Electric Power
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AP Physics 2 › Electric Power
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
larger, because a smaller current implies a larger resistance and thus larger power
larger, because the device has been on longer and used more energy
smaller, because $P=IV$ and $I$ is smaller while $V$ is constant
unchanged, because the voltage source is constant
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.
A resistor is connected across a constant-voltage source for $4\ \text{s}$. The source voltage is increased from $V$ to $2V$ while the resistor’s resistance stays constant. Compared to before, the resistor’s power dissipation is
half as large, because higher voltage reduces current draw
unchanged, because resistance is unchanged
four times as large, because $P=\dfrac{V^2}{R}$ and $R$ is constant
twice as large, because doubling voltage doubles power
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.
A $3.0\ \Omega$ resistor and a $6.0\ \Omega$ resistor are connected in parallel to a battery for several seconds. Each resistor has the same voltage $V$ across it. Compared to the $6.0\ \Omega$ resistor, the $3.0\ \Omega$ resistor’s power dissipation is
twice as large, because $P=\dfrac{V^2}{R}$ at the same voltage
four times as large, because its current is twice and power scales as $I^2$
the same, because both are connected to the same battery
half as large, because its resistance is half
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.
A resistor is connected to an ideal $9.0,\text{V}$ battery for $15,\text{s}$. If the resistance doubles while the battery voltage stays the same, the resistor’s power becomes
unchanged because the battery voltage is fixed
four times as large
half as large
twice as large
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.
A resistor is replaced by another so that the current stays the same but the voltage across the resistor doubles. Compared to before, the resistor’s power is
doubled, because with constant $I$ the power $P=IV$ doubles with $V$
quadrupled, because doubling voltage always quadruples power
halved, because higher voltage means fewer charges flow each second
unchanged, because the current is unchanged
Explanation
This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = IV. When the voltage doubles while current remains constant, the new power becomes P_new = I(2V) = 2IV = 2P_original. The power doubles because it is directly proportional to voltage when current is held constant. Choice D incorrectly claims doubling voltage always quadruples power, which would only be true if resistance were constant (P = V²/R) and current also doubled. When current is constant and voltage changes, power changes by the same factor as the voltage: use P = IV.
Two resistors are in series for $20,\text{s}$: $R_1=2.0,\Omega$ and $R_2=8.0,\Omega$. The same current $I$ flows through each. Which resistor dissipates more power?
Both, because they carry the same current
$R_2$, because it has larger resistance
Both, because series elements share the same voltage
$R_1$, because it has smaller resistance
Explanation
This question tests understanding of electric power. Electric power represents the rate of electrical energy transfer, and for resistors carrying current I, it's calculated as P = I²R. In series circuits, all elements carry the same current, so the resistor with larger resistance dissipates more power. For R₁: P₁ = I²(2.0 Ω) = 2.0I² W, while for R₂: P₂ = I²(8.0 Ω) = 8.0I² W, making R₂ dissipate four times more power than R₁. Choice C incorrectly claims series elements share the same voltage—they actually share the same current while voltages differ. When resistors in series carry the same current, remember that P = I²R shows larger resistance means more power dissipation.
A $10,\Omega$ resistor has $2.0,\text{A}$ through it for $12,\text{s}$. A $5.0,\Omega$ resistor has $1.0,\text{A}$ through it for the same time. Which dissipates more power?
They dissipate the same power because both are resistors
The $5.0,\Omega$ resistor, because lower resistance means more power
They dissipate the same power because the time is the same
The $10,\Omega$ resistor, because $P=I^2R$
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.
A motor operates so that its current decreases while its voltage remains constant. Compared to before, the motor’s electric power is
smaller, because with constant $V$ the power $P=IV$ decreases with $I$
unchanged, because constant voltage means constant power
greater, because less current means less energy wasted
smaller, because power depends only on time of operation
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.
Two devices run for the same time: Device 1 has $P=30\ \text{W}$, Device 2 has $P=10\ \text{W}$. Which transfers more electrical energy?
Device 1, because energy transferred in equal time is larger for larger power
They transfer equal energy because time is the same
Device 2, because lower power means energy is conserved better
They transfer equal energy because watts already include seconds
Explanation
This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, meaning energy = power × time (E = Pt). Device 1 with P = 30 W transfers E₁ = 30t joules in time t, while Device 2 with P = 10 W transfers E₂ = 10t joules in the same time. Since 30t > 10t for any positive time t, Device 1 transfers three times more energy than Device 2. Choice B incorrectly assumes lower power conserves energy better, confusing efficiency with total energy transfer. Remember: higher power means more energy transferred per unit time, so for equal operating times, higher power always means more total energy transferred.
A heater draws a constant current of $2.0\ \text{A}$ from a $12\ \text{V}$ battery for $30\ \text{s}$. Which statement about the heater’s power is correct?
It is $12\ \text{W}$ because the voltage sets the energy per charge only
It is $360\ \text{W}$ because power equals energy divided by time
It is $6\ \text{W}$ because the current is spread over $30\ \text{s}$
It is $24\ \text{W}$ because $P=IV$ and both $I$ and $V$ are constant
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.