Electric Power - AP Physics 2

A resistor $R$ is connected to an ideal battery. The battery voltage doubles while $R$ stays constant, so the current doubles. Compared to before, the resistor's power dissipation becomes
Answer: four times as large because $P=IV$ and both doubled This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = IV. When both voltage V and current I double, the new power becomes P_new = (2I)(2V) = 4IV = 4P_original. This can also be verified using P = V²/R: if V doubles and R is constant, P increases by a factor of 2² = 4. Choice A incorrectly thinks doubling both quantities only doubles the power, missing that power is the product of voltage and current. When analyzing power changes, remember that P = IV means power scales with the product of changes in both V and I.
A resistor's voltage doubles while its current also doubles for a brief interval. Compared to before, the resistor's instantaneous power is
Answer: four times as large, because $P=IV$ and both factors double This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = IV. When both voltage and current double, the new power becomes P_new = (2I)(2V) = 4IV = 4P_original. This quadrupling occurs because power depends on the product of two factors that each double, resulting in a factor of 2 × 2 = 4 increase. Choice A incorrectly considers only the voltage doubling, missing that current also doubles. To analyze power changes, always check how both voltage AND current change, then multiply the factors together.
A $3\ \Omega$ resistor has a constant current of $2\ \text{A}$ for $4\ \text{s}$ . The voltage across it is $V=IR$ and the power is $P=IV$ . Which statement about its power is correct?
Answer: The power is $12\ \text{W}$ This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, not the total energy transferred. Using V = IR, the voltage is V = (2 A)(3 Ω) = 6 V, and using P = IV, the power is P = (2 A)(6 V) = 12 W. Power has units of watts (W), which represents joules per second, not joules. Choices C and D incorrectly give answers in joules, confusing power with energy (energy would be P × t = 12 W × 4 s = 48 J). Remember: power is measured in watts and represents the rate of energy transfer, not the total energy.
A device is connected to a constant $6.0\ \text{V}$ source for several seconds. During this interval, its current decreases from $2.0\ \text{A}$ to $1.0\ \text{A}$ . Compared to the start, the device's electric power is
Answer: smaller, because $P=IV$ and $I$ is smaller while $V$ is constant 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.
A battery provides a constant terminal voltage of $6\ \text{V}$ . In case X the current is $1\ \text{A}$ ; in case Y it is $3\ \text{A}$ . Compared to X, the power in Y is
Answer: three times as large, because $P=IV$ and $V$ is constant This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = IV. With the battery providing constant voltage of 6 V, case X has $P_X$ = (6 V)(1 A) = 6 W, while case Y has $P_Y$ = (6 V)(3 A) = 18 W. Since 18 W is three times 6 W, the power in case Y is three times larger than in case X. Choice C incorrectly claims power is always proportional to I², which only applies when resistance is constant (P = I²R), not when voltage is constant. When voltage is held constant, power increases linearly with current: P = IV.
A heating element is connected to a $120\ \text{V}$ outlet. For the first minute, the current is $5.0\ \text{A}$ ; for the second minute, the current is $2.5\ \text{A}$ . The voltage remains $120\ \text{V}$ throughout. During the second minute, the power delivered to the element is
Answer: half as large, because $P=IV$ and $I$ is half while $V$ is constant This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = IV. During the first minute, power is P₁ = 5.0 A × 120 V = 600 W, while during the second minute, power is P₂ = 2.5 A × 120 V = 300 W. Since the current halved while voltage remained constant, the power also halved from 600 W to 300 W. Choice D incorrectly claims power is quartered, confusing the P = I²R formula (which applies when resistance is constant) with this scenario where voltage is constant. To find power changes, identify what's constant (here, voltage) and apply the appropriate power formula.
Two resistors, $R$ and $2R$ , are connected in parallel to the same battery for $10\,\text{s}$ . Compared to the $2R$ resistor, the power dissipated by the $R$ resistor is
Answer: twice as large This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and for resistors in parallel with the same voltage V, power is P = V²/R. The R resistor dissipates $P_R$ = V²/R, while the 2R resistor dissipates P_2R = V²/(2R) = (V²/R)/2 = $P_R$ /2. Therefore, the R resistor dissipates twice as much power as the 2R resistor. Choice A incorrectly reverses this relationship, suggesting the smaller resistor dissipates less power when it actually dissipates more. In parallel circuits with constant voltage, power is inversely proportional to resistance: half the resistance means double the power.
A motor operates so that its current decreases while its voltage remains constant. Compared to before, the motor's electric power is
Answer: smaller, because with constant $V$ the power $P=IV$ decreases with $I$ 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.
A device operates at constant voltage $V$ for $5\,\text{s}$ . Its current decreases from $3.0\,\text{A}$ to $1.0\,\text{A}$ . Compared to the start, the device's power is now
Answer: one-third as large This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = VI where V is voltage and I is current. At constant voltage V, the initial power is P₁ = V(3.0 A) = 3.0V W, while the final power is P₂ = V(1.0 A) = 1.0V W. The ratio P₂/P₁ = 1.0V/3.0V = 1/3, so the power is one-third as large. Choice B incorrectly assumes power stays constant with voltage—but power depends on both voltage and current, not voltage alone. When voltage is constant, power is directly proportional to current: if current drops to one-third, power also drops to one-third.
Two resistors $R_A$ and $R_B$ are connected in parallel across a $10\ \text{V}$ battery for $5\ \text{s}$ . The branch currents are $I_A=0.20\ \text{A}$ and $I_B=0.50\ \text{A}$ , so each resistor has the same voltage $10\ \text{V}$ . Which resistor dissipates the most power?
Answer: $R_B$ , because it has the larger current at the same voltage This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = IV. In parallel circuits, all branches share the same voltage but can have different currents. With both resistors at 10 V, we find $P_A$ = 0.20 A × 10 V = 2.0 W and $P_B$ = 0.50 A × 10 V = 5.0 W, so $R_B$ dissipates more power due to its larger current. Choice C incorrectly assumes equal voltage means equal power, missing that current also matters in the power calculation. When comparing power in parallel branches, focus on current differences since voltage is the same.
Two resistors are in series for $20\ \text{s}$ : $R_1=2\ \Omega$ and $R_2=6\ \Omega$ , with the same current through both. Which dissipates more power?
Answer: $R_2$ , because with equal current $P=I^2R$ is larger for larger $R$ 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.
In a circuit, element X has a constant potential difference $V_X$ while the current through it increases from $I$ to $2I$ over a short time interval. Which statement about X's instantaneous power is correct?
Answer: It doubles because $P=VI$ and $V_X$ is constant This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = VI, where V is voltage and I is current. Since element X maintains a constant potential difference $V_X$ while the current doubles from I to 2I, the power becomes P = $V_X$ (2I) = 2V_X I, which is twice the original power. Choice B incorrectly assumes power depends only on voltage, ignoring that current also changed. When analyzing power problems, remember that power depends on the product of voltage AND current, so changing either factor affects the total power.
A resistor is connected to a constant current source so that the current stays at $I$ . Over time, the resistor heats up and its resistance increases. The power dissipated by the resistor
Answer: increases, because $P=I^2R$ This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and when current is held constant, power is calculated as P = I²R. A constant current source maintains the same current I regardless of resistance changes. As the resistor heats up and R increases while I stays constant, the power P = I²R must increase proportionally with resistance. Choice B incorrectly assumes higher resistance always means less power, but this is only true for constant voltage sources (P = V²/R), not constant current sources. When dealing with constant current sources, remember that power increases with resistance according to P = I²R.
Two identical bulbs are connected in series to an ideal battery and glow for $20\ \text{s}$ . The same current flows through each bulb, and each bulb has the same resistance $R$ . Compared to one bulb alone connected to the same battery, the power in each series bulb is
Answer: less because the current is smaller 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.
A device transfers $120\text{ J}$ of electrical energy to thermal energy in $4.0\text{ s}$ while the current remains steady. Which statement about the device's average power is correct?
Answer: It is $30\text{ W}$ because $P=\dfrac{\Delta E}{\Delta t}$ This question tests understanding of electric power. Electric power is the rate of energy transfer, defined as P = ΔE/Δt where ΔE is the energy transferred and Δt is the time interval. The device transfers 120 J in 4.0 s, so the average power is P = 120 J / 4.0 s = 30 W. Choice B incorrectly multiplies energy by time (480 J·s), confusing the power formula and producing units of J·s instead of watts. Remember that power equals energy divided by time, not multiplied, and always check that your answer has units of watts (W) or joules per second (J/s).
A $12\ \text{V}$ battery is connected to two resistors, $R_1=6\ \Omega$ and $R_2=3\ \Omega$ , in series for $10\ \text{s}$ . The same current $I$ flows through both resistors, and the voltage drops satisfy $V_1=IR_1$ and $V_2=IR_2$ with $V_1+V_2=12\ \text{V}$ . Which element dissipates the most power during the interval?
Answer: $R_1$ dissipates more power than $R_2$ 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 drops are different: V₁ = IR₁ and V₂ = IR₂. Since R₁ = 6 Ω is larger than R₂ = 3 Ω, and they share the same current, R₁ has a larger voltage drop (V₁ = 2V₂). Using P = I²R, we see that R₁ dissipates twice the power of R₂ because its resistance is twice as large. Choice D incorrectly assumes no power dissipation when current flows through resistance. Remember: in series circuits with equal current, power is proportional to resistance.
A motor draws a constant current of $0.50\ \text{A}$ . Its terminal voltage decreases from $12\ \text{V}$ to $9\ \text{V}$ over time. Compared to initially, the motor's electrical power input is
Answer: smaller because $P=IV$ and $V$ decreased This question tests understanding of electric power. Electric power is the rate at which electrical energy is transferred, calculated as P = IV. With constant current I = 0.50 A and voltage decreasing from 12 V to 9 V, the power changes from P_initial = (0.50 A)(12 V) = 6.0 W to P_final = (0.50 A)(9 V) = 4.5 W, which is 75% of the initial power (smaller). Choice C incorrectly claims power depends only on current, ignoring that P = IV requires both current and voltage. When current is constant, power is directly proportional to voltage: decreasing voltage means decreasing power.
A resistor is connected to a variable power supply for $8\ \text{s}$ . The supply is adjusted so that the current through the resistor doubles while the resistor's resistance stays constant. Which statement about the resistor's power dissipation is correct?
Answer: It quadruples, because $P=I^2R$ and $R$ is constant This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and when resistance is constant, we use P = I²R. If current doubles from I to 2I while resistance R stays constant, the new power becomes P_new = (2I)²R = 4I²R, which is four times the original power P = I²R. This quadrupling occurs because power depends on the square of current when resistance is fixed. Choice A incorrectly assumes a linear relationship between power and current, missing the squared dependence. When resistance is constant, remember that power scales with the square of current, not linearly.
Two bulbs A and B operate for the same time interval. Bulb A has instantaneous values $V_A=10\text{ V}$ and $I_A=0.40\text{ A}$ ; bulb B has $V_B=5\text{ V}$ and $I_B=0.80\text{ A}$ . Which statement is correct?
Answer: They dissipate the same power because $V_AI_A=V_BI_B$ 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.
Two identical bulbs are connected to the same battery in different trials for $30\ \text{s}$ . In Trial 1, a single bulb is connected. In Trial 2, two bulbs are connected in series. Which statement about the power of one bulb is correct?
Answer: It is smaller in Trial 2, because the series current is smaller This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, and depends on both voltage and current through P = IV. In Trial 1, one bulb gets the full battery voltage V and draws current I₁ = V/R. In Trial 2, two bulbs in series share the voltage, so each gets V/2, and the series current is I₂ = V/(2R) = I₁/2. The power for one bulb in Trial 2 is P₂ = (I₁/2) × (V/2) = I₁V/4, which is one-fourth the power in Trial 1. Choice A incorrectly focuses on total resistance rather than individual bulb conditions. In series connections, adding components reduces the power to each individual component.

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