AP Physics 2: Algebra-Based, AP exams, 15–18%

Unit 11: Electric Circuits

Unit 11 covers steady-current circuits and the conservation principles behind them. Resistivity, resistance, and Ohm's law lead into series and parallel analysis, and Kirchhoff's rules formalize conservation of charge and energy in a circuit. The revised framework gives circuits a fuller treatment, including capacitors.

Position: 3 of 7 units
Published weighting: 15–18% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics 2: Algebra-Based. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Electric current and charge conservation: Current as rate of charge flow, and the direction conventions used in circuit analysis.
Resistivity and resistance: How material, length, and cross-sectional area determine resistance, and Ohm's law as a model.
Series and parallel circuits: Equivalent resistance, and how current and potential difference distribute in each arrangement.
Kirchhoff's rules: The junction rule from conservation of charge and the loop rule from conservation of energy.
Power in circuits: Energy dissipation in resistors and the several equivalent expressions for electrical power.
Capacitors in circuits: Capacitance, stored charge and energy, and the behavior of capacitors in series and parallel.

Study tips

How to practice Electric Circuits.

Section-specific guidance that follows from how this material is usually tested.

Redraw every complicated circuit in a simpler equivalent form before applying any rule.
Predict qualitatively what happens when a bulb is added or removed, then verify with calculation.
Practice Kirchhoff's loop rule with careful sign tracking around the loop; sign errors dominate mistakes here.

Where this fits

Unit 11: Electric Circuits is 3 of 7 units in the AP Physics 2: Algebra-Based guide, and it carries a published weighting of 15–18% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics 2: Algebra-Based guide Official exam information

Sample Electric Circuits questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A 4.0 ohm resistor is connected across an ideal battery that maintains a potential difference of 12 V. What is the current through the resistor?
1. A0.33 A
2. B3.0 A
3. C8.0 A
4. D48 A
Show answer
Answer: B. 3.0 A
Ohm's law gives I = V/R = 12 V / 4.0 ohms = 3.0 A. The 48 A choice multiplies voltage by resistance instead of dividing, and 0.33 A inverts the ratio.
Key insight: For a fixed potential difference, current and resistance are inversely related.
Question 2Foundational
A 6.0 ohm resistor and a 3.0 ohm resistor are connected in parallel with each other. What is the equivalent resistance of the combination?
1. A0.50 ohms
2. B2.0 ohms
3. C4.5 ohms
4. D9.0 ohms
Show answer
Answer: B. 2.0 ohms
For two resistors in parallel the equivalent resistance is the product over the sum: (6.0)(3.0)/(6.0 + 3.0) = 18/9 = 2.0 ohms. The 9.0 ohm choice adds the resistances, which applies to a series connection instead.
Key insight: Adding a parallel path always lowers total resistance below the smallest branch, because it gives charge another route.
Question 3Intermediate
A single resistor dissipates 24 W of power when connected across an ideal 12 V battery. What is the resistance of the resistor?
1. A0.50 ohms
2. B2.0 ohms
3. C6.0 ohms
4. D288 ohms
Show answer
Answer: C. 6.0 ohms
Using P = V^2/R gives R = V^2/P = 144/24 = 6.0 ohms, and the current is 2.0 A as a check, since (12 V)(2.0 A) = 24 W. The 288 ohm choice multiplies the power by the voltage instead of dividing the squared voltage by the power.
Key insight: Three equivalent power expressions, IV, I^2R, and V^2/R, let you start from whichever two quantities are given.
Question 4Intermediate
Resistors of 2.0 ohms, 3.0 ohms, and 5.0 ohms are connected in series across an ideal 20 V battery. What is the potential difference across the 3.0 ohm resistor?
1. A2.0 V
2. B6.0 V
3. C6.7 V
4. D20 V
Show answer
Answer: B. 6.0 V
The series resistance is 10 ohms, so the current everywhere is 20 V / 10 ohms = 2.0 A, and the 3.0 ohm resistor drops (2.0 A)(3.0 ohms) = 6.0 V. The 6.7 V choice splits the 20 V evenly among the three resistors, ignoring that a larger resistance takes a larger share.
Key insight: Series elements share the same current, so potential difference divides in proportion to resistance.
Question 5Advanced
A 4.0 x 10^-6 F capacitor is fully charged by a 50 V source. How much energy is stored in the capacitor?
1. A1.0 x 10^-4 J
2. B2.0 x 10^-4 J
3. C5.0 x 10^-3 J
4. D1.0 x 10^-2 J
Show answer
Answer: C. 5.0 x 10^-3 J
Stored energy is (1/2)CV^2 = (1/2)(4.0 x 10^-6)(50)^2 = 5.0 x 10^-3 J. The 2.0 x 10^-4 choice is the stored charge CV in coulombs, and 1.0 x 10^-2 J omits the factor of one half.
Key insight: Only half of the energy a battery supplies while charging a capacitor ends up stored in it; the rest is dissipated in the circuit.

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Unit 11: Electric Circuits