Power is how fast electrical energy is being transferred, converted, or dissipated—energy per unit time. Energy (measured in joules) is the total amount; power (watts, 1 W = 1 J/s) is the rate. In circuits AP focuses on power in a component using P = I ΔV (Essential Knowledge 11.4.A.1). Using Ohm’s law you get the useful forms P = I^2R or P = (ΔV)^2/R. Example: a 60 W bulb converts electrical energy to light/heat at 60 J each second. Higher power → brighter bulb (qualitative prediction for brightness is in the CED). Remember power depends on both current and potential difference across the element, so two identical bulbs can have different brightness in different circuit positions. For a clearer walkthrough and practice problems tied to the CED, check the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and thousands of practice questions at (https://library.fiveable.me/practice/ap-physics-2-revised).

P = IΔV just says “power = rate energy is transferred” across a circuit element. Think physically: each charge q that passes drops ΔV of electric potential, so the energy given up per charge is qΔV. If a current I flows, the amount of charge passing per second is I (since I = Δq/Δt). Multiply energy per charge by charge per second: (qΔV)·(Δq/Δt)/q → IΔV, which is energy per second (watts). So P = IΔV tells you how fast electrical energy is being converted (to light, heat, mechanical work) by that element. Using Ohm’s law you get the useful forms P = I^2R and P = ΔV^2/R used to predict bulb brightness or heating. This is exactly what the CED expects in 11.4.A (rate energy transfer depends on I and ΔV). For extra practice, check the Topic 11.4 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and try problems from the practice bank (https://library.fiveable.me/practice/ap-physics-2-revised).

Use P = IΔV as the fundamental statement: power delivered to (or removed from) any circuit element equals the current through it times the potential difference across it (CED 11.4.A.1). The other two forms come from Ohm’s law ΔV = IR, so choose them when you only know I or ΔV and R: - If you know both I and ΔV (or want a general expression): P = IΔV. - If you know I and the element’s resistance R (common for resistors): P = I^2R (substitute ΔV = IR). - If you know ΔV across the resistor and R: P = ΔV^2 / R (substitute I = ΔV / R). Quick tips: for a resistor in a circuit use any form that matches the given values; for a battery use P = εI where I is the circuit current; if I = 0 (open circuit) power = 0; if ΔV = 0 (ideal short) power dissipated in that element = 0. For AP-style practice and examples, see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and try problems from the AP unit page (https://library.fiveable.me/ap-physics-2-revised/unit-11) or the practice set (https://library.fiveable.me/practice/ap-physics-2-revised).

Power dissipated vs. power delivered is mostly about sign and where energy goes. - Power dissipated: energy converted to heat/light inside a passive element (like a resistor or bulb). Use P = I ΔV and the common forms P = I^2R = ΔV^2 / R. For a resistor ΔV is the drop across it and P is positive (energy leaves electrical form and becomes thermal/light). - Power delivered: energy supplied to the rest of the circuit by an active element (like a battery or a current source). For an element, P = I ΔV still applies, but if I flows from low to high potential through the element (ΔV > 0 while conventional current enters the negative terminal), P can be negative—meaning the element is delivering power to the circuit. Example: an ideal battery delivers P = ε I (positive delivered to circuit); a real battery delivers P_out = ε I − I^2 r, where I^2 r is power dissipated in its internal resistance. On the AP, use P = I ΔV and the derived forms to calculate and compare brightness or energy flow (Topic 11.4). For more review, check the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and unit resources (https://library.fiveable.me/ap-physics-2-revised/unit-11). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Brightness is directly related to the electrical power the bulb dissipates: greater power → brighter bulb (CED 11.4.A.2). Use P = I ΔV (and the forms P = I^2R = ΔV^2 / R) to compare bulbs. Examples you can use on the exam: - For a single bulb with fixed R, raising the voltage increases power fast (P ∝ ΔV^2), so it gets much brighter. - In series, bulbs share the battery voltage, so each gets less ΔV and lower P → dimmer than the same bulb on full supply. - In parallel, each bulb sees the full supply ΔV, so each dissipates its full power and stays brighter. On AP problems, you’ll be asked to use these formulas to predict relative brightness (qualitatively or with numbers). For a quick topic review, check the Topic 11.4 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

When you add more identical lightbulbs in series, the total resistance rises (R_total = nR). With the same battery, the circuit current drops: I = V/(nR). Power delivered to each bulb depends on that current: P_bulb = I^2R = V^2/(n^2R). So each bulb gets much less power as you add bulbs, and brightness (which increases with power per the CED 11.4.A.2) falls—roughly like 1/n^2 for identical bulbs. You can also say the battery’s fixed ΔV is shared across more bulbs, so the potential difference across each bulb is smaller, lowering P = I ΔV as well. This is a classic Topic 11.4 application (compare scenarios and use P = I^2R). For extra review, check the Topic 11.4 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and more practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Power = I ΔV because power is how fast electrical energy moves through a circuit element. If a charge Q moves across a potential difference ΔV, the energy transferred is Q·ΔV. Rate of energy transfer is energy divided by time: P = (Q·ΔV)/Δt. But current I is charge per time (I = Q/Δt), so P = I·ΔV. That’s the CED essential-knowledge statement: the rate energy is transferred depends on current and potential difference (11.4.A.1). Two quick consequences you’ll see on the AP: using Ohm’s law (ΔV = IR) gives P = I²R or P = ΔV²/R, and higher power means a brighter bulb (11.4.A.2). For a short refresher and more examples, check the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and unit review (https://library.fiveable.me/ap-physics-2-revised/unit-11). Practice problems are great for this concept (https://library.fiveable.me/practice/ap-physics-2-revised).

Short answer: it depends what’s held constant. Use the CED power formulas: P = I ΔV, and the derived forms P = I^2R = (ΔV)^2 / R. - If the voltage across the element is fixed (an ideal battery across a resistor), P = V^2/R—increasing R decreases power. Example: V = 10 V, R = 2 Ω → P = 50 W; R = 4 Ω → P = 25 W. - If the current through the element is fixed (a current source), P = I^2R—increasing R increases power. - In typical circuits (resistors with a fixed battery), changing one resistor changes the total current, so you must combine the relations: I = ε / R_total, then P_resistor = I^2R (use algebra to see how it changes). For AP exam questions, show which quantity is held constant, write P = IΔV and substitute I = ε/R_total or ΔV = IR as needed, and compare cases algebraically or with numbers. For a targeted review, see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and more practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Think of energy in a circuit like water pushed through pipes: the battery (or power source) converts chemical energy into electrical potential energy (an emf) and the electric field does work on charges so they move. The rate that energy moves is power: P = I ΔV (and P = I^2R = ΔV^2/R). Where that energy goes depends on the element: - Resistors: electrical energy → thermal (heating) (lightbulbs are resistors that emit visible light + heat). - Capacitors: energy stored as electric potential energy in the field (½CV^2) and released later. - Motors: electrical → mechanical work (plus some heat). - Wires ideally carry current with negligible dissipation; real wires still dissipate some heat. On the AP exam you should describe these transfers using P = IΔV and compare powers (e.g., brightness of bulbs from power). For a focused review check the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and the Unit 11 overview (https://library.fiveable.me/ap-physics-2-revised/unit-11). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Total power in a circuit with multiple resistors is just the sum of the powers dissipated by each element. Use P = I ΔV (or the derived forms P = I^2R = ΔV^2 / R) for each resistor, then add: - Series: the same current I flows through every resistor. Find I = Vbattery / Req, compute Pk = I^2 Rk for each resistor, then Ptotal = Σ I^2 Rk. (Also Ptotal = I^2 Req = Vbattery · I.) - Parallel: the voltage across each branch equals the battery voltage V. For each resistor, Pk = V^2 / Rk. Sum: Ptotal = Σ V^2 / Rk. (Also Ptotal = V^2 / Req = V · Itotal.) On the AP exam you should show which form you use and justify with Ohm’s law / equivalent resistance per the CED (11.4.A.1). For practice and worked examples, check the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and more problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

A 100 W bulb shines brighter than a 60 W bulb because it converts electrical energy to light (and heat) at a higher rate—i.e., it has greater power. Power in a circuit is P = I ΔV (and equivalent forms P = I^2R = ΔV^2/R). For two bulbs designed to run at the same supply voltage, the bulb with the higher rated power either draws more current (if its resistance is lower) or dissipates more energy per second across itself. Because brightness increases with power (CED 11.4.A.2), the 100 W lamp emits more luminous energy per second than the 60 W lamp, so it appears brighter. This is exactly the kind of qualitative reasoning AP asks for on Topic 11.4. For a quick review, see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Power is the rate electrical energy is converted per second. In circuits use P = I ΔV (and the common forms P = I^2R = ΔV^2/R). In a resistor those formulas tell you how much electrical energy becomes thermal energy (heat) per second—that’s Joule heating (dissipation). So larger current or larger voltage across a resistor → more power → more heating (and a bulb gets brighter because it’s dissipating more power as light + heat). The CED specifically gives P = I ΔV and derived forms (11.4.A.1); AP Physics 2 expects you to connect power to energy transfer/dissipation and to use those equations on the exam. AP Physics C only stresses mechanical/electrical energy transfer but you should still know that electrical energy can be lost as thermal energy. For a quick review, see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and the Unit 11 overview (https://library.fiveable.me/ap-physics-2-revised/unit-11). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Find the power dissipated in each bulb—the one with the largest power is brightest. Use P = I ΔV (and the forms P = I^2R or P = ΔV^2/R) from the CED. Steps: (1) reduce the circuit (identify series vs. parallel) or use Kirchhoff’s rules to get currents and voltages for each branch. (2) For each bulb, plug the current through it and/or the voltage across it into P = IΔV (or P = I^2R or P = V^2/R). (3) Rank the bulbs by power; higher power → brighter. Quick shortcuts: identical bulbs in series share the same current (compare P = I^2R), identical bulbs in parallel share the same voltage (compare P = V^2/R). On the AP exam you’ll be asked to compare qualitatively or compute numerically, so show the circuit reduction or Kirchhoff work and state which formula you used. For extra practice and worked examples see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m), the Unit 11 overview (https://library.fiveable.me/ap-physics-2-revised/unit-11), and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

You were supposed to see that a bulb’s brightness tracks the electrical power dissipated in it. Use P = IΔV (and the forms P = I^2R or P = ΔV^2 / R) to predict brightness: more power → brighter bulb (CED 11.4.A.1–2). Typical lab observations: - Identical bulbs in series look much dimmer than when each is in parallel with the battery. In series the current is smaller so P per bulb is lower; in parallel each bulb gets the full ΔV and so dissipates more power → brighter. - If you change resistance (different bulbs or added resistors), bulbs with higher P (from higher ΔV or I) glow brighter. - Shorting part of the circuit can make one bulb go out or another brighten (current/ΔV redistribute). This is exactly the qualitative reasoning AP asks for: relate current and potential difference to power and use power to rank brightness (Topic 11.4). Review the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised) if you want quick examples and questions.

Short answer: pick whichever quantity you can get directly from the circuit or from the givens, then use one of the equivalent power formulas. The CED gives P = I ΔV and the derived forms P = I^2R = (ΔV)^2 / R—use whichever fits the knowns. Quick checklist: - If the problem gives voltage across the element and its resistance (or you can find ΔV with Kirchhoff), use P = (ΔV)^2 / R or find current with I = ΔV / R and use P = IΔV. - If you know current (from an ammeter reading, series current, or from solving the circuit) and R or ΔV, use P = IΔV or P = I^2R. - If the circuit is more complex, first use series/parallel rules or Kirchhoff’s laws (Topic 11.5 & 11.6) to get either I or ΔV for that element, then calculate power. On the AP exam, show your pathway: state the relation you use (P = IΔV or Ohm’s law) and why you solved for I or ΔV first. For extra practice and worked examples, see the Topic 11.4 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/4-electric-power/study-guide/XlqNl6VwuwtsWe2m) and Unit 11 review (https://library.fiveable.me/ap-physics-2-revised/unit-11).