This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, calculated as P = VI for any electrical device. With constant voltage of 24 V, the initial power is P₁ = (24 V)(1.0 A) = 24 W, and the final power is P₂ = (24 V)(2.0 A) = 48 W. The ratio P₂/P₁ = 48 W/24 W = 2, so the power is twice as large. Choice B incorrectly assumes power remains constant with fixed voltage, ignoring that current also affects power through P = VI. When voltage is constant, power is directly proportional to current: double the current means double the power consumption.

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.

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.

This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, which can be calculated using P = V²/R when voltage and resistance are known. For this 4 Ω resistor with 8 V across it, P = (8 V)²/(4 Ω) = 64/4 = 16 W. This can be verified using P = IV, where I = V/R = 8 V/4 Ω = 2 A, giving P = (2 A)(8 V) = 16 W. Choice D incorrectly uses P = VR, which has the wrong units (V·Ω = V²/A, not watts) and represents a nonsensical formula. When given voltage and resistance, use P = V²/R or first find current then use P = IV.

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.

This question tests understanding of electric power. Electric power is the rate of electrical energy transfer, given by P = I²R for a resistor with constant resistance. As current I decreases steadily over time while resistance R remains constant, the power P = I²R decreases because it depends on the square of the current. Choice C incorrectly assumes constant resistance means constant power, forgetting that power depends on both current and resistance. For a fixed resistance, power varies with the square of current: halving current reduces power to one-fourth, demonstrating why decreasing current means decreasing power.

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 voltage V is held constant across a resistor and current decreases from 2.0 A to 1.0 A (halved), the power changes from P_initial = (2.0 A)(V) to P_final = (1.0 A)(V) = 0.5(2.0 A)(V) = 0.5P_initial. Choice D incorrectly applies P = I²R, forgetting that when V is constant and I changes, R must also change (the resistor is not fixed). The key strategy is to identify which quantities are constant and use the appropriate power formula: when V is constant, use P = IV directly.

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.

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.

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.