2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A 1.5 Ω resistor is connected to an ideal battery so that the potential difference across it is 1.5 V, and it stays connected for 4.0 s. At what rate is energy dissipated in the resistor?
Answer and reasoning
A6.0 W A student who does not distinguish energy from power picks this: (1.5 W)(4.0 s) = 6.0 J is the energy dissipated in 4.0 s, reported in watts. The question asks for a rate, the power, 1.5 W.
B1.0 W A student who takes the current as the power picks this: ΔV/R = 1.0 A. That is the current; the power is IΔV = (1.0 A)(1.5 V) = 1.5 W.
C3.4 W A student who uses the potential difference in place of the current in P = I²R picks this: (1.5)²(1.5) = 3.4. The current is ΔV/R = 1.0 A, so P = I²R = (1.0 A)²(1.5 Ω) = 1.5 W.
D1.5 WCorrect The rate of energy dissipation is the power. With ΔV and R given, P = (ΔV)²/R = (1.5 V)²/1.5 Ω = 1.5 W. (Equivalently, I = ΔV/R = 1.0 A and P = IΔV = 1.5 W.) The 4.0 s is not needed for a rate.
Working P = (ΔV)²/R = (1.5 V)²/(1.5 Ω) = 1.5 W. Check: I = ΔV/R = 1.0 A; P = IΔV = (1.0 A)(1.5 V) = 1.5 W. The time interval is not needed for the rate.
Two lightbulbs in different circuits are compared. Which quantity must be greater for the brighter of the two bulbs?
Answer and reasoning
AThe current in the bulb, whatever the ΔV across it A student who thinks brightness depends on current alone picks this. A bulb with a smaller current can be brighter if the potential difference across it is large enough to give it more power.
BThe ΔV across the bulb, whatever current it carries A student who thinks brightness depends on potential difference alone picks this. A bulb with a larger ΔV but a very small current can have less power and be dimmer.
CThe share of the current that the bulb uses up A student who thinks a bulb glows by using up current picks this. No current is used up: the current leaving a bulb equals the current entering it. Brightness follows the power.
DThe rate at which it transfers electrical energyCorrect The brightness of a bulb increases with its power, the rate at which it transfers energy, P = IΔV. The brighter bulb has the greater power, whatever the individual values of I and ΔV.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.4.A.1 Electric power, P Fix
Electric power, P
The rate at which energy is transferred, converted or dissipated by a circuit element: P = IΔV, where I is the current in the element and ΔV the potential difference across it. Unit: watt (W), where 1 W = 1 J/s = 1 A·V.
Power in a resistor
Using ΔV = IR, the power dissipated in a resistor is P = I²R = (ΔV)²/R. P = I²R is convenient when the current is known or is the same for the elements compared; P = (ΔV)²/R when the potential difference is known or is the same.
Energy transferred
For a constant power, the energy transferred in a time interval Δt is ΔE = PΔt; for a changing power, it is the area under the power–time graph. Unit: joule (J).
Students often think Power and energy are the same quantity, so an energy can be reported as a power, or a power as an energy, without using the time interval. In fact No. Power is the rate at which energy is transferred, P = ΔE/Δt, measured in watts (J/s). The energy transferred in a time interval is the power multiplied by the time interval (for a changing power, the area under the power–time graph), measured in joules.
Students often think The current in a device is its power: the rate at which charge flows through it is the rate at which it transfers energy. In fact No. Current is the rate at which charge passes a point, in amperes; power is the rate at which energy is transferred, in watts. For a resistor, P = IΔV = I²R = (ΔV)²/R, so the power depends on the potential difference or the resistance as well as on the current.
11.4.A.2 Brightness of a bulb Fix
Brightness of a bulb
The brightness of a lightbulb increases with the power it dissipates, so comparing the bulbs' powers, P = IΔV, predicts qualitatively which bulb is brighter.
Students often think The brightness of a bulb is set by the current in it alone: the bulb with the largest current is the brightest. In fact No. Brightness increases with power, P = IΔV, which depends on the potential difference across the bulb as well as on the current in it. A bulb with a smaller current can be brighter if the potential difference across it is large enough.
Students often think The brightness of a bulb is set by the potential difference across it alone: the bulb with the largest ΔV is the brightest. In fact No. Brightness increases with power, P = IΔV. A bulb with a large potential difference across it but a very small current can have less power, and be dimmer, than a bulb with a smaller potential difference.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The potential difference across an ohmic resistor is increased so that the current in the resistor becomes three times as large. What is the ratio of the new power dissipated in the resistor, P₂, to the original power, P₁?
Answer and reasoning
AP₂/P₁ = 3 A student who uses P = IΔV with ΔV treated as fixed picks this. For the current in an ohmic resistor to triple, the potential difference across it must triple too, so P = IΔV increases by 3 × 3 = 9.
BP₂/P₁ = 6 A student who 'squares' the factor 3 by doubling it picks this: 3 × 2 = 6. Squaring means multiplying the factor by itself: 3² = 9.
CP₂/P₁ = 9Correct The resistor is ohmic, so R is unchanged. With P = I²R, tripling I multiplies P by 3² = 9. (Equivalently, ΔV = IR also triples, so P = IΔV increases by 3 × 3 = 9.)
DP₂/P₁ = 1 A student who thinks a resistor dissipates a fixed power picks this. The power depends on the current: P = I²R rises as the current rises.
Working Ohmic: R constant. P = I²R, so P₂/P₁ = (I₂/I₁)² = 3² = 9. (Errors: P ∝ I with ΔV fixed gives 3; 3² taken as 3 × 2 gives 6; fixed power gives 1.)
A heating element is a wire of length ℓ and diameter d, connected to an ideal battery so that the potential difference across it is ΔV. It is replaced by a wire of the same material with length ℓ/2 and diameter d/2, connected to the same battery. How does the power P₂ dissipated in the new wire compare with the power P₁ dissipated in the first wire?
Answer and reasoning
AP₂ = 0.50P₁Correct The resistance is R = ρℓ/A with A = πd²/4. Halving the length halves R, and halving the diameter makes A one-fourth as large, which multiplies R by 4, so R₂ = 2R₁. With the same ΔV, P = (ΔV)²/R, so P₂ = 0.50P₁.
BP₂ = 8.00P₁ A student who thinks a thicker wire has more resistance, R = ρℓA, picks this: the new wire's 'resistance' is (1/2)(1/4) = 1/8 of the first, so P = (ΔV)²/R is 8 times as large. Resistance is inversely proportional to the area, R = ρℓ/A, so the thinner wire has more resistance: R₂ = 2R₁ and P₂ = 0.50P₁.
CP₂ = 2.00P₁ A student who finds R₂ = 2R₁ correctly but thinks a larger resistance always dissipates more power, from P = I²R, picks this. The battery keeps ΔV fixed, so doubling R halves the current: P = (ΔV)²/R, and P₂ = 0.50P₁.
DP₂ = 1.00P₁ A student who takes the cross-sectional area to be proportional to the diameter picks this: halving both ℓ and 'A' leaves R, and so P, unchanged. A = πd²/4, so halving d makes A one-fourth as large; R₂ = 2R₁ and P₂ = 0.50P₁.
Working R = ρℓ/A with A = πd²/4, so R ∝ ℓ/d². New wire: R₂ = R₁ × (1/2)/(1/2)² = 2R₁. With ΔV fixed, P = (ΔV)²/R, so P₂ = P₁/2 = 0.50P₁. (m09, R ∝ ℓA: R₂ = R₁ × (1/2)(1/4) = R₁/8, P₂ = 8.00P₁; m07, P ∝ R: R doubles so P₂ = 2.00P₁; new misconception, A ∝ d: R₂ = R₁ × (1/2)/(1/2) = R₁, P₂ = 1.00P₁.)
Three different lightbulbs, X, Y and Z, are each connected to a different battery. The graph shows the potential difference ΔV across each bulb and the current I in it. Which ranks the bulbs from brightest to dimmest?
Answer and reasoning
AX > Y > Z A student who ranks brightness by current alone picks this: X has the largest current. Power depends on the potential difference as well: X's small ΔV gives it 1.2 W, less than Y's 2.0 W.
BY > X > ZCorrect Brightness increases with power, P = IΔV. X: (0.60 A)(2.0 V) = 1.2 W; Y: (0.40 A)(5.0 V) = 2.0 W; Z: (0.10 A)(4.0 V) = 0.40 W. So Y is the brightest and Z the dimmest.
CY > Z > X A student who ranks brightness by potential difference alone picks this: Z has a larger ΔV than X. Z's current is so small that its power, 0.40 W, is the least of the three.
DZ > Y > X A student who thinks the bulb with the largest resistance is the brightest picks this: ΔV/I is 40 Ω for Z, 13 Ω for Y and 3.3 Ω for X. The bulbs have neither the same current nor the same potential difference, so resistance alone does not rank them; power, P = IΔV, does.
Working P = IΔV. X: 0.60 × 2.0 = 1.2 W. Y: 0.40 × 5.0 = 2.0 W. Z: 0.10 × 4.0 = 0.40 W. Ranking Y > X > Z. (By I: X > Y > Z. By ΔV: Y > Z > X. By R = ΔV/I: Z (40 Ω) > Y (13 Ω) > X (3.3 Ω).)
The potential difference across a resistor is reduced so that the power dissipated in it decreases steadily, as shown in the graph of power P against time t. How much electrical energy is converted to thermal energy in the resistor between t = 0 and t = 2.0 s?
Answer and reasoning
A6.0 JCorrect Power is the rate of energy transfer, so the energy transferred is the area under the power–time graph. The area is a trapezoid: ½(4.0 W + 2.0 W)(2.0 s) = 6.0 J.
B8.0 J A student who multiplies the starting power by the time interval picks this: (4.0 W)(2.0 s) = 8.0 J. ΔE = PΔt holds only for a constant power; here the power falls from 4.0 W to 2.0 W, so the energy is the area under the line, 6.0 J.
C3.0 J A student who does not distinguish power from energy picks this, reporting the average power, ½(4.0 W + 2.0 W) = 3.0 W, as the energy. Energy is power multiplied by time: (3.0 W)(2.0 s) = 6.0 J.
D1.0 J A student who uses the slope of the graph instead of the area picks this: (4.0 W − 2.0 W)/(2.0 s) = 1.0 W/s. The slope shows how fast the power changes; the energy is the area under the graph.
Working E = area under the P–t graph = ½(4.0 W + 2.0 W)(2.0 s) = 6.0 J.
Two ohmic resistors, of resistance R and 3R, are compared in two situations. In situation 1, each resistor carries the same current I. In situation 2, each resistor has the same potential difference ΔV across it. In which situation or situations does the 3R resistor dissipate more power than the R resistor?
Answer and reasoning
AIn situation 2 only, where the ΔV values are equal A student who treats current and potential difference as interchangeable picks this, using P = I²R for the equal-ΔV case and P = (ΔV)²/R for the equal-current case. With equal ΔV, the 3R resistor carries one-third of the current and dissipates less power.
BIn situation 1 only, where the currents are equalCorrect With equal currents, P = I²R, so the 3R resistor dissipates three times as much power. With equal potential differences, P = (ΔV)²/R, so it dissipates one-third as much. Which resistor dissipates more depends on which quantity is the same.
CIn both situations, since a larger R means more power A student who thinks a larger resistance always dissipates more power picks this. That holds when the currents are equal; with equal ΔV, P = (ΔV)²/R is smaller for the larger resistance.
DIn neither situation, as a larger R means less power A student who thinks a larger resistance always dissipates less power picks this. That holds when the potential differences are equal; with equal currents, P = I²R is larger for the larger resistance.
Working Situation 1 (equal I): P = I²R, so P3R/PR = 3. Situation 2 (equal ΔV): P = (ΔV)²/R, so P3R/PR = 1/3. The 3R resistor dissipates more power only in situation 1.
An ohmic heating element of resistance R has a steady current I₀ in it and transfers energy ΔE to its surroundings as thermal energy in a time interval Δt. The current is then changed to a new steady value so that the element transfers the same energy ΔE in a time interval Δt/2. What is the new current?
Answer and reasoning
A1.00I₀ A student who treats power and energy as the same quantity sets I²R = ΔE and picks this: the energy is unchanged, so the current seems unchanged. Power is energy per unit time, I²R = ΔE/Δt, and halving Δt doubles the power, so I = 1.41I₀.
B2.00I₀ A student who takes the power to be proportional to the current, as if ΔV stayed the same, picks this: twice the power needs twice the current. In an ohmic resistor ΔV rises with I, so P = I²R; doubling P multiplies I by √2, giving 1.41I₀.
C1.41I₀Correct The same energy in half the time means twice the power: P = ΔE/Δt doubles. For a given resistor P = I²R, so I² must double and the current becomes √2 × I₀ = 1.41I₀.
D4.00I₀ A student who squares the factor again when working back from I² picks this: the power, and so I², must double, and squaring the 2 gives 4. The current changes by the square root of the power factor: √2 ≈ 1.41, so the new current is 1.41I₀.
Working The power is P = ΔE/Δt; transferring ΔE in Δt/2 doubles it. For a given resistor P = I²R, so I² doubles and I = √2·I₀ = 1.41I₀. (m04, P ∝ I: 2.00I₀; m01, power and energy the same, I²R = ΔE with ΔE unchanged: 1.00I₀; new misconception, factor squared again instead of square-rooted: 2² = 4, 4.00I₀.)
An ohmic heating element dissipates power P₁ when the potential difference across it is ΔV₁. The element is then connected so that the potential difference across it is ΔV₁/k, where k > 1, and it is used to heat a mass m of water of specific heat c. All the energy dissipated in the element goes into the water. Which expression gives the time needed to raise the temperature of the water from Ti to Tf?
Answer and reasoning
Ak²mc(Tf − Ti)/P₁Correct The element is ohmic, so its resistance R = (ΔV₁)²/P₁ does not change. At the new potential difference it dissipates P₂ = (ΔV₁/k)²/R = P₁/k². The water needs energy mc(Tf − Ti), and P₂Δt = mc(Tf − Ti) gives Δt = k²mc(Tf − Ti)/P₁.
Bmc(Tf − Ti)/P₁ A student who treats the power P₁ as a fixed property of the element picks this. P₁ is the power only at ΔV₁; what stays fixed is the element's resistance, so at ΔV₁/k the power is P₁/k² and the time is k² times as long.
Ckmc(Tf − Ti)/P₁ A student who takes the power to be proportional to the potential difference, as if the current did not change, picks this: ΔV falls by the factor k, so the power seems to fall by k. For an ohmic element the current falls by k too, so P = IΔV falls by k², and the time is k² times as long.
Dkmc(Tf − Ti)/(2P₁) A student who 'squares' the factor 1/k by doubling it takes the new power as 2P₁/k and picks this. Squaring multiplies the factor by itself: P₂ = P₁(1/k)² = P₁/k², so the time is k² times as long as at P₁.
Working The element is ohmic, so its resistance R = (ΔV₁)²/P₁ is unchanged. New power: P₂ = (ΔV₁/k)²/R = P₁/k². Energy needed: Q = mc(Tf − Ti) = P₂Δt, so Δt = mc(Tf − Ti)/P₂ = k²mc(Tf − Ti)/P₁. (m06, fixed power P₁: mc(Tf − Ti)/P₁; m18, power scaled with the potential difference alone, current taken as unchanged, by 1/k: kmc(Tf − Ti)/P₁; m05, (1/k)² taken as 2(1/k), so P₂ = 2P₁/k: kmc(Tf − Ti)/(2P₁).)
A 2.0 Ω resistor carries a steady current, and 9.0 J of electrical energy is converted to thermal energy in the resistor in 2.0 s. How much charge passes through the resistor in this time?
Answer and reasoning
A4.2 C A student who puts the energy, 9.0 J, into P = I²R as if it were the power picks this: I = √(9.0/2.0) = 2.1 A and q = 4.2 C. The power is the energy per unit time, 4.5 W, which gives I = 1.5 A and q = 3.0 C.
B6.0 C A student who uses P = (ΔV)²/R with the current in place of the potential difference picks this: I² = PR = 9.0, so I = 3.0 A and q = 6.0 C. For a current, P = I²R, so I² = P/R = 2.25 A² and I = 1.5 A.
C3.0 CCorrect The power is ΔE/Δt = 9.0 J/2.0 s = 4.5 W. From P = I²R, I = √(4.5 W/2.0 Ω) = 1.5 A, and the charge that passes in 2.0 s is q = IΔt = (1.5 A)(2.0 s) = 3.0 C.
D4.5 C A student who solves P = I²R as I = P/R, dropping the square, picks this: 4.5 W/2.0 Ω gives 2.25 'A' and q = 4.5 C. P/R is I², so I = √(2.25 A²) = 1.5 A and q = 3.0 C.
Working P = ΔE/Δt = 9.0 J/2.0 s = 4.5 W. P = I²R gives I = √(P/R) = √(4.5 W/2.0 Ω) = 1.5 A. q = IΔt = (1.5 A)(2.0 s) = 3.0 C. (m01, the energy used as the power: I = √(9.0/2.0) = 2.1 A, q = 4.2 C; m03, P = (ΔV)²/R used with the current: I = √(PR) = 3.0 A, q = 6.0 C; m17, square dropped: I = P/R = 2.25 A, q = 4.5 C.)
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
