United StatesPhysics 2Syllabus dot point

How fast does a circuit deliver and dissipate energy, and what sets a device's power?

Topic 11.4 Electric Power: calculate the power delivered or dissipated in a circuit using P = IV, P = I squared R and P = V squared over R.

A focused answer to AP Physics 2 Topic 11.4, covering electric power as the rate of energy transfer, the three equivalent power formulas, the power dissipated in a resistor, energy used over time, and how to choose the right formula, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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Quick answer

Electric power is the rate at which a circuit transfers or dissipates energy, $P = IV$ (current times voltage), measured in watts (W $=$ J/s). For a resistor, substituting Ohm's law gives two more forms: $P = I^2 R$ and $P = \dfrac{V^2}{R}$ . Choosing the right one depends on what is held constant: at fixed current (a series element), power rises with resistance ( $P = I^2 R$ ); at fixed voltage (across a source), power falls with resistance ( $P = V^2/R$ ). The energy delivered in a time $t$ is $E = Pt$ . A source delivers power $\mathcal{E} I$ to the circuit; resistors dissipate it as heat (and bulbs as light). Electric power is the bridge from circuits back to the energy ideas of mechanics: it is how much electrical energy is converted each second, which is exactly what an electricity bill charges for.

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What this topic is asking
What electric power is
The three power formulas
Energy, heat and the strategic role
Try this

What this topic is asking

The College Board (Topic 11.4) wants you to calculate the electric power delivered or dissipated in a circuit, using $P = IV$ and its forms $P = I^2 R$ and $P = \dfrac{V^2}{R}$ , and to find the energy used over time.

What electric power is

Power is energy per second, the rate of energy conversion. The base formula $P = IV$ comes straight from the energy-per-charge idea: each coulomb crossing a voltage $V$ delivers energy $V$ joules (from Topic 10.7), and $I$ coulombs cross each second, so the power is $IV$ . This is the most general form and applies to any component, not just resistors.

The three power formulas

The three forms are the same physics written differently, and picking the right one avoids confusion. The classic trap is asking which resistor dissipates more power. The answer depends on what is fixed: at fixed current, more resistance means more power ( $P = I^2 R$ , the case in a series circuit); at fixed voltage, more resistance means less power ( $P = V^2/R$ , the case across a battery). Identifying whether the current or the voltage is the shared, fixed quantity is the key decision.

Energy, heat and the strategic role

Power dissipated in a resistor becomes heat (and light, in a bulb): the electrical energy converts to thermal energy at the rate $P$ . Over a time $t$ , the total energy is $E = Pt$ , which is what utilities bill (often in kilowatt-hours). The source's power output is $\mathcal{E} I$ , and by conservation of energy this equals the total power dissipated and stored in the circuit, the energy bookkeeping that closes every circuit problem. The strategic point is that power ties the circuit back to energy conservation: the voltage-times-current that drives the circuit is delivered by the source and dissipated by the resistances, so summing the powers checks an analysis. Combined with the series and parallel rules of Topic 11.5, power lets you find which components run hot and how much energy a circuit uses, the practical payoff of the whole unit.

Power and energy in a light bulb

A bulb is rated to dissipate $60$ W when connected to a $120$ V supply. Find the current it draws, its resistance, and the energy it uses in $5.0$ hours.

step 1 Current from P = IV

$I = \dfrac{P}{V} = \dfrac{60}{120} = 0.50$ A.

step 2 Resistance from Ohm's law

$R = \dfrac{V}{I} = \dfrac{120}{0.50} = 240$ ohms (or $R = V^2/P = 120^2/60 = 240$ ohms).

step 3 Energy used over time

$t = 5.0 \times 3600 = 1.8 \times 10^4$ s, so $E = Pt = (60)(1.8 \times 10^4) = 1.08 \times 10^6$ J.

step 4 Interpret

The bulb converts $60$ J of electrical energy each second into light and heat, totalling about $1.1$ million joules over five hours (about $0.30$ kilowatt-hours on a bill).

Try this

Q1. A device draws $2.0$ A at $24$ V. Calculate its power. [2 points]

Cue. $P = IV = (2.0)(24) = 48$ W.

Q2. State which power formula to use to compare two resistors carrying the same current. [1 point]

Cue. $P = I^2 R$ (at fixed current, the larger resistance dissipates more power).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (short FRQ). A resistor of

8.0

ohms carries a current of

1.5

A. (a) Calculate the power dissipated in the resistor. (b) Calculate the voltage across it, then verify the power using P = IV. (c) Calculate the energy dissipated in

2.0

minutes.

Show worked answer →

A 6-point FRQ on electric power.

(a) Power (2 points): $P = I^2 R = (1.5)^2 (8.0) = (2.25)(8.0) = 18$ W.
(b) Voltage and check (2 points): $V = IR = (1.5)(8.0) = 12$ V; $P = IV = (1.5)(12) = 18$ W, which matches.
(c) Energy (2 points): $E = Pt = (18)(2.0 \times 60) = (18)(120) = 2160$ J.

Markers reward the power calculation, the cross-check with $P = IV$ , and the energy as power times time.

AP 2023 (style)1 marksSection I (multiple choice). Two resistors are connected across the same fixed voltage. Which dissipates more power? (A) the one with the larger resistance (B) the one with the smaller resistance (C) they dissipate equal power (D) it cannot be determined. Justify your reasoning.

Show worked answer →

A 1-point MCQ on choosing the power formula. The answer is (B).

Across a fixed voltage, $P = \dfrac{V^2}{R}$ , so power is inversely proportional to resistance: the smaller resistance dissipates more power. The trap is (A): that would be correct only for a fixed current ( $P = I^2 R$ ), but here the voltage is fixed, so the smaller resistance wins.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)