Definition of a Circuit

AP Physics 2· AP Physics 2 CED — Electric Circuits· 14 min read

1. Core Definition of an Electric Circuit★☆☆☆☆⏱ 3 min

An electric circuit is a network of conducting components connected to allow continuous flow of electric charge (current) driven by a potential difference. This topic is the foundation of Unit 4, which accounts for 15-19% of your total AP Physics 2 exam score, and is tested in both MCQ and FRQ sections.

📘 Definition

Electric Circuit

A closed conducting network that enables continuous flow of electric charge driven by a source of potential difference

Example:

A battery connected to a light bulb with two copper wires forms a complete electric circuit

2. Open vs Closed Circuits★★☆☆☆⏱ 4 min

Every functional steady-current circuit requires three core components: (1) a source of potential difference (emf) to drive charge flow; (2) a continuous closed conducting path connecting one source terminal back to the other; (3) a load that converts electrical energy to other forms (the source's internal resistance can act as a load if no external load is present).

A closed circuit has no breaks in the path, so steady current can flow. An open circuit has at least one break, so no steady current can flow. Even in an open circuit, potential difference exists across the break: with no current flowing, there is no voltage drop across connected components, so the full emf of the source appears across the open gap.

📐 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, so this is an open circuit with no steady current.
2
For any connected conducting component carrying zero current, Ohm's law gives a voltage drop of:
3
$V = IR = 0 \times R = 0$
4
By conservation of energy around the path, the total potential difference from the positive to negative battery terminal equals the battery's emf. All of this potential difference appears across the open gap.
5
Final Answer: 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★★★☆☆⏱ 5 min

All real sources of electrical energy (batteries, generators) have internal resistance $r$ , which comes from resistance to charge flow within the source material itself.

📘 Definition

Emf

\varepsilon

The work done per unit charge by non-electrostatic forces inside the source to separate charge. It is an intrinsic property equal to terminal potential difference when no current is drawn.

Example:

A new AA battery has a constant emf of 1.5 V

📘 Definition

Terminal Potential Difference

V

The actual potential difference measured across the source terminals when current $I$ flows through the circuit.

Example:

Terminal potential difference is always less than emf when current flows, due to voltage drop across internal resistance

The relationship between emf and terminal potential difference comes from Kirchhoff's loop rule:

V = \varepsilon - Ir

📐 Worked Example

A battery with emf $ε = 9.0$ V and internal resistance $r = 0.50$ $Ω$ is connected to a 4.0 $Ω$ 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:
2
$R_{\text{total}} = R + r = 4.0 \, \Omega + 0.50 \, \Omega = 4.5 \, \Omega$
3
Use Ohm's law to find the circuit current:
4
$I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9.0 \, \text{V}}{4.5 \, \Omega} = 2.0 \, \text{A}$
5
Apply the terminal potential difference formula:
6
$V = \varepsilon - Ir = 9.0 \, \text{V} - (2.0 \, \text{A})(0.50 \, \Omega) = 8.0 \, \text{V}$
7
Double-check by calculating voltage across the external resistor (which equals terminal PD):
8
$V = IR = (2.0 \, \text{A})(4.0 \, \Omega) = 8.0 \, \text{V}$
9
Final Answer: Terminal potential difference = 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★★☆☆☆⏱ 3 min

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):

\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), 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.

📐 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 after the battery's positive terminal. All 5.0 A from the battery enters the junction, and splits into three currents.
2
Apply the junction rule (charge conservation for steady current):
3
$I_{\text{total}} = I_1 + I_2 + I_3$
4
Rearrange to solve for the unknown current:
5
$I_3 = I_{\text{total}} - I_1 - I_2 = 5.0 \, \text{A} - 2.0 \, \text{A} - 1.5 \, \text{A} = 1.5 \, \text{A}$
6
Confirm at the return junction: sum of currents equals 5.0 A, matching the total returning to the battery, so charge is conserved.
7
Final Answer: Current through the third resistor = 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. Concept Check★★☆☆☆⏱ 3 min

✓ Quick check

Test your understanding of core circuit definitions with this AP-style multiple choice question:

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?
- Current = 0 A, Potential Difference = 0 V
- Current = 0 A, Potential Difference = 12 V
- Current = 12 A, Potential Difference = 0 V
- Current = 12 A, Potential Difference = 12 V
Reveal answer
1 —
The open end means there is no continuous conducting path, so current is 0 A. For ideal wires with zero current, all 12 V of the battery's emf appears across the open gap, so this option is correct.

6. Common Pitfalls

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 $V = I R$ and incorrectly concluding $I = 0$ implies $V = 0$ .

Correct move:

Remember $V = I R$ applies only to conducting components. For an open gap with effectively infinite resistance, use the loop rule: all the source emf drops across the gap, so $V = ε$ .

Wrong move:

Using emf $ε$ 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 explicitly given.

Correct move:

If the problem provides an internal resistance, calculate $V = ε - I r$ 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.

7. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Closed Circuit Requirements	Source of emf + closed conducting path + load	Short circuits have no external load, but internal resistance acts as the load
Open Circuit Steady Current	$I = 0$	No continuous path means no steady current flow
Open Circuit Terminal Voltage	$V = ε$	Full source emf appears across the open gap when $I = 0$
Terminal Potential Difference	$V = ε - I r$	Applies to all real sources with internal resistance $r$
Junction Rule (Charge Conservation)	$\sum I_{in} = \sum I_{out}$	Valid for all junctions in steady-state circuits
Series Circuit Current	$I_{1} = I_{2} = I_{total}$	No junctions mean current is constant by charge conservation
Ohm's Law	$V = I R$	Only applies to conducting components, not open gaps

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Physics 2
Conceptual open circuit voltage MCQ
2022 · AP Physics 2
FRQ terminal PD calculation

Going deeper

official syllabusAP Physics 2 Course and Exam DescriptionOfficial reference for AP Physics 2 exam topics

What's Next

This module 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.