AP Physics 2 Quiz: Electric Power

What this quiz covers

This quiz focuses on Electric Power, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. smaller, because $P=IV$ and $I$ is smaller while $V$ is constant (correct answer)
2. larger, because the device has been on longer and used more energy
3. unchanged, because the voltage source is constant
4. larger, because a smaller current implies a larger resistance and thus larger power

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = IV where I is current and V is voltage. When connected to a constant 6.0 V source, the initial power is P₁ = 2.0 A × 6.0 V = 12 W, while the final power is P₂ = 1.0 A × 6.0 V = 6 W, showing the power decreased by half. The device's changing current indicates its resistance increased over time, but this doesn't affect the P = IV calculation directly. Choice D incorrectly claims larger resistance means larger power, a common misconception that ignores how resistance affects current. Remember that power depends on the product of current and voltage, not on resistance alone.

Question 2

1. twice as large, because doubling voltage doubles power
2. four times as large, because $P=\dfrac{V^2}{R}$ and $R$ is constant (correct answer)
3. unchanged, because resistance is unchanged
4. half as large, because higher voltage reduces current draw

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and when resistance is constant, we use P = V²/R. If voltage doubles from V to 2V while resistance R remains constant, the new power becomes P_new = (2V)²/R = 4V²/R, which is four times the original power P = V²/R. This quadrupling occurs because power depends on the square of voltage when resistance is fixed. Choice A incorrectly assumes a linear relationship between power and voltage, missing the squared dependence. When resistance is constant, power scales with the square of voltage, not linearly.

Question 3

1. half as large, because its resistance is half
2. twice as large, because $P=\dfrac{V^2}{R}$ at the same voltage (correct answer)
3. the same, because both are connected to the same battery
4. four times as large, because its current is twice and power scales as $I^2$

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and for resistors at the same voltage, we use P = V²/R. The 3.0 Ω resistor has power P₁ = V²/3.0, while the 6.0 Ω resistor has P₂ = V²/6.0. Since P₁/P₂ = (V²/3.0)/(V²/6.0) = 6.0/3.0 = 2, the 3.0 Ω resistor dissipates twice the power. This occurs because at constant voltage, smaller resistance allows more current and thus more power. Choice A incorrectly assumes power is proportional to resistance, reversing the actual inverse relationship. For parallel resistors, smaller resistance means larger power dissipation.

Question 4

1. unchanged because the battery voltage is fixed
2. twice as large
3. four times as large
4. half as large (correct answer)

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = V²/R when voltage is constant. With the battery maintaining 9.0 V, the initial power is P₁ = (9.0 V)²/R, and when resistance doubles to 2R, the new power becomes P₂ = (9.0 V)²/(2R) = P₁/2. Doubling the resistance halves the power when voltage is constant. Choice D incorrectly assumes power cannot change with fixed voltage, but power depends on both voltage and resistance through the relationship P = V²/R. For constant voltage sources, power is inversely proportional to resistance: double the resistance means half the power.

Question 5

1. The $10\,\Omega$ resistor, because $P=I^2R$ (correct answer)
2. They dissipate the same power because the time is the same
3. The $5.0\,\Omega$ resistor, because lower resistance means more power
4. They dissipate the same power because both are resistors

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = I²R for resistors carrying current. For the 10 Ω resistor: P₁ = (2.0 A)²(10 Ω) = 40 W, while for the 5.0 Ω resistor: P₂ = (1.0 A)²(5.0 Ω) = 5.0 W. The 10 Ω resistor dissipates 40 W compared to only 5.0 W for the 5.0 Ω resistor, making it dissipate eight times more power. Choice A incorrectly assumes lower resistance always means more power, but this depends on the current—with different currents, you must calculate P = I²R for each case. When comparing power in resistors with different currents, always calculate P = I²R individually rather than relying on resistance alone.

Question 6

1. It is $360\ \text{W}$ because power equals energy divided by time
2. It is $24\ \text{W}$ because $P=IV$ and both $I$ and $V$ are constant (correct answer)
3. It is $12\ \text{W}$ because the voltage sets the energy per charge only
4. It is $6\ \text{W}$ because the current is spread over $30\ \text{s}$

Explanation: This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, measured in watts (W), and is calculated as the product of voltage and current: P = IV. For this heater, with a constant current of 2.0 A and voltage of 12 V, the power is P = (2.0 A)(12 V) = 24 W. The power remains constant at 24 W throughout the 30-second period because both current and voltage are constant. Choice C incorrectly divides the current by time, revealing the misconception that power depends on the duration of operation. Remember: power is an instantaneous rate (energy per unit time), not affected by how long a device operates.

Question 7

1. They dissipate the same energy because their currents are different
2. Bulb B dissipates more power because it has the higher current
3. They dissipate the same power because $V_AI_A=V_BI_B$ (correct answer)
4. Bulb A dissipates more power because it has the higher voltage

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = VI for any electrical component. For bulb A: P_A = (10 V)(0.40 A) = 4.0 W, and for bulb B: P_B = (5 V)(0.80 A) = 4.0 W, showing both bulbs dissipate the same power despite having different voltage and current values. Choice B incorrectly focuses only on current magnitude, ignoring that power depends on the product of both voltage and current. When comparing power between components, always calculate P = VI for each rather than comparing individual V or I values alone.

Question 8

1. It increases because higher resistance always means higher power
2. It stays constant because the battery fixes the power
3. It becomes zero because current must stop immediately
4. It decreases because with constant $V$, $P=\dfrac{V^2}{R}$ (correct answer)

Explanation: This question tests understanding of electric power. Electric power is the rate of energy dissipation, which for a resistor with constant voltage can be expressed as P = V²/R. As the heating element's resistance R increases while voltage V remains constant (fixed by the battery), the power P = V²/R decreases because R appears in the denominator. Choice A incorrectly assumes higher resistance always means higher power, which is only true for constant current situations, not constant voltage. For constant voltage scenarios, remember that power is inversely proportional to resistance, so increasing resistance reduces power dissipation.

Question 9

1. $R_1$ dissipates more power because it has smaller resistance
2. $R_2$ dissipates more power because it has larger resistance (correct answer)
3. Both dissipate the same power because the current is the same
4. Both dissipate the same power because the total energy is fixed

Explanation: This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = IV where I is current and V is voltage. In a series circuit, the same current flows through both resistors, but the voltage drop across each resistor depends on its resistance according to Ohm's law (V = IR). Since R₂ = 2R₁ and they have the same current, V₂ = 2V₁, making the power in R₂ equal to P₂ = IV₂ = I(2V₁) = 2IV₁ = 2P₁. Choice C incorrectly assumes equal power because of equal current, missing that voltage also matters in the power formula. When resistors are in series with the same current, the one with larger resistance (and thus larger voltage drop) dissipates more power.

Question 10

1. $R_2$, because it has the smaller resistance
2. $R_1$, because it has the larger voltage drop at the same current (correct answer)
3. $R_1$ and $R_2$ dissipate equal power because they are in series
4. $R_2$, because it transfers more energy in $10\ \text{s}$

Explanation: This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = IV where I is current and V is voltage. In a series circuit, the same current flows through both resistors, but they can have different voltage drops. Since R₁ has V₁ = 9 V and R₂ has V₂ = 3 V, and both carry the same current I, we calculate power as P₁ = I × 9 V and P₂ = I × 3 V, making P₁ three times larger than P₂. Choice C incorrectly assumes equal power in series circuits, reflecting the misconception that series elements always share power equally. To solve power problems, identify which quantities are the same (here, current) and which differ (here, voltage), then apply P = IV.

Question 11

1. $R_1$, because lower resistance always means greater power
2. $R_2$, because with equal current $P=I^2R$ is larger for larger $R$ (correct answer)
3. They dissipate equal power because the same current flows through each
4. $R_1$, because it has the smaller voltage drop in series

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and for resistors in series with the same current, power is calculated using P = I²R. Since the same current flows through both resistors in a series circuit, the resistor with larger resistance dissipates more power. With R₂ = 6 Ω being three times larger than R₁ = 2 Ω, resistor R₂ dissipates three times more power than R₁. Choice A incorrectly assumes lower resistance always means greater power, which is only true when voltage is constant (parallel circuits). For series circuits with constant current, use P = I²R to see that larger resistance means more power dissipation.

Question 12

1. doubled
2. unchanged
3. halved
4. quadrupled (correct answer)

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and for a resistor it can be calculated as P = I²R. When the current through a resistor doubles (from I to 2I) while resistance remains constant, the new power becomes P_new = (2I)²R = 4I²R = 4P_original. The power increases by a factor of four because power is proportional to the square of the current. Choice A incorrectly assumes power is directly proportional to current rather than to current squared. Remember: when using P = I²R, doubling the current quadruples the power.

Question 13

1. Resistor 1 dissipates three times the power of resistor 2 (correct answer)
2. Resistor 2 dissipates three times the power of resistor 1
3. Both resistors dissipate the same power
4. Both resistors dissipate the same energy but different power

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = IV. In parallel circuits, all components have the same voltage across them (9 V in this case), but currents can differ. Since I₁ = 3I₂ and both resistors have the same voltage, using P = IV gives P₁ = (9 V)(3I₂) = 3(9 V)(I₂) = 3P₂. Choice C incorrectly assumes equal power when only voltage is equal, ignoring that current differs. Remember: in parallel circuits with equal voltage, power is proportional to current.

Question 14

1. $R_2$ dissipates more power because it has higher resistance
2. $R_1$ dissipates more power because it draws more current (correct answer)
3. $R_1$ and $R_2$ dissipate equal power because voltage is the same
4. Neither dissipates power because the source is ideal

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = IV or P = V²/R. In parallel circuits, both resistors have the same voltage (4 V), but different currents: I₁ = V/R₁ = 4/2 = 2 A and I₂ = V/R₂ = 4/8 = 0.5 A. Using P = V²/R, we get P₁ = 16/2 = 8 W and P₂ = 16/8 = 2 W, so R₁ dissipates more power. Choice A incorrectly assumes higher resistance means more power in parallel circuits, when actually lower resistance allows more current and thus more power at constant voltage. Remember: in parallel circuits with equal voltage, lower resistance means higher power.

Question 15

1. greater because the total resistance is greater
2. less because the current is smaller (correct answer)
3. the same because the bulbs are identical
4. zero because the voltage is shared

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = I²R or P = V²/R. When two identical bulbs are in series, the total resistance doubles, causing the current to halve (I_series = V/2R compared to I_single = V/R). Each bulb now has half the original current, so using P = I²R, each bulb's power becomes P_series = (I/2)²R = I²R/4, which is one-fourth the power of a single bulb. Choice C incorrectly assumes identical bulbs always have the same power, ignoring that the circuit configuration affects current. Remember: in series circuits, increased total resistance reduces current, which reduces power in each component.

Question 16

1. Element A, because series elements share voltage equally
2. They dissipate the same power because the current is the same
3. Element B, because $P=IV$ and both have the same $I$ (correct answer)
4. Element A, because lower voltage means higher power

Explanation: This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, given by P = IV where I is current and V is voltage. Since elements A and B are in series, they carry the same current I, but element B has twice the voltage (6.0 V vs 3.0 V), so its power is P_B = I(6.0 V) = 2I(3.0 V) = 2P_A, making element B dissipate more power. Choice D incorrectly assumes equal current means equal power, ignoring the role of voltage differences. When comparing power in series circuits, remember that elements with higher voltage drops dissipate more power since they share the same current.

Question 17

1. the same because the current is constant
2. half as large because $P=VI$ and $V$ halves (correct answer)
3. twice as large because the time interval is $10\text{ s}$
4. one-fourth as large because voltage decreased by $6\text{ V}$

Explanation: This question tests understanding of electric power. Electric power is the instantaneous rate of electrical energy transfer, calculated as P = VI where V is voltage and I is current. With constant current I and voltage dropping from 12 V to 6 V, the instantaneous power at the end is P = (6 V)I = 0.5(12 V)I, which is half the initial power. Choice C incorrectly assumes constant current means constant power, ignoring that voltage also affects power. When analyzing time-varying circuits, focus on the instantaneous values of V and I at specific moments to calculate the power at those instants.

Question 18

1. greater, because less current means less energy wasted
2. smaller, because power depends only on time of operation
3. smaller, because with constant $V$ the power $P=IV$ decreases with $I$ (correct answer)
4. unchanged, because constant voltage means constant power

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = IV, where I is current and V is voltage. When the motor's current decreases while voltage remains constant, the product IV becomes smaller, resulting in decreased power. This makes sense because power represents energy transferred per unit time, and with less current (fewer charges per second) at the same voltage (energy per charge), less total energy is transferred per second. Choice A incorrectly suggests less current means less energy wasted, confusing efficiency with total power. Remember: when using P = IV, if one factor decreases while the other stays constant, power decreases proportionally.

Question 19

1. one-third as large because the voltage increased
2. three times as large because $P=IV$ and $I$ is constant (correct answer)
3. unchanged because the current is constant
4. five times as large because time increases energy

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = IV where I is current and V is voltage. With constant current I and voltage increasing from 2 V to 6 V (tripling), the power at t = 5 s is P_final = I(6 V) = 3I(2 V) = 3P_initial. Choice C incorrectly assumes power is constant when current is constant, forgetting that power depends on both current and voltage. The strategy is to recognize that power P = IV requires considering both factors; when one quantity changes while the other is constant, power changes proportionally.

Question 20

1. doubled because the resistance doubled
2. doubled because $P=I^2R$ and $R$ doubled (correct answer)
3. unchanged because the current is the same
4. halved because the current stays constant

Explanation: This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, which can be calculated using P = I²R when current is constant. With constant current I and resistance changing from R to 2R, the power changes from P_original = I²R to P_new = I²(2R) = 2I²R = 2P_original. This can be verified using P = IV: since V = IR doubles when R doubles at constant I, power P = I(2V) = 2IV doubles. Choice B incorrectly thinks constant current means constant power, missing that power also depends on resistance (or equivalently, voltage). When current is held constant, use P = I²R to see that power is proportional to resistance.