Electric Circuits — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Current, voltage, resistance, Ohm's law, Kirchhoff's laws, series and parallel resistors, RC circuit charging/discharging, and power dissipation for the Electric Circuits syllabus section.

You should already know: AP Physics 1 or equivalent.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Electric Circuits?

An electric circuit is a closed, continuous path for net electric charge to flow, converting electrical energy to other forms (heat, light, mechanical work) as charge moves from a high-potential energy source to a lower-potential sink. This topic accounts for 18-22% of your total AP Physics 2 exam score across multiple-choice and free-response sections, per the official CED. Common synonyms tested on the exam include DC circuits, resistive-capacitive circuits, and ohmic circuits.

2. Current, voltage, resistance, Ohm's law

All circuit analysis builds on three core quantities and their relationship in Ohm's law:

1. Current ($I$): The rate of net charge flow past a fixed point in a circuit, measured in Amperes (A). The formula is $I=\frac{\Delta Q}{\Delta t}$, where $\Delta Q$ is total charge passing the point in time $\Delta t$. Note that conventional current (used for all AP problems) describes the flow of positive charge, opposite to the actual direction of electron motion.
2. Voltage ($V$, or emf $\epsilon$ for sources): The difference in electric potential energy per unit charge between two points in a circuit, measured in Volts (V). It is the "driving force" that pushes charge through the circuit, defined as $V=\frac{\Delta U}{q}$ where $\Delta U$ is potential energy change for charge $q$.
3. Resistance ($R$): A material's opposition to charge flow, measured in Ohms ($\Omega$). For a uniform conductor, resistance depends on material resistivity $\rho$, length $L$, and cross-sectional area $A$: $R=\rho \frac{L}{A}$.

Ohm's law describes the linear relationship between these three quantities for ohmic materials (materials with constant resistance at fixed temperature): $$V=IR$$

Worked Example: A 3V button battery powers a 150Ω ohmic resistor in a hearing aid. Calculate the current through the resistor, and total charge that flows over 1 hour of use.

1. Rearrange Ohm's law: $I=\frac{V}{R}=\frac{3V}{150\Omega }=0.02A=20mA$
2. Rearrange current formula: $\Delta Q=I\Delta t=0.02A\ast 3600s=72C$

Exam tip: Examiners often test non-ohmic materials (e.g., heated tungsten filaments, diodes) with non-linear I-V curves: never apply Ohm's law to these, read values directly from the provided graph instead.

3. Kirchhoff's laws

For complex multi-loop circuits that cannot be simplified with series/parallel rules, use Kirchhoff's two conservation laws:

1. Kirchhoff's Junction Rule (KCL): Derived from conservation of charge, the sum of all currents entering a junction equals the sum of all currents leaving the junction: $$\sum I_{in}=\sum I_{out}$$
2. Kirchhoff's Loop Rule (KVL): Derived from conservation of energy, the sum of all potential differences around any closed loop in a circuit equals zero: $$\sum V_{loop}=0$$

To avoid sign errors with KVL, follow these conventions:

• When moving from the negative to positive terminal of a battery, add the emf ($+\epsilon$)
• When moving from the positive to negative terminal of a battery, subtract the emf ($-\epsilon$)
• When moving in the direction of labeled current through a resistor, subtract the voltage drop ($-IR$)
• When moving opposite the direction of labeled current through a resistor, add the voltage drop ($+IR$)

Worked Example: A closed loop has a 9V battery, 1Ω internal resistance, 3Ω load resistor, and 5Ω load resistor in series. Use KVL to find the total current in the loop.

1. Label current $I$ flowing clockwise through the loop
2. Write the loop equation: $9V-I(1\Omega )-I(3\Omega )-I(5\Omega )=0$
3. Rearrange to solve for $I$: $9=9I\to I=1A$

Note: If you guess the wrong direction for current, the final value of $I$ will be negative, indicating current flows opposite to your initial label.

4. Series and parallel resistors

Most simple circuits can be simplified by combining groups of resistors in series or parallel to find a single equivalent resistance $R_{eq}$:

Series resistors

Resistors connected end-to-end share the same current, and total voltage across the group is the sum of individual voltage drops. Equivalent resistance is: $$R_{series}=R_1+R_2+...+R_n$$ The equivalent resistance is always larger than the largest individual resistor in the group.

Parallel resistors

Resistors connected across the same two points share the same voltage, and total current through the group is the sum of individual currents. Equivalent resistance is: $$\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$$ For two parallel resistors, this simplifies to $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$. The equivalent resistance is always smaller than the smallest individual resistor in the group.

Worked Example: You have four resistors: 2Ω, 4Ω, 4Ω, and 8Ω. The two 4Ω resistors are in parallel, and this group is in series with the 2Ω and 8Ω resistors. Calculate total equivalent resistance.

1. Combine the parallel 4Ω resistors first: $\frac{1}{R_p}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\to R_p=2\Omega$
2. Combine all series resistors: $R_{eq}=2\Omega +2\Omega +8\Omega =12\Omega$

Exam tip: Always combine resistors from the innermost group outwards, never skip steps to avoid arithmetic errors.

5. RC circuits — charging and discharging

An RC circuit consists of a resistor $R$ and capacitor $C$ connected in series with a battery and switch, with predictable transient behavior when the switch is flipped. The core parameter for RC circuits is the time constant $\tau$, the time required for the capacitor to charge to 63% of its maximum voltage or discharge to 37% of its initial voltage: $$\tau =RC$$ Units of $\tau$ are seconds, since $1F\ast 1\Omega =1s$.

Charging an RC circuit

When the switch closes to connect an uncharged capacitor to the battery:

• At $t=0$, the capacitor has 0 voltage and acts like a short circuit, so current is maximum: $I_0=\frac{\epsilon }{R}$
• As the capacitor charges, voltage across it rises exponentially, and current falls exponentially: $$V_c(t)=\epsilon (1-e^{-t/\tau })$$ $$I(t)=I_0{e}^{-t/\tau }$$
• At $t\to \infty$, the capacitor is fully charged, current drops to 0, and the capacitor acts like an open circuit.

Discharging an RC circuit

When the battery is removed and the capacitor is connected across a resistor:

• Voltage across the capacitor and current both fall exponentially: $$V_c(t)=V_0{e}^{-t/\tau }$$ $$I(t)=\frac{V_0}{R}{e}^{-t/\tau }$$
• After $5\tau$, the capacitor is ~99% discharged, treated as fully discharged for AP exam purposes.

Worked Example: A 50μF uncharged capacitor, 20kΩ resistor, and 10V battery form an RC circuit. Calculate the time constant, voltage across the capacitor after 1 second, and initial current.

1. Time constant: $\tau =50e-6F\ast 20e3\Omega =1s$
2. Voltage at $t=1s$: $V_c=10(1-e^{-1})\approx 10\ast 0.632=6.32V$
3. Initial current: $I_0=\frac{10V}{20e3\Omega }=5e-4A=500\mu A$

6. Power dissipation

Power is the rate of energy transfer in a circuit, measured in Watts (W). All power dissipated by resistors is converted to heat (Joule heating). The base formula for power is: $$P=VI$$ Substitute Ohm's law to get two alternative forms for resistive loads: $$P=I^{2}R\text{(best for series circuits, where current is constant)}$$ $$P=\frac{V^2}{R}\text{(best for parallel circuits, where voltage is constant)}$$

For batteries with internal resistance $r$, the total power supplied by the battery is $P_{total}=\epsilon I$, and the power lost to internal heating is $P_{lost}=I^{2}r$. The remaining power is delivered to the external load.

Worked Example: A 12V car battery has 0.2Ω internal resistance, connected to a 5.8Ω headlight load. Calculate total power supplied by the battery, power delivered to the headlight, and power lost as heat.

1. Total circuit resistance: $0.2\Omega +5.8\Omega =6\Omega$, current $I=\frac{12V}{6\Omega }=2A$
2. Total power supplied: $P_{total}=12V\ast 2A=24W$
3. Power to headlight: $P_{load}=I^{2}R=(2{)}^2\ast 5.8=23.2W$
4. Power lost to internal resistance: $P_{lost}=(2{)}^2\ast 0.2=0.8W$ Check: $23.2W+0.8W=24W$, matching total power supplied.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Applying Ohm's law to non-ohmic materials (diodes, heated filaments) with non-linear I-V curves. Why: Students memorize $V=IR$ without context. Correct move: Only use Ohm's law for explicitly labeled ohmic materials; read voltage/current values directly from provided I-V graphs for non-ohmic components.
• Pitfall 2: Misapplying KVL sign conventions, adding voltage drops instead of subtracting them when traversing a loop. Why: Students forget to label current directions before writing loop equations. Correct move: Explicitly label all current directions at junctions first, follow a single consistent direction around each loop, and subtract $IR$ when moving with the current, add when moving against it.
• Pitfall 3: Adding parallel resistors directly like series resistors (e.g., calculating 2Ω || 3Ω as 5Ω). Why: Confusion between series and parallel resistance rules. Correct move: Remember that parallel equivalent resistance is always smaller than the smallest resistor in the group; if your result is larger than any individual parallel resistor, it is immediately wrong.
• Pitfall 4: Mixing up RC circuit transient behavior, treating a capacitor as an open circuit at $t=0$ or a short at $t\to \infty$. Why: Students memorize rules without understanding the underlying physics. Correct move: Use the rule "capacitors resist changes in voltage": at $t=0$, voltage is equal to its initial value (0 for uncharged, so acts as short), at $t\to \infty$, voltage stops changing so current is 0 (acts as open).
• Pitfall 5: Using $P=\frac{V^2}{R}$ with the battery emf to calculate load power, forgetting internal resistance. Why: Students confuse emf with terminal voltage. Correct move: Only use terminal voltage (voltage across the load) for load power calculations; use $P=\epsilon I$ for total power supplied by the battery.

8. Practice Questions (AP Physics 2 Style)

Question 1

A student builds a circuit with a 12V battery (negligible internal resistance), a 3Ω resistor, a 6Ω resistor, and a 6Ω resistor. The two 6Ω resistors are in parallel, and this combination is in series with the 3Ω resistor. Calculate: (a) Total equivalent resistance of the circuit (b) Current through the 3Ω resistor (c) Voltage across one of the 6Ω resistors (d) Total power dissipated in the circuit

Solution

(a) First combine the parallel 6Ω resistors: $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\to R_p=3\Omega$. Total resistance: $R_{eq}=3\Omega +3\Omega =6\Omega$. (b) Total current from the battery: $I=\frac{V}{R_{eq}}=\frac{12V}{6\Omega }=2A$. This is the current through the series 3Ω resistor. (c) Voltage across the parallel group: $V_p=I\ast R_p=2A\ast 3\Omega =6V$, which is equal to the voltage across each 6Ω resistor. (d) Total power: $P=VI=12V\ast 2A=24W$, or $P=\frac{V^2}{R}=\frac{144}{6}=24W$.

Question 2

A multi-loop circuit has a junction J where 1.2A enters from the left, 0.8A enters from below, and one current $I$ leaves to the right. A closed loop in the same circuit has a 15V battery, 2Ω resistor, 4Ω resistor, and an unknown voltage drop across a fan. If the loop current is 1.5A clockwise, calculate the value of $I$ and the voltage drop across the fan.

Solution

1. Apply KCL at junction J: Sum of incoming currents = sum of outgoing currents. $1.2A+0.8A=I\to I=2A$.
2. Apply KVL around the loop: $15V-(1.5A\ast 2\Omega )-(1.5A\ast 4\Omega )-V_{fan}=0$. Rearrange: $15-3-6=V_{fan}\to V_{fan}=6V$.

Question 3

An RC circuit has a 10μF uncharged capacitor, 100kΩ resistor, and 9V battery connected in series with a switch closed at $t=0$. Calculate: (a) The time constant of the circuit (b) Current in the circuit immediately after the switch is closed (c) Voltage across the capacitor after 2 time constants have passed (d) Approximate time for the capacitor to reach 99% of its maximum charge

Solution

(a) Time constant: $\tau =RC=10e-6F\ast 100e3\Omega =1s$. (b) At $t=0$, the capacitor acts as a short circuit: $I_0=\frac{V}{R}=\frac{9V}{100e3\Omega }=9e-5A=90\mu A$. (c) $V_c(t)=\epsilon (1-e^{-t/\tau })=9(1-e^{-2})\approx 9\ast 0.865=7.79V$. (d) 99% charge is reached after ~5 time constants, so $t\approx 5\ast 1s=5s$.

9. Quick Reference Cheatsheet

Rules of Thumb

• Uncharged capacitor = short at $t=0$, open at $t\to \infty$
• 5τ = ~99% charge/discharge for RC circuits
• Parallel $R_{eq}$ < smallest resistor in the parallel group

10. What's Next

Electric Circuits is a foundational topic for the rest of the AP Physics 2 electricity and magnetism sequence: you will apply these rules to analyze magnetic induction circuits, AC circuits, and electromagnetic wave transmitters later in the syllabus. Your understanding of RC circuit transient behavior will also be directly tested in free-response questions that combine electrostatics, capacitance, and circuit analysis, so mastering these rules now will cut down on study time for later units and reduce errors on multi-concept exam questions.

If you struggle with any of the concepts, practice questions, or sign conventions covered in this guide, you can ask Ollie, our AI tutor, for personalized explanations, extra practice problems, or step-by-step walkthroughs tailored to your learning gaps. You can also find more AP Physics 2 study guides and official College Board practice resources on the homepage to continue your exam prep and track your progress towards a 5.