathrm{A}$ and $I_{0.70} = 1.5/2.57 = 0.58 \, \mathrm{A}$ , so $P = I \Delta V$ gives $(0.298)(1.5) = 0.45 \, \mathrm{W}$ and $(0.584)(1.5) = 0.88 \, \mathrm{W}$ . Both match.
The ratio is cleaner than either number. At fixed $\Delta V$ , $P \propto 1/R$ , and Topic 11.3 put the two resistances in the ratio 1.96, so $\dfrac{P_{0.70}}{P_{0.50}} = 1.96$ .
You can get that ratio straight from the geometry without ever computing a resistance. At fixed $\Delta V$ , substituting $R = \rho \ell / A$ gives $P = \dfrac{(\Delta V)^2 A}{\rho \ell}$ , so for the same material and length $P \propto A \propto d^2$ , and $(0.70/0.50)^2 = 1.96$ .
Turn the power ratio into a time. Delivering the same energy to the water takes $\Delta t = \Delta E / P$ , so the thick lead needs $1/1.96 = 0.51$ times as long.

The 0.70 mm lead wins, dissipating 0.88 W against 0.45 W, a factor of 1.96, and heating a given quantity of water in about 0.51 of the time. Student A's conclusion was right and Student B's claim about resistance was also right: at a fixed potential difference, the lower resistance is the one that dissipates more power. Notice how much the wiring is doing. Put the two leads in series instead, and the current would be shared, $P = I^2 R$ would take over, and the thin lead would be the hotter one.

Factors of change in a resistor's power

An ohmic resistor dissipates 12.0 W when the potential difference across it is 6.00 V. (a) Find its resistance and the current in it. (b) The potential difference is raised to 9.00 V. Find the new power. (c) Instead, at the original 6.00 V, the resistor is replaced by one of three times the resistance. Find the new power. (d) Instead, the original resistor is placed in a circuit that doubles the current in it. Find the new power.

(a) You have a power and a potential difference, so pick the form built from those two. $P = (\Delta V)^2 / R$ rearranges to $R = \dfrac{(\Delta V)^2}{P} = \dfrac{(6.00 \, \mathrm{V})^2}{12.0 \, \mathrm{W}} = \dfrac{36.0}{12.0} = 3.00 \, \Omega$ .
The current comes from the printed equation: $I = \dfrac{P}{\Delta V} = \dfrac{12.0 \, \mathrm{W}}{6.00 \, \mathrm{V}} = 2.00 \, \mathrm{A}$ . Cross-check with Ohm's law: $6.00 / 3.00 = 2.00 \, \mathrm{A}$ .
(b) The resistance is unchanged and the potential difference varies, so $P \propto (\Delta V)^2$ . The factor is $\left(\dfrac{9.00}{6.00}\right)^2 = 1.50^2 = 2.25$ , giving $P = (12.0)(2.25) = 27.0 \, \mathrm{W}$ .
Confirm by direct substitution: $P = \dfrac{(9.00 \, \mathrm{V})^2}{3.00 \, \Omega} = \dfrac{81.0}{3.00} = 27.0 \, \mathrm{W}$ .
(c) Now the potential difference is fixed and the resistance changes, so $P \propto 1/R$ and tripling $R$ cuts the power to a third: $P = 12.0 / 3 = 4.00 \, \mathrm{W}$ . Directly, $P = \dfrac{(6.00 \, \mathrm{V})^2}{9.00 \, \Omega} = 4.00 \, \mathrm{W}$ .
(d) This time the current is what is specified, so switch to $P = I^2 R$ . The resistance is unchanged, so $P \propto I^2$ and doubling the current quadruples the power: $P = (2)^2 (12.0) = 48.0 \, \mathrm{W}$ . Directly, $P = (4.00 \, \mathrm{A})^2 (3.00 \, \Omega) = 48.0 \, \mathrm{W}$ .
Compare (c) and (d). Tripling the resistance at fixed potential difference cut the power, while doubling the current at fixed resistance raised it. Neither result is about resistance or current on its own. Both are about which quantity the circuit held still.

(a) $R = 3.00 \, \Omega$ and $I = 2.00 \, \mathrm{A}$ . (b) 27.0 W. (c) 4.00 W. (d) 48.0 W. Every part was faster as a ratio than as a fresh substitution, which is what suggested skill 2.D trains. The one thing to state each time is which quantity is held fixed, because parts (c) and (d) push the same equation in opposite directions.

A 120 W bulb and a 40 W bulb, in series and in parallel

Two bulbs are rated 120 W and 40 W, each at 120 V. Treat both as ohmic, as the AP Physics 2 exam conventions direct, and take the battery and wires as ideal. (a) Find each bulb's resistance at its rating. (b) Connect the two in series across 120 V and find the current and each bulb's power. (c) Connect the two in parallel across 120 V and find each bulb's power. (d) Which bulb is brighter in each case?

(a) A rating pairs a power with a potential difference, so use $P = (\Delta V)^2 / R$ rearranged to $R = (\Delta V)^2 / P$ . The 120 W bulb: $R_1 = \dfrac{(120 \, \mathrm{V})^2}{120 \, \mathrm{W}} = \dfrac{14400}{120} = 120 \, \Omega$ . The 40 W bulb: $R_2 = \dfrac{14400}{40} = 360 \, \Omega$ .
Notice the inversion before going further. The higher-rated bulb has the lower resistance, by a factor of exactly 3. That single fact is the source of the surprise in part (d).
(b) In series the resistances add, so $R_{\text{eq}} = 120 \, \Omega + 360 \, \Omega = 480 \, \Omega$ , and the current is $I = \dfrac{120 \, \mathrm{V}}{480 \, \Omega} = 0.250 \, \mathrm{A}$ , the same in both bulbs.
With the current fixed, use $P = I^2 R$ . The 120 W bulb: $P_1 = (0.250 \, \mathrm{A})^2 (120 \, \Omega) = (0.0625)(120) = 7.50 \, \mathrm{W}$ . The 40 W bulb: $P_2 = (0.0625)(360) = 22.5 \, \mathrm{W}$ .
Check the total with the printed equation: $P_{\text{total}} = I \Delta V = (0.250 \, \mathrm{A})(120 \, \mathrm{V}) = 30.0 \, \mathrm{W}$ , matching $7.50 + 22.5$ . The potential differences check too: $(0.250)(120) = 30.0 \, \mathrm{V}$ and $(0.250)(360) = 90.0 \, \mathrm{V}$ , summing to the 120 V supplied. Neither bulb reaches its rating here, because neither has 120 V across it.
(c) In parallel each bulb has the full 120 V across it, its rated potential difference, so each runs at its rating: $14400 / 120 = 120 \, \mathrm{W}$ and $14400 / 360 = 40 \, \mathrm{W}$ . The total is 160 W, against 30.0 W for the series arrangement.
(d) Brightness tracks power, per 11.4.A.2. In series the 40 W bulb dissipates 22.5 W against 7.50 W, three times as much, so it is the brighter of the two. In parallel the 120 W bulb dissipates three times as much the other way, so it is the brighter one.

(a) $120 \, \Omega$ for the 120 W bulb and $360 \, \Omega$ for the 40 W bulb. (b) Series: $I = 0.250 \, \mathrm{A}$ , with 7.50 W in the 120 W bulb and 22.5 W in the 40 W bulb. (c) Parallel: 120 W and 40 W, each at its rating. (d) The 40 W bulb is brighter in series, the 120 W bulb is brighter in parallel, three times the power each way. So Unit 11's essential question about making a 120 watt bulb brighter than a 40 watt bulb is answered by wiring them in parallel, and the question is worth asking because the series answer runs the other way. Brightness ranks with power, but 11.4.A.2 only promises a ranking: three times the power is not three times the perceived brightness.

Frequently asked questions

What is the formula for electric power in AP Physics 2?

P = IΔV, current times the potential difference across the element. That is the only power equation for circuits printed on the AP Physics 2 equation sheet, in the Electricity group. The AP Physics 2 CED gives it as the relevant equation for essential knowledge 11.4.A.1 and separately lists P = I²R and P = (ΔV)²/R as derived equations, each obtained by substituting Ohm's law, I = ΔV/R, into the printed form. Power is measured in watts, where one watt is one joule per second.

Is P = I²R on the AP Physics 2 equation sheet?

No. The Electricity group of the AP Physics 2 sheet contains 20 equations, and the only power equation among them is P = IΔV. Both P = I²R and P = (ΔV)²/R are absent from the sheet, and the CED labels them derived equations under essential knowledge 11.4.A.1. You are expected to produce them yourself in one substitution each, which is suggested skill 2.A for this topic. Nothing stops you from using them once you have derived them.

Which resistor dissipates more power, the larger one or the smaller one?

It depends on what the circuit holds fixed. In series, all elements carry the same current, so P = I²R applies and the larger resistance dissipates more power. In parallel, all branches share the same potential difference, so P = (ΔV)²/R applies and the smaller resistance dissipates more. Identify the connection first, then choose the form, then rank.

Why is a 40 W bulb brighter than a 120 W bulb when they are connected in series?

Because a rating describes one operating point, not the bulb in every circuit. Rated at the same voltage, the 120 W bulb has the lower resistance: R = (ΔV)²/P gives 120 ohms for the 120 W bulb and 360 ohms for the 40 W bulb at 120 V. In series the two carry the same current, so P = I²R applies and the larger resistance dissipates more. With 120 V across the pair the current is 0.250 A, and the 40 W bulb dissipates 22.5 W against the other bulb's 7.50 W. Wire them in parallel instead and each runs at its rating, so the 120 W bulb is brighter.

How do you find the energy a resistor dissipates?

Multiply the power by the elapsed time: ΔE = PΔt, the definition of average power rearranged. Find the power first from P = IΔV, or from P = I²R or P = (ΔV)²/R if that matches what you know, then multiply by the time in seconds. A resistor dissipating 3.0 W for 40 s converts 120 J of electrical energy to thermal energy. Convert minutes and hours to seconds first, since a watt is already a joule per second.

Does a light bulb's power rating hold in every circuit?

No. A rating states the power the bulb dissipates at one particular potential difference, usually the supply voltage it was designed for. Put that bulb in series with anything else and it has less than the rated potential difference across it, so it dissipates less than its rated power. Two bulbs rated 120 W and 40 W at 120 V dissipate only 7.50 W and 22.5 W when placed in series across 120 V. Treat a rating as a pair of numbers, power and voltage together, never as a property of the bulb alone.

What is the difference between electric power and potential difference?

Potential difference is energy per unit charge, measured in volts, and it is a property of two points in a circuit. Power is energy per unit time, measured in watts, and it is a rate at which a circuit element transfers, converts, or dissipates energy. The two are linked by the current: P = IΔV, since current is charge per unit time and multiplying joules per coulomb by coulombs per second gives joules per second. Two elements with the same potential difference across them dissipate different powers if they carry different currents.