Definition of a Circuit — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: core requirements for a functional electric circuit, open vs closed circuits, emf vs terminal potential difference, steady current definition, charge conservation in circuits, and identification of complete vs incomplete conducting paths.

You should already know: Electric charge and conventional current direction. Electric potential and potential difference. Ohm's law for resistors.

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 / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Definition of a Circuit?

An electric circuit is a network of conducting components connected to allow continuous flow of electric charge (current) driven by a potential difference. According to the AP Physics 2 Course and Exam Description (CED), this topic is the foundation of Unit 4: Electric Circuits, which accounts for 15-19% of the total AP exam score. Definition of a circuit concepts appear in both MCQ and FRQ sections: they are often tested as standalone conceptual questions, and form the base reasoning for all more complex circuit analysis problems. Synonyms used in exam questions include "electric circuit", "conducting network", and "current loop". Standard notation conventions use $\epsilon$ for emf, $V$ for terminal potential difference, $I$ for current, and $r$ for internal source resistance. The key distinguishing feature of a circuit, unlike a random collection of components, is a closed continuous path that allows charge to flow without accumulating or depleting at any point for steady current. Mistakes in basic circuit classification cascade into errors on all subsequent circuit problems, so this foundational topic requires mastery.

2. Open vs Closed Circuits: Core Requirements

Every functional steady-current circuit requires three core components: (1) a source of potential difference (emf) to create the electric field that drives charge flow; (2) a continuous closed conducting path connecting one terminal of the source back to the other; and (3) a load (a component that dissipates or converts electrical energy, which can be the internal resistance of the source itself if no external load is present). A closed circuit has no breaks in the conducting path, so steady current can flow. An open circuit has at least one break in the path, so no continuous steady current can flow. Even in an open circuit, there is still a potential difference across the break: because no current flows, there is no voltage drop across any connected components, so the full emf of the source appears across the open gap. A useful intuition is to compare a circuit to a closed loop water pipe system: the emf source is a pump that creates pressure difference, the pipe is the conducting path, and the load is a water wheel that extracts energy from the flow. A break in the pipe (open circuit) stops all flow, but the full pump pressure still appears across the break, just as full emf appears across an open switch.

Worked Example

A student builds a setup with a 1.5 V ideal AA battery, a copper wire connecting the positive battery terminal to one lead of a light bulb, and a second copper wire connecting the other bulb lead to a point 1 cm away from the negative battery terminal, leaving an open gap. What is the potential difference across the open gap, assuming ideal wires and an ideal battery?

1. The setup has a break in the conducting path between the end of the second wire and the negative terminal, so this is an open circuit.
2. For an open circuit, no current flows through any connected components (wires, bulb filament). For ideal wires with zero resistance, the voltage drop across any component carrying zero current is $V=IR=0\times R=0$.
3. By conservation of energy around the path, the total potential difference from the positive to negative battery terminal must equal the battery's emf. All of this potential difference appears across the only break in the path.
4. The battery emf is 1.5 V, so the potential difference across the open gap is 1.5 V.

Exam tip: On AP MCQ questions asking for voltage across a single open switch in a series circuit, the answer is almost always equal to the source emf, as long as there are no other breaks in the circuit.

3. Emf vs Terminal Potential Difference

All real sources of electrical energy (batteries, generators, charging stations) have internal resistance $r$, which comes from resistance to charge flow within the source material itself. Emf ($\epsilon$) is defined as the work done per unit charge by the non-electrostatic force inside the source (e.g., chemical reactions in a battery) to separate charge and create a potential difference. Emf is an intrinsic property of the source, and equals the terminal potential difference when no current is drawn. Terminal potential difference ($V$) is the actual potential difference measured across the source terminals when current $I$ flows through the circuit. The relationship between the two quantities comes from Kirchhoff's loop rule, which accounts for the voltage drop across the internal resistance: $$V=\epsilon -Ir$$ This formula matches our earlier observation for open circuits: when $I=0$, $V=\epsilon$, which makes sense. For a short circuit (zero external resistance), $I=\epsilon /r$, so $V=\epsilon -(\epsilon /r)r=0$, which also matches the observation that terminal voltage drops to zero in a short circuit. This distinction is frequently tested on the AP exam, as students often mistakenly use emf instead of terminal voltage for power calculations.

Worked Example

A battery with emf $\epsilon =9.0$ V and internal resistance $r=0.50$ $\Omega$ is connected to a 4.0 $\Omega$ external resistor in a closed series circuit. What is the terminal potential difference of the battery?

1. First calculate total resistance of the closed circuit: $R_{\text{total}}=R+r=4.0\Omega +0.50\Omega =4.5\Omega$.
2. Use Ohm's law to find the circuit current: $I=\frac{\epsilon }{R_{\text{total}}}=\frac{9.0\text{V}}{4.5\Omega }=2.0\text{A}$.
3. Apply the terminal potential difference formula: $V=\epsilon -Ir=9.0\text{V}-(2.0\text{A})(0.50\Omega )=8.0\text{V}$.
4. Double-check by calculating voltage across the external resistor: $V=IR=(2.0\text{A})(4.0\Omega )=8.0\text{V}$, which matches the result. The terminal potential difference is 8.0 V.

Exam tip: Always check if the problem gives an internal resistance for the source. If it does, never use emf directly to calculate power delivered to the external load — always use terminal voltage first.

4. Charge Conservation and Steady Current

For a steady-state circuit (current does not change with time), charge cannot be created, destroyed, or accumulated at any point in the circuit. This fundamental principle leads to Kirchhoff's Current Law (Junction Rule), which states that the sum of currents entering any junction in a circuit equals the sum of currents leaving the junction: $$\sum I_{\text{in}}=\sum I_{\text{out}}$$ A common misconception is that current is "used up" by resistors: in reality, electrical energy is converted to other forms (heat, light) by components, but charge is conserved. The same amount of charge that enters a resistor must leave it, so current is constant at all points in a series circuit with no junctions. This principle is the foundation of all circuit analysis, and AP exam graders frequently require explicit reference to charge conservation for conceptual reasoning questions.

Worked Example

Three resistors are connected in parallel to a 12 V battery. The current through the first resistor is 2.0 A, and the current through the second resistor is 1.5 A. The total current leaving the positive terminal of the battery is 5.0 A. What is the current through the third resistor?

1. All three parallel resistors connect to a common junction right after the battery's positive terminal. All 5.0 A from the battery enters the junction, and splits into three currents going to each resistor.
2. Apply the junction rule (charge conservation for steady current): $I_{\text{total}}=I_1+I_2+I_3$.
3. Rearrange to solve for the unknown current: $I_3=I_{\text{total}}-I_1-I_2=5.0\text{A}-2.0\text{A}-1.5\text{A}=1.5\text{A}$.
4. Confirm at the return junction to the negative terminal: the sum of currents from all three resistors is $2.0+1.5+1.5=5.0\text{A}$, which matches the total current returning to the battery, so charge is conserved. The current through the third resistor is 1.5 A.

Exam tip: When asked to explain why current is constant in a series circuit, always explicitly cite charge conservation for steady current — generic statements about "charge flowing through" will not earn full points on FRQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Stating that the potential difference across an open switch in a series circuit is 0 V. Why: Students confuse zero current with zero voltage, recalling $V=IR$ and incorrectly concluding $I=0$ implies $V=0$. Correct move: Remember $V=IR$ applies only to the conducting component. For an open gap with effectively infinite resistance, use the loop rule: all the source emf drops across the gap, so $V=\epsilon$.
• Wrong move: Using emf $\epsilon$ instead of terminal potential difference $V$ when calculating power delivered to the external load of a battery. Why: Students default to ideal battery assumptions even when internal resistance is given. Correct move: If the problem provides an internal resistance, calculate $V=\epsilon -Ir$ before calculating power or voltage for the external circuit.
• Wrong move: Claiming that less current leaves a resistor than enters it, because current is "used up" to power the component. Why: Students confuse energy consumption with charge consumption. Correct move: Always apply charge conservation for steady current: the current entering any component equals the current leaving it; only energy is reduced, not charge.
• Wrong move: Classifying a short circuit (battery terminals connected by a zero-resistance wire) as "not a circuit" because it has no external load. Why: Students memorize that circuits require loads, so they incorrectly exclude short circuits. Correct move: Remember that the battery's internal resistance always acts as a load, so a short circuit is a valid closed circuit with very high current.
• Wrong move: Assuming any setup with a battery is automatically a closed circuit, ignoring visible breaks in the conducting path. Why: Students associate batteries with working circuits, so they rush past checks for continuity. Correct move: Always confirm there is a continuous closed path from the positive terminal back to the negative terminal before classifying a setup as a closed circuit.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A student connects one end of a copper wire to the positive terminal of a 12 V ideal battery, and leaves the other end of the wire hanging freely, not connected to anything else. Which of the following correctly describes the current in the wire and the potential difference between the end of the wire and the negative terminal of the battery? A) Current = 0 A, Potential Difference = 0 V B) Current = 0 A, Potential Difference = 12 V C) Current = 12 A, Potential Difference = 0 V D) Current = 12 A, Potential Difference = 12 V

Worked Solution: The hanging open end means there is no continuous path from the positive terminal back to the negative terminal, so this is an open circuit. No steady current can flow, so current is 0 A, eliminating options C and D. For an ideal battery and ideal wires, there is no voltage drop across the wire because current is zero, so the full emf of 12 V appears across the open gap between the wire end and the negative terminal. The correct answer is B.

Question 2 (Free Response)

A student constructs two setups: Setup 1: A 6.0 V battery with internal resistance 0.20 Ω connected to a 1.0 Ω resistor and a 1.8 Ω resistor in series, with all connections complete. Setup 2: Same as Setup 1, but the wire between the two resistors is cut, leaving an open gap. (a) Classify each setup as open or closed circuit. (b) Calculate the total current flowing in Setup 1, and find the terminal potential difference of the battery. (c) Determine the potential difference across the open gap in Setup 2. Justify your answer.

Worked Solution: (a) Setup 1 has a continuous closed conducting path, so it is a closed circuit. Setup 2 has a break in the path from the cut wire, so it is an open circuit. (b) Total resistance of the circuit: $R_{\text{total}}=r+R_1+R_2=0.20\Omega +1.0\Omega +1.8\Omega =3.0\Omega$. By Ohm's law, $I=\frac{\epsilon }{R_{\text{total}}}=\frac{6.0\text{V}}{3.0\Omega }=2.0\text{A}$. Terminal potential difference: $V=\epsilon -Ir=6.0\text{V}-(2.0\text{A})(0.20\Omega )=5.6\text{V}$. (c) In Setup 2, no steady current can flow ($I=0$), so the voltage drop across all resistors and internal resistance is $V=IR=0$. By Kirchhoff's loop rule, the full emf of the battery must appear across the open gap, so the potential difference across the gap is 6.0 V.

Question 3 (Application / Real-World Style)

A faulty electric vehicle charging connector has an open air gap between the 400 V output positive terminal of the charging station and the corresponding terminal on the car. The charging station has an emf of 400 V and internal resistance of 0.10 Ω. Air breaks down (becomes conducting) when the potential difference across it exceeds 3000 V. What is the potential difference across the open air gap, what is the current flowing before breakdown, and will the air break down in this scenario?

Worked Solution:

1. The open gap creates an open circuit, so no steady current can flow before breakdown: $I=0\text{A}$.
2. Calculate the potential difference across the gap using the terminal voltage formula: $V=\epsilon -Ir=400\text{V}-(0\text{A})(0.10\Omega )=400\text{V}$.
3. Compare to the breakdown voltage: $400\text{V}<3000\text{V}$, so air does not reach breakdown. In context: The open gap remains non-conducting, so no current flows to charge the car's battery, and the charger will not operate until the loose connection is repaired.

7. Quick Reference Cheatsheet

8. What's Next

This chapter establishes the foundational definitions and core physical principles you will use for all subsequent circuit analysis in AP Physics 2 Unit 4. Next you will apply the core concepts from this chapter — distinguishing open/closed circuits, charge conservation, and the emf/terminal voltage distinction — to analyze resistors in series and parallel, then extend this to full Kirchhoff's rules for complex multi-loop circuits. Without mastering the core ideas here, you will make consistent, cascading mistakes in all later circuit problems, from equivalent resistance calculations to power dissipation and transient RC circuits. This topic also feeds into the broader study of electromagnetism, where you will analyze induced currents in conducting loops.

• Resistors in Series and Parallel
• Kirchhoff's Circuit Rules
• RC Circuits