AP Physics 2: Algebra-Based Practice Quiz: Electric Power

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

In the context of an electric circuit, which of the following best describes electric power?

A)The total amount of energy stored in the circuit.B)The rate at which energy is transferred or dissipated.C)The total charge that passes a point in the circuit.D)The opposition to the flow of electric current.

All Questions (10)

In the context of an electric circuit, which of the following best describes electric power?

A) The total amount of energy stored in the circuit.

B) The rate at which energy is transferred or dissipated.

C) The total charge that passes a point in the circuit.

D) The opposition to the flow of electric current.

Correct Answer: B

Based on the provided content, power is defined as 'the rate at which energy is transferred, converted, or dissipated by a circuit element.' This means it is a measure of energy per unit time, not a total amount of energy or charge.

A small motor operates with a current of 0.5 A when connected to a 12 V battery. What is the power consumed by the motor?

A) 6 W

B) 12.5 W

C) 24 W

D) 0.042 W

Correct Answer: A

Using the fundamental equation for electric power, P = IΔV. Plugging in the given values: P = (0.5 A)(12 V) = 6 W.

Three light bulbs are operating in a circuit. Bulb 1 dissipates 40 W, Bulb 2 dissipates 100 W, and Bulb 3 dissipates 60 W. Which statement correctly describes their relative brightness?

A) Bulb 1 is the brightest.

B) Bulb 2 is the brightest.

C) Bulb 3 is the dimmest.

D) All bulbs have the same brightness.

Correct Answer: B

The content states that 'the brightness of a bulb increases with power.' Therefore, the bulb with the highest power dissipation will be the brightest. Bulb 2 dissipates 100 W, which is more than Bulb 1 (40 W) and Bulb 3 (60 W).

A heating coil with a resistance of 20 Ω has a steady current of 5 A flowing through it. At what rate is energy dissipated by the coil?

A) 100 W

B) 4 W

C) 500 W

D) 25 W

Correct Answer: C

The rate of energy dissipation is power. Using the derived equation P = I²R, we can calculate the power: P = (5 A)² * (20 Ω) = 25 * 20 = 500 W.

A resistor with a resistance of 150 Ω is connected across a 30 V potential difference. What is the power dissipated in the resistor?

A) 0.2 W

B) 5.0 W

C) 6.0 W

D) 4500 W

Correct Answer: C

Using the derived equation P = (ΔV)²/R, we can calculate the power: P = (30 V)² / (150 Ω) = 900 / 150 = 6.0 W.

A resistor is connected to a variable power supply. If the potential difference across the resistor is halved, by what factor does the power dissipated by the resistor change?

A) It is quartered.

B) It is halved.

C) It is doubled.

D) It is quadrupled.

Correct Answer: A

Power is related to potential difference by the equation P = (ΔV)²/R. Since power is proportional to the square of the potential difference, halving the voltage (ΔV/2) results in a new power of (ΔV/2)²/R = (ΔV)²/(4R) = P/4. The power is quartered.

If the current flowing through a light bulb of fixed resistance is doubled, how does the power dissipated by the bulb change?

A) The power is halved.

B) The power is doubled.

C) The power is quadrupled.

D) The power remains the same.

Correct Answer: C

Power is related to current by the equation P = I²R. Since power is proportional to the square of the current, doubling the current (2I) results in a new power of (2I)²R = 4I²R = 4P. The power is quadrupled.

Two resistors, R₁ = 10 Ω and R₂ = 20 Ω, are connected in parallel to a 10 V battery. Which resistor dissipates more power, and how can this be used to predict the brightness if they were light bulbs?

A) R₁ dissipates more power, so it would be brighter.

B) R₂ dissipates more power, so it would be brighter.

C) Both dissipate the same power and would have equal brightness.

D) It's impossible to determine without knowing the current.

Correct Answer: A

When connected in parallel, both resistors have the same potential difference (ΔV = 10 V). Using P = (ΔV)²/R, power is inversely proportional to resistance. The resistor with the lower resistance (R₁) will dissipate more power. For R₁, P = (10V)²/10Ω = 10 W. For R₂, P = (10V)²/20Ω = 5 W. Since brightness increases with power, R₁ would be brighter.

Two light bulbs, Bulb A with resistance R and Bulb B with resistance 2R, are connected in series to a power source. Which statement correctly compares their brightness?

A) Bulb A is brighter than Bulb B.

B) Bulb B is brighter than Bulb A.

C) Both bulbs have the same brightness.

D) Their relative brightness depends on the voltage of the source.

Correct Answer: B

When connected in series, both bulbs have the same current (I) flowing through them. The power dissipated can be calculated using P = I²R. Since the current is the same for both, power is directly proportional to resistance. The bulb with the higher resistance (Bulb B) will dissipate more power and therefore be brighter.

Which of the following expressions is NOT a valid equation for calculating electric power (P) in a resistive circuit element with resistance R, current I, and potential difference ΔV?

A) P = IΔV

B) P = I²R

C) P = (ΔV)²/R

D) P = I/R

Correct Answer: D

The provided content lists three valid equations for power: P = IΔV, P = I²R, and P = (ΔV)²/R. The expression P = I/R is not a valid equation for power; it does not have the correct units and does not represent the physical relationship.