AP Physics 2 • Electric Circuits

Simple Circuits

Understanding how voltage, current, and resistance govern energy transfer in series and parallel configurations.

An atom with wave and electric energy symbols.
Inside this lesson 9 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The study of electric circuits grew from centuries of inquiry into the nature of electricity. Early investigators observed static sparks and the attraction of amber to light objects, but a systematic understanding of steady current flow only became possible once chemical batteries could supply a continuous source of charge. The mathematical relationships that govern simple circuits—linking voltage, current, and resistance—were forged through careful experimentation in the early nineteenth century, and they remain the backbone of every electrical system today, from smartphone chargers to hospital defibrillators.

1800 • Volta's Pile

Alessandro Volta constructs the first true battery—stacked zinc and copper disks separated by brine-soaked cloth—providing the first reliable source of steady electric current for laboratory experiments.

1827 • Ohm's Law Published

Georg Simon Ohm publishes his empirical finding that current through a conductor is directly proportional to voltage and inversely proportional to resistance, V = IR, laying the quantitative foundation for circuit analysis.

1845 • Kirchhoff's Rules

Gustav Kirchhoff formulates the junction rule (conservation of charge) and the loop rule (conservation of energy), enabling analysis of complex multi-loop circuits beyond simple series and parallel arrangements.

1882 • Edison's DC Power Grid

Thomas Edison opens the Pearl Street Station in New York City, demonstrating a practical DC distribution system that illuminates homes and businesses using the very principles Ohm and Kirchhoff established.

The central question these pioneers answered is deceptively simple: How does the arrangement of components in a circuit determine the current through each element and the voltage across it? Answering that question requires understanding series and parallel topologies, the conservation laws that constrain charge and energy flow, and the mathematical tools—Ohm's law and Kirchhoff's rules—that translate physical principles into solvable equations.

SECTION 2

Core Principles & Definitions

A simple circuit is any closed conducting path through which charge can flow, driven by a source of electromotive force (emf) such as a battery. The behavior of every simple circuit rests on a handful of foundational ideas that connect the macroscopic quantities of voltage, current, and resistance to the microscopic motion of charge carriers.

1

Current (I)

The rate of charge flow past a point, measured in amperes (A). Conventional current flows from high to low potential. 1 A = 1 C/s.
2

Voltage (V)

The electric potential difference between two points, measured in volts (V). It represents the energy per unit charge available to drive current through a component.
3

Resistance (R)

The opposition a material offers to current flow, measured in ohms (Ω). Higher resistance means less current for a given voltage, as governed by Ohm's law.
4

Ohm's Law

V = IR relates the voltage across a resistor to the current through it and its resistance. This linear relationship holds for ohmic materials at constant temperature.
5

Kirchhoff's Rules

The junction rule (ΣI_in = ΣI_out) enforces charge conservation; the loop rule (ΣΔV = 0) enforces energy conservation around any closed path.

SECTION 3

Visual Explanation — Series & Parallel Topologies

Series CircuitParallel CircuitεR₁R₂R₃Same current I through all elementsε = V₁ + V₂ + V₃R_eq = R₁ + R₂ + R₃εR₁R₂Same voltage ε across each branchI_total = I₁ + I₂1/R_eq = 1/R₁ + 1/R₂Quick ComparisonSeries: current is the same everywhereVoltages add: ε = ΣVᵢR_eq increases with each resistorParallel: voltage is the same across branchesCurrents add: I_total = ΣIᵢR_eq decreases with each resistor
Left: a series circuit where R₁, R₂, and R₃ carry the same current and the battery emf equals the sum of all voltage drops. Right: a parallel circuit where each branch shares the same voltage and the total current splits among branches. The summary box at the bottom contrasts the key properties of each topology.

The diagram above captures the two fundamental ways to arrange resistors. In a series configuration, there is only one path for charge to follow, so the current I is identical through every element. Ohm's law applied to each resistor gives individual voltage drops V₁ = IR₁, V₂ = IR₂, and V₃ = IR₃, which must sum to the battery emf by Kirchhoff's loop rule. In a parallel configuration, multiple paths exist between the same two nodes, so each branch experiences the full battery voltage. The junction rule then requires the total current entering a node to equal the sum of branch currents, I₁ + I₂. These two patterns form the building blocks for analyzing any circuit, no matter how complex.

An independently authored battery bench with a sealed battery, finite load, series ammeter and parallel voltmeter, with explicitly joined leads.During discharge, internal resistance makes terminal voltage smaller than the emf. The source power divides between useful load power and internal heating.

A loaded battery has distinct current and voltage paths. During discharge, internal resistance makes terminal voltage smaller than the emf. The source power divides between useful load power and internal heating.During discharge, internal resistance makes terminal voltage smaller than the emf. The source power divides between useful load power and internal heating.

SECTION 4

Mathematical Framework

OHM'S LAW

V=IRV = I R

V = potential difference across the resistor (V), I = current through the resistor (A), R = resistance (Ω). This is a linear relation valid for ohmic materials.

SERIES EQUIVALENT RESISTANCE

Req=R1+R2+R3+⋯R_eq = R₁ + R₂ + R₃ + ⋯

Resistors in series add directly. The equivalent resistance is always larger than the largest individual resistor, so adding series resistors decreases total current.

PARALLEL EQUIVALENT RESISTANCE

1/Req=1/R1+1/R2+1/R3+⋯1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ⋯

The reciprocal of the equivalent resistance equals the sum of the reciprocals of each branch resistance. R_eq is always less than the smallest individual resistor because additional branches provide extra paths for current.

POWER DISSIPATED

P=IV=I2R=V2/RP = IV = I²R = V²/R

P = power dissipated (W). The three equivalent forms follow from substituting Ohm's law. Choose the form that matches the known quantities: P = IV is universal; P = I²R is convenient when current is known; P = V²/R is convenient when voltage is known.

The loop rule (Kirchhoff's voltage law) is a statement of conservation of energy: as you traverse any closed loop, the algebraic sum of all potential gains (from batteries) and drops (across resistors) equals zero. The junction rule (Kirchhoff's current law) is a statement of conservation of charge: current entering a node must equal current leaving it, because charge cannot accumulate in steady state. Together these two rules, combined with Ohm's law, provide enough independent equations to solve for every unknown current and voltage in any resistor network.

SECTION 5

Detailed Breakdown — Combining Series & Parallel

Many real circuits contain combination (series-parallel) networks. The strategy for analyzing them is to simplify from the inside out: identify groups of resistors that are purely in series or purely in parallel, replace each group with its equivalent resistance, and repeat until the entire network reduces to a single equivalent resistor connected to the battery. Once you know the total current from the battery, you can 'unpack' the reductions in reverse order to find the current through and voltage across every individual resistor.

Step-by-Step Reduction of a Combination CircuitStep 1: Original Circuit12V4Ω6Ω3Ω6Ω ∥ 3Ω in parallelStep 2: Replace parallel pair12V4Ω2Ω(6×3)/(6+3) = 2ΩNow 4Ω + 2Ω in seriesStep 3: Final equivalent12V6Ω4 + 2 = 6ΩResultsI_total = 12V / 6Ω = 2 AV across 4Ω = 8V, V across ∥ pair = 4V
A combination circuit reduced in three steps. Step 1 shows the original network: a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω. Step 2 replaces the parallel pair with its equivalent 2 Ω resistance. Step 3 adds the remaining series resistors to yield R_eq = 6 Ω. The total current from the 12 V battery is 2 A.
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Strategy Tip

SECTION 6

Worked Example

Worked example

  1. 1

    Step 1 — Identify Given Values

    A battery with emf ε = 24 V (assume negligible internal resistance) is connected to R₁ = 8 Ω in series with a parallel combination of R₂ = 12 Ω and R₃ = 6 Ω. We seek the total current, the current through each resistor, and the power dissipated by R₂.
  2. 2

    Step 2 — Find the Parallel Equivalent

    Apply the parallel resistance formula: 1/R_∥ = 1/R₂ + 1/R₃ = 1/12 + 1/6 = 1/12 + 2/12 = 3/12.
    R_∥ = 12/3 = 4 Ω
  3. 3

    Step 3 — Find Total Equivalent Resistance

    R₁ is in series with the parallel combination, so R_eq = R₁ + R_∥ = 8 + 4.
    R_eq = 12 Ω
  4. 4

    Step 4 — Find Total Current

    Apply Ohm's law: I_total = ε / R_eq = 24 V / 12 Ω.
    I_total = 2 A
  5. 5

    Step 5 — Find Voltage Across the Parallel Section

    The voltage drop across R₁ is V₁ = I_total × R₁ = 2 × 8 = 16 V. By Kirchhoff's loop rule, the voltage across the parallel pair is V_∥ = ε − V₁ = 24 − 16 = 8 V. Equivalently, V_∥ = I_total × R_∥ = 2 × 4 = 8 V, which confirms the result.
    V_∥ = 8 V
  6. 6

    Step 6 — Find Branch Currents

    Each parallel branch sees V_∥ = 8 V. Thus I₂ = V_∥ / R₂ = 8/12 ≈ 0.667 A and I₃ = V_∥ / R₃ = 8/6 ≈ 1.333 A. Check: I₂ + I₃ = 0.667 + 1.333 = 2.0 A = I_total. ✓
    I₂ ≈ 0.67 A, I₃ ≈ 1.33 A
  7. 7

    Step 7 — Power Dissipated by R₂

    Using P = V²/R is convenient here since we know the voltage across R₂. P₂ = (8)²/12 = 64/12.
    P₂ ≈ 5.33 W

SECTION 7

Series vs. Parallel — Strengths & Limitations

PropertySeriesParallel
CurrentSame through all elementsSplits among branches; total = sum of branch currents
VoltageDivides among elements; sum = emfSame across every branch
R_eqSum of all resistances (always > largest R)Reciprocal sum (always < smallest R)
Effect of removing one elementBreaks the entire circuit — all current stopsOther branches continue to operate
Practical useVoltage dividers, current-limiting circuitsHousehold wiring, ensuring independent device operation

Comparison of series and parallel circuit properties

SECTION 8

Connection to Advanced Circuit Analysis

Simple resistor circuits form the conceptual scaffolding for more complex topics you will encounter in AP Physics 2 and beyond. The same Kirchhoff laws that let you analyze a two-resistor loop also govern circuits containing capacitors (RC circuits), where current is not constant but decays exponentially as the capacitor charges. Internal resistance of real batteries, explored later in the course, adds a series resistor inside the battery itself, causing the terminal voltage to sag under load. Understanding simple DC circuits also prepares you for AC circuit analysis in more advanced courses, where impedance replaces resistance and phasors replace scalar voltages.

Simple Circuits (This Lesson)Advanced Extensions
Ideal battery with fixed emf εReal battery: ε − Ir, where r is internal resistance
Resistors only; steady-state DCRC circuits: time-varying current, exponential charging/discharging
Kirchhoff's rules applied to resistor networksMulti-loop circuits requiring simultaneous equations
Power P = IV dissipated as thermal energyEnergy stored in capacitors: U = ½CV²

How simple circuit concepts extend to advanced topics

On the AP Physics 2 exam, circuit problems frequently ask you to predict what happens when a switch is opened or closed, a resistor is added or removed, or a battery's internal resistance is non-negligible. Mastering the simple circuit principles in this lesson—particularly the conservation-based reasoning encoded in Kirchhoff's rules—gives you the analytical toolkit needed for those more nuanced scenarios.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two identical resistors are connected in parallel to a battery. A third identical resistor is then added in parallel. Which of the following correctly describes what happens to the total current drawn from the battery and the voltage across each resistor?
  • A. Total current increases; voltage across each resistor increases
  • B. Total current increases; voltage across each resistor stays the same
  • C. Total current stays the same; voltage across each resistor decreases
  • D. Total current decreases; voltage across each resistor stays the same
Show answer (B)
B is correct. In a parallel circuit with an ideal battery, each branch is connected directly across the battery terminals, so every resistor experiences the full battery voltage regardless of how many branches exist. Adding a third parallel resistor creates an additional current path, which decreases R_eq and therefore increases the total current by Ohm's law (I = ε/R_eq). Choice A is wrong because the voltage across each resistor is fixed at ε for an ideal battery, not increased. Choice C is wrong because total current increases, not stays the same (R_eq decreases when a parallel branch is added). Choice D is wrong because total current increases rather than decreases.

Question 2: BASIC CALCULATION

A 9 V battery is connected to three resistors in series: R₁ = 3 Ω, R₂ = 4 Ω, and R₃ = 2 Ω. What is the current through R₂?
  • A. 2.25 A
  • B. 1.5 A
  • C. 1.0 A
  • D. 0.5 A
Show answer (C)
C is correct. In a series circuit, the current is the same through every element. The total resistance is R_eq = 3 + 4 + 2 = 9 Ω. By Ohm's law, I = ε/R_eq = 9 V / 9 Ω = 1.0 A. This is the current through R₂ (and through R₁ and R₃ as well). Choice A (2.25 A) would result from incorrectly using only R₂ = 4 Ω as the denominator. Choice B (1.5 A) would result from dividing 9 V by 6 Ω, perhaps omitting R₃. Choice D (0.5 A) has no straightforward derivation from these values and is simply too low.

Question 3: INTERMEDIATE

A circuit consists of a 20 V battery connected to R₁ = 10 Ω in series with a parallel combination of R₂ = 15 Ω and R₃ = 30 Ω. What is the power dissipated by R₁?
  • A. 10.0 W
  • B. 6.67 W
  • C. 40.0 W
  • D. 20.0 W
Show answer (A)
A is correct. First find the parallel equivalent: 1/R_∥ = 1/15 + 1/30 = 2/30 + 1/30 = 3/30, so R_∥ = 10 Ω. The total resistance is R_eq = R₁ + R_∥ = 10 + 10 = 20 Ω. The total current is I = 20 V / 20 Ω = 1 A. The power dissipated by R₁ is P₁ = I²R₁ = (1)²(10) = 10.0 W. Choice B (6.67 W) might result from an incorrect parallel calculation. Choice C (40 W) equals P = ε²/R₁ = 400/10, which incorrectly applies the full battery voltage across R₁ alone. Choice D (20 W) equals the total power from the battery (P = εI = 20 × 1), not the power through R₁ alone.

Question 4: APPLIED

A student has a battery of unknown emf (but known to be between 3 V and 12 V), an ammeter, a voltmeter, a set of five resistors with marked values (10 Ω, 20 Ω, 47 Ω, 100 Ω, 220 Ω), connecting wires, and a switch. Design an experiment to determine the emf of the battery and its internal resistance. In your response: (a) describe the circuit setup, (b) describe the experimental procedure including what quantities are measured, (c) explain how the data should be analyzed to extract emf and internal resistance, and (d) identify one source of systematic error and how it could affect results.
Show answer
Sample response: (a) Connect the battery in series with the ammeter, the switch, and one external resistor. Place the voltmeter across the battery terminals to measure terminal voltage VTV_T. (b) Close the switch and record the ammeter reading I and voltmeter reading VTV_T. Open the switch. Replace the external resistor with a different value and repeat, collecting at least five (I, VTV_T) data pairs using different resistors. (c) The terminal voltage obeys VTV_T = ε − Ir. Plotting VTV_T on the vertical axis against I on the horizontal axis yields a straight line with y-intercept equal to ε and slope equal to −r. A best-fit line through the data points allows extraction of both quantities. (d) One systematic error is contact resistance at wire junctions and clips, which acts as an additional series resistance indistinguishable from internal resistance. This would make the measured slope steeper (larger apparent r). To mitigate this, use clean, tight connections and soldered joints where possible.

Question 5: CRITICAL THINKING

A circuit consists of a 12 V ideal battery connected to three resistors. R₁ = 6 Ω is connected in series with a parallel combination of R₂ = 4 Ω and R₃ = 12 Ω. (a) Calculate the equivalent resistance of the circuit. (b) Determine the current through each resistor. (c) If R₃ is replaced by a wire (short-circuited), explain qualitatively and quantitatively what happens to the current through R₂ and the total power delivered by the battery.
Show answer
(a) R_∥ = (4 × 12)/(4 + 12) = 48/16 = 3 Ω. R_eq = R₁ + R_∥ = 6 + 3 = 9 Ω. (b) I_total = 12 V / 9 Ω = 4/3 A ≈ 1.33 A. The voltage across the parallel section is V_∥ = I_total × R_∥ = (4/3)(3) = 4 V. Therefore I₂ = 4 V / 4 Ω = 1 A and I₃ = 4 V / 12 Ω = 1/3 A ≈ 0.33 A. Check: 1 + 1/3 = 4/3 ✓. (c) Replacing R₃ with a wire creates a zero-resistance path in parallel with R₂. Since 1/R_∥ = 1/4 + 1/0 → ∞, R_∥ = 0 Ω, meaning the parallel combination is a short circuit. The entire voltage of the parallel section drops to 0 V, so I₂ = 0 V / 4 Ω = 0 A — no current flows through R₂ because all current takes the zero-resistance path. The new total resistance is just R₁ = 6 Ω, giving I_total = 12/6 = 2 A. The total power is now P = εI = 12 × 2 = 24 W, compared to the original 12 × (4/3) = 16 W, a 50% increase. All 24 W is dissipated in R₁.Simple-circuit preset: 12 V, series 6 Ω, parallel 4/12 Ω; Req=9 Ω, total 4/3 A, branches 1 and 1/3 A.Separate simple-circuit preset: 6 Ω in series with 4 Ω and 12 Ω branches.Opening the 4 Ω branch leaves 6 Ω and 12 Ω in series: 2/3 A, branch voltage 8 V.Opening the 4 Ω branch leaves the 12 Ω branch carrying current. A bridge between A and B bypasses both parallel resistors while the 6 Ω series resistor remains.Bridging the parallel section sets VAB=0; only 6 Ω limits 12 V, giving total 2 A and zero resistor-branch currents.A bridge across nodes A and B bypasses both parallel resistors.

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