AP Physics C Electricity and Magnetism Quiz: Electric Power

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Electric PowerQuestion 1 of 20

A single resistor $R_1$ is connected to an ideal battery, dissipating a total power $P$ . A second resistor $R_2$ is then connected in series with $R_1$ . How does the new total power dissipated by the circuit compare to $P$ ?

The new total power is greater than

P

because the total resistance has increased.

The new total power is less than

P

because the total resistance has increased.

The new total power is equal to

P

because the battery's voltage remains constant.

The change in total power depends on whether

R_2

is greater or less than

R_1

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Power

Practice Electric Power in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 C Electricity and Magnetism.

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.

All questions

Question 1

The new total power is greater than $P$ because the total resistance has increased.
The new total power is less than $P$ because the total resistance has increased. (correct answer)
The new total power is equal to $P$ because the battery's voltage remains constant.
The change in total power depends on whether $R_2$ is greater or less than $R_1$ .

Explanation: The initial power is

P = V^2/R_1

. When

R_2

is added in series, the new total resistance is

R_{\text{series}} = R_1 + R_2

. Since

R_1+R_2 > R_1

, the new total resistance is greater than the original resistance. The new total power is

P_{\text{new}} = V^2/R_{\text{series}}

. Because the denominator has increased, the new total power is less than the original power

P

Question 2

A cylindrical wire has length $L$ , radius $r$ , and is made of a material with resistivity $\rho$ . If a potential difference $V$ is applied between the ends of the wire, what is the power dissipated in the wire?

$\frac{V^2 \rho L}{\pi r^2}$
$\frac{V \pi r^2}{\rho L}$
$\frac{V^2 L}{\rho \pi r^2}$
$\frac{V^2 \pi r^2}{\rho L}$ (correct answer)

Explanation: First, find the resistance of the wire using the formula

R = \frac{\rho L}{A}

, where the cross-sectional area

A = \pi r^2

. So,

R = \frac{\rho L}{\pi r^2}

. Then, use the power formula

P = \frac{V^2}{R}

. Substituting the expression for R gives

P = \frac{V^2}{\rho L / (\pi r^2)} = \frac{V^2 \pi r^2}{\rho L}

Question 3

A student analyzes a series circuit with a battery and two different resistors, $R_1$ and $R_2$ . The student claims, "The total power dissipated by the two resistors in series is the sum of the powers each would dissipate if connected individually to the same battery." Which of the following is a correct evaluation of this claim?

The claim is correct, because power is a scalar quantity and is conserved in the circuit.
The claim is incorrect, because the current in the series circuit is different from the current through each resistor when connected individually. (correct answer)
The claim is correct, because total resistance is the sum of individual resistances in a series circuit.
The claim is incorrect, because the voltage from the battery is split between the two resistors in series.

Explanation: The claim is incorrect. Let the battery voltage be

V

. The power dissipated by

R_1

alone is

P_1 = V^2/R_1

, and by

R_2

alone is

P_2 = V^2/R_2

. Their sum is

V^2(1/R_1 + 1/R_2)

. In series, the total resistance is

R_s = R_1 + R_2

, and the total power is

P_s = V^2/(R_1 + R_2)

. These expressions are not equal. The physical reason is that the current drawn from the battery changes. In the series circuit, the current is

I_s = V/(R_1+R_2)

, which is smaller than the current through either resistor when connected alone.

Question 4

A resistor of resistance $R$ is connected to Battery A, which has emf $\mathcal{E}$ and internal resistance $r_A$ , dissipating power $P_A$ . The same resistor is then connected to Battery B, which also has emf $\mathcal{E}$ but a smaller internal resistance $r_B < r_A$ . The power dissipated in R is now $P_B$ . How does $P_A$ compare to $P_B$ ?

$P_A < P_B$ (correct answer)
$P_A > P_B$
$P_A = P_B$
The relationship cannot be determined without knowing the value of $R$ .

Explanation: The power dissipated in the external resistor is

P = I^2 R = \left(\frac{\mathcal{E}}{R+r}\right)^2 R

. Since

\mathcal{E}

and

R

are the same for both cases, power depends on the total resistance

R+r

. As

r_B < r_A

, the total resistance of the circuit with Battery B,

R+r_B

, is less than that with Battery A,

R+r_A

. A smaller total resistance results in a larger current. Since power is proportional to the square of the current,

P_B > P_A

Question 5

A capacitor of capacitance $C$ is charged to an initial potential difference $V_0$ . At time $t=0$ , it is connected across a resistor $R$ and begins to discharge. What is the total energy that will be dissipated as heat by the resistor during the entire discharging process?

$C V_0^2$
$\frac{1}{2} C V_0^2$ (correct answer)
$\frac{V_0^2}{R}$
Zero, as the energy returns to the circuit.

Explanation: By the principle of conservation of energy, the total energy dissipated by the resistor must be equal to the total energy initially stored in the capacitor. The initial energy stored in the capacitor is given by

U_C = \frac{1}{2} C V_0^2

. This entire amount of energy is converted into thermal energy in the resistor as the capacitor discharges to zero potential.

Question 6

Consider a circuit with two loops. The first loop contains a battery $\mathcal{E}_1$ and two resistors, $R_1$ and $R_2$ . The second loop contains a battery $\mathcal{E}_2$ and two resistors, $R_2$ and $R_3$ . Resistor $R_2$ is common to both loops. All resistances and EMFs are positive.

To determine which resistor in the circuit dissipates the most power, which of the following provides the sufficient and necessary information?

The current through each resistor, because the resistor with the largest current will dissipate the most power.
The potential difference across each resistor, because the resistor with the largest potential difference will dissipate the most power.
The resistance of each resistor, because the resistor with the largest resistance will dissipate the most power.
The product of the current through and the potential difference across each resistor. (correct answer)

Explanation: The power dissipated by a resistor is given by

P = IV = I^2R = V^2/R

. Since the resistors are not identical and the currents and voltages are generally different for each, knowing only the current (A), voltage (B), or resistance (C) is insufficient. For example, a large current in a small resistor may dissipate less power than a smaller current in a large resistor. One must calculate the power for each resistor using one of the formulas, which requires knowing

I

and

V

, or

I

and

R

, or

V

and

R

. The product

IV

is the definition of power and is always sufficient.

Question 7

A 18.0 V battery powers a portable circuit with two resistors in parallel, $R_1=9.0\,\Omega$ and $R_2=18.0\,\Omega$ , giving total resistance $R_\text{eq}=6.0\,\Omega$ . The total current drawn from the battery is $I=3.0\,\text{A}$ . Use $P=IV$ to compute the power supplied by the battery. Report power in watts (W). Calculate the power supplied by the battery.

6.0 W
54 W (correct answer)
162 W
54 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an 18.0 V battery with total current I=3.0 A, requiring calculation of power supplied by the battery. Choice B (54 W) is correct because it applies the formula P=IV using values I=3.0 A and V=18.0 V, resulting in P=(3.0 A)(18.0 V)=54 W. Choice D (54 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that battery power output is always P=IV where I is total current drawn and V is battery voltage. Practice with parallel circuits to understand how total current relates to individual branch currents.

Question 8

A 9.0 V battery powers a wearable device modeled as two resistors in series: $R_1=3.0\,\Omega$ and $R_2=6.0\,\Omega$ , so the total resistance is $9.0\,\Omega$ . The same current flows through both resistors and equals $1.0\,\text{A}$ . Determine the power dissipated by $R_2$ using $P=I^2R$ . Give your answer in watts (W). What is the power dissipated by the resistor in the circuit?

6.0 W (correct answer)
3.0 W
9.0 W
6.0 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes two resistors in series with R₂=6.0 Ω and current I=1.0 A, requiring calculation of power dissipated by R₂. Choice A (6.0 W) is correct because it applies the formula P=I²R using values I=1.0 A and R=6.0 Ω, resulting in P=(1.0 A)²(6.0 Ω)=6.0 W. Choice D (6.0 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize unit consistency and that power dissipation in a resistor is always P=I²R when current is known. Practice recognizing series circuits where the same current flows through all components.

Question 9

A long extension cord is modeled as a $R=0.20\,\Omega$ resistor in series with a load, and the current through the cord is $I=8.0\,\text{A}$ . The cord warms during operation, indicating resistive heating in the wire. Use $P=I^2R$ to determine the power lost as heat in the cord. Give your answer in watts (W). How much power is lost in the transmission line?

1.6 W
12.8 W (correct answer)
40 W
12.8 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an extension cord with R=0.20 Ω carrying I=8.0 A, requiring calculation of resistive power loss. Choice B (12.8 W) is correct because it applies the formula P=I²R using values I=8.0 A and R=0.20 Ω, resulting in P=(8.0 A)²(0.20 Ω)=64×0.20=12.8 W. Choice D (12.8 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that power loss in wires is continuous dissipation measured in watts. Practice calculating wire losses to understand why thick wires are used for high currents.

Question 10

A reading lamp uses a bulb modeled as a $R=60\,\Omega$ resistor connected to a $12\,\text{V}$ supply in a simple lighting circuit. The voltage across the bulb remains $12\,\text{V}$ during operation. Use $P=\frac{V^2}{R}$ to compute the bulb's power usage. Report the result in watts (W). What is the power usage of the light bulb?

2.4 W (correct answer)
0.20 W
24 W
2.4 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 60 Ω bulb connected across 12 V, requiring calculation of power using P=V²/R. Choice A (2.4 W) is correct because it applies the formula P=V²/R using values V=12 V and R=60 Ω, resulting in P=(12 V)²/(60 Ω)=144/60=2.4 W. Choice D (2.4 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize consistent unit usage and that low-voltage bulbs typically consume a few watts. Practice with various voltage and resistance combinations to build computational fluency.

Question 11

A transmission line is modeled as a resistor of $R=0.80\,\Omega$ delivering power to a remote load, and the line current is $I=15\,\text{A}$ . The line's heating is a safety concern, so you calculate resistive loss. Use $P=I^2R$ with current in amperes and resistance in ohms. Give the power loss in watts (W). How much power is lost in the transmission line?

12 W
180 W (correct answer)
225 W
180 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a transmission line with R=0.80 Ω carrying I=15 A, requiring calculation of resistive power loss. Choice B (180 W) is correct because it applies the formula P=I²R using values I=15 A and R=0.80 Ω, resulting in P=(15 A)²(0.80 Ω)=225×0.80=180 W. Choice D (180 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that transmission line losses are calculated using P=I²R and represent continuous power dissipation. Practice with real-world examples to understand why minimizing transmission losses is important.

Question 12

A household night-light is modeled as a single resistor of $R=240\,\Omega$ connected directly across a $120\,\text{V}$ outlet. The circuit is steady-state and the resistor is the only load. Use $P=\frac{V^2}{R}$ to find the electrical power converted to thermal energy and light. Report the result in watts (W). Determine the power consumed by the appliance.

0.50 W
60 W (correct answer)
30 W
60 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 240 Ω resistor connected across 120 V, requiring calculation of power using P=V²/R. Choice B (60 W) is correct because it applies the formula P=V²/R using values V=120 V and R=240 Ω, resulting in P=(120 V)²/(240 Ω)=14400/240=60 W. Choice A (0.50 W) is incorrect due to likely calculating I=V/R=0.5 A without completing the power calculation, often a result of stopping midway through the problem. To help students, emphasize completing all steps and choosing the appropriate power formula based on given quantities. Practice with household appliance examples to build intuition for typical power values.

Question 13

A space heater is modeled as a single resistor of $R=24\,\Omega$ connected across a $120\,\text{V}$ outlet. The heater is the only load, so the voltage across the resistor equals the source voltage. Use $P=\frac{V^2}{R}$ to compute the electrical power converted to heat. Report the power in watts (W). Determine the power consumed by the appliance.

600 W (correct answer)
5.0 W
300 W
600 J

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 24 Ω heater connected across 120 V, requiring calculation of power using P=V²/R. Choice A (600 W) is correct because it applies the formula P=V²/R using values V=120 V and R=24 Ω, resulting in P=(120 V)²/(24 Ω)=14400/24=600 W. Choice D (600 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that space heaters typically consume hundreds of watts and always check units. Practice with common household appliances to build intuition for realistic power values.

Question 14

A desk lamp uses a bulb modeled as a resistor of $R=144\,\Omega$ connected across a $12.0\,\text{V}$ DC supply. The lamp is used in a low-voltage lighting circuit with negligible wire resistance. Use $P=\frac{V^2}{R}$ to determine the bulb's power usage. Report the answer in watts (W). What is the power usage of the light bulb?

1.0 W (correct answer)
12 W
0.083 W
144 W

Explanation: This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 144 Ω bulb connected across 12.0 V DC, requiring calculation of power using P=V²/R. Choice A (1.0 W) is correct because it applies the formula P=V²/R using values V=12.0 V and R=144 Ω, resulting in P=(12.0 V)²/(144 Ω)=144/144=1.0 W. Choice D (144 W) is incorrect due to using V² without dividing by R, often a result of incomplete formula application. To help students, emphasize careful formula application and unit checking. Practice with low-voltage DC circuits to build familiarity with typical power values.

Question 15

A $50 \, \Omega$ resistor is connected to a power source, resulting in a constant current of $0.50 \, \text{A}$ . How much total thermal energy is dissipated in the resistor during a $2.0$ -minute interval?

$12.5 \, \text{J}$
$25.0 \, \text{J}$
$750 \, \text{J}$
$1500 \, \text{J}$ (correct answer)

Explanation: First, calculate the power dissipated:

P = I^2R = (0.50 \, \text{A})^2 (50 \, \Omega) = 12.5 \, \text{W}

. Next, convert the time interval to seconds:

t = 2.0 \, \text{min} \times 60 \, \text{s/min} = 120 \, \text{s}

. The total energy dissipated is

E = P \times t = 12.5 \, \text{W} \times 120 \, \text{s} = 1500 \, \text{J}

Question 16

A single resistor $R_1$ is connected to an ideal battery, resulting in a total power $P$ delivered by the battery. A second resistor $R_2$ is then connected in parallel with $R_1$ . How does the new total power delivered by the battery compare to $P$ ?

The new total power is greater than $P$ because the equivalent resistance has decreased. (correct answer)
The new total power is less than $P$ because the equivalent resistance has decreased.
The new total power is equal to $P$ because the battery's voltage is applied to the circuit.
The change in total power depends on whether $R_2$ is greater or less than $R_1$ .

Explanation: The initial power is

P = V^2/R_1

. When

R_2

is added in parallel, the new equivalent resistance is

R_{\text{parallel}} = (1/R_1 + 1/R_2)^{-1}

. This equivalent resistance is always less than

R_1

. The new total power is

P_{\text{new}} = V^2/R_{\text{parallel}}

. Because the denominator (the equivalent resistance) has decreased, the new total power delivered by the battery is greater than the original power

P

Question 17

The current $I$ flowing through a resistor $R$ is given as a function of time $t$ by the expression $I(t) = I_0 e^{-t/\tau}$ , where $I_0$ and $\tau$ are positive constants. What is the instantaneous power $P(t)$ being dissipated by the resistor?

$P(t) = I_0^2 R e^{-t/\tau}$
$P(t) = I_0 R e^{-t/\tau}$
$P(t) = I_0^2 R (1 - e^{-t/\tau})$
$P(t) = I_0^2 R e^{-2t/\tau}$ (correct answer)

Explanation: Instantaneous power is given by the formula

P(t) = [I(t)]^2 R

. Substituting the given expression for the current:

P(t) = (I_0 e^{-t/\tau})^2 R = I_0^2 (e^{-t/\tau})^2 R = I_0^2 R e^{-2t/\tau}

Question 18

Two identical light bulbs, each with resistance $R$ , are connected to an ideal battery with voltage $V$ . What is the ratio of the total power consumed when they are in parallel to the total power consumed when they are in series, $P_{\text{parallel}}/P_{\text{series}}$ ?

$1/4$
$1/2$
$2$
$4$ (correct answer)

Explanation: In series, the total resistance is

R_{\text{series}} = R + R = 2R

. The total power is

P_{\text{series}} = V^2 / R_{\text{series}} = V^2 / (2R)

. In parallel, the equivalent resistance is

R_{\text{parallel}} = (1/R + 1/R)^{-1} = R/2

. The total power is

P_{\text{parallel}} = V^2 / R_{\text{parallel}} = V^2 / (R/2) = 2V^2/R

. The ratio is

\frac{P_{\text{parallel}}}{P_{\text{series}}} = \frac{2V^2/R}{V^2/(2R)} = 4

Question 19

In a series RC circuit connected to a battery of emf $\mathcal{E}$ , the switch is closed at $t=0$ . Which statement correctly describes the energy transformation in the circuit during the charging process?

The power delivered by the battery is constant and is equally split between the resistor and capacitor.
The total power delivered by the battery is converted entirely into thermal energy in the resistor.
The instantaneous power delivered by the battery equals the sum of the rate of energy dissipation in the resistor and the rate of energy storage in the capacitor. (correct answer)
The rate of energy storage in the capacitor is constant, while the rate of dissipation in the resistor decreases exponentially.

Explanation: The power delivered by the battery is

P_{batt} = I(t)\mathcal{E}

. This power is used for two purposes: dissipating energy as heat in the resistor at a rate

P_R = I(t)^2 R

, and storing energy in the capacitor's electric field at a rate

P_C = dU_C/dt = V_C(t)I(t)

. By conservation of energy,

P_{batt} = P_R + P_C

. Since the current

I(t)

changes with time, these are all instantaneous rates.

Question 20

An inductor $L$ and a resistor $R$ are connected in series with a battery of emf $\mathcal{E}$ and a switch. The switch is closed at time $t=0$ . What is the power dissipated by the resistor as a function of time $t$ ?

$\frac{\mathcal{E}^2}{R} e^{-2Rt/L}$
$\frac{\mathcal{E}^2}{R} (1 - e^{-Rt/L})$
$\frac{\mathcal{E}^2}{R} (1 - e^{-Rt/L})^2$ (correct answer)
$\frac{\mathcal{E}^2}{R}$

Explanation: The current in a charging LR circuit is

I(t) = \frac{\mathcal{E}}{R}(1 - e^{-Rt/L})

. The power dissipated in the resistor is

P_R(t) = (I(t))^2 R

. Substituting the expression for

I(t)

gives

P_R(t) = \left[\frac{\mathcal{E}}{R}(1 - e^{-Rt/L})\right]^2 R = \frac{\mathcal{E}^2}{R^2}(1 - e^{-Rt/L})^2 R = \frac{\mathcal{E}^2}{R}(1 - e^{-Rt/L})^2