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  1. AP Physics 2
  2. Compound Direct Current (DC) Circuits

AP PHYSICS 2: ALGEBRA-BASED • ELECTRIC CIRCUITS

Compound Direct Current (DC) Circuits

Master the analysis of circuits containing both series and parallel combinations of resistors.

SECTION 1

Historical Context & Motivation

Real-world electrical systems—from the wiring inside your phone to the power grid feeding your home—almost never consist of purely series or purely parallel arrangements. Instead, they combine both topologies into what engineers and physicists call compound (or combination) circuits. Understanding how to analyze these networks required centuries of foundational work in electrostatics, current flow, and energy conservation, culminating in the elegant circuit laws we rely on today.

1800
Volta's Pile
Alessandro Volta constructed the first true battery, providing a steady source of direct current and making sustained circuit experiments possible for the first time.
1827
Ohm's Law Published
Georg Simon Ohm formalized the proportional relationship V = IR between voltage, current, and resistance, giving physicists the key tool for quantitative circuit analysis.
1845
Kirchhoff's Circuit Laws
Gustav Kirchhoff introduced the junction rule (conservation of charge) and the loop rule (conservation of energy), enabling systematic analysis of any circuit topology, no matter how complex.
1883
Edison's DC Power Grid
Thomas Edison's Pearl Street Station in New York demonstrated a practical compound DC network, wiring hundreds of lamps in parallel across series distribution lines.

The central question this lesson addresses is: given a circuit where resistors appear in both series and parallel groupings, how do we systematically find the equivalent resistance, the current through each element, and the voltage drop across each element? Mastering this technique is essential for AP Physics 2, where compound circuits appear regularly in both MCQ and FRQ contexts.

SECTION 2

Core Principles & Definitions

Before tackling compound circuits, you need a firm grasp of the rules governing the two fundamental resistor configurations and the conservation laws that bind every circuit together. The following principles form the analytical toolkit you will apply repeatedly throughout this lesson.

1

Series Resistors

Resistors in series share the same current. Their resistances add directly: Req = R₁ + R₂ + … . The total voltage splits among them proportionally to each resistance.
2

Parallel Resistors

Resistors in parallel share the same voltage. Their reciprocals add: 1/Req = 1/R₁ + 1/R₂ + … . The total current splits, with more current flowing through the smaller resistance.
3

Kirchhoff's Junction Rule

At any junction (node), the sum of currents entering equals the sum of currents leaving. This is a direct consequence of conservation of electric charge.
4

Kirchhoff's Loop Rule

Around any closed loop, the sum of all voltage gains and drops equals zero. This reflects conservation of energy per unit charge around a complete path.
5

Compound (Combination) Circuit

A circuit containing both series and parallel sub-networks. Analysis proceeds by identifying reducible sub-groups, replacing them with equivalent resistances, and working inward then outward.
✦ KEY TAKEAWAY
KEY TAKEAWAY
SECTION 3

Visual Explanation — Anatomy of a Compound Circuit

Compound DC Circuit — Series-Parallel Networkε12 VR₁ = 4 ΩAR₂ = 6 ΩR₃ = 12 ΩBIR₂ and R₃ are in parallel (same nodes A and B).R₁ is in series with the parallel combination.Series sub-circuitSame current through R₁Parallel sub-circuitSame voltage across R₂ & R₃
A 12 V battery drives current through R₁ (4 Ω) in series with a parallel pair: R₂ (6 Ω) and R₃ (12 Ω). Nodes A and B mark where the current splits and recombines.

The diagram above illustrates the most common type of compound circuit you will encounter on the AP exam. Notice how the current has only one path through R₁, making it a series element. At node A the current forks: some flows through R₂ and the rest through R₃. Because R₂ and R₃ share nodes A and B, they form a parallel sub-network. The strategy is to reduce the parallel pair to a single equivalent resistance, then add that equivalent in series with R₁ to obtain the total circuit resistance.

SECTION 4

Mathematical Framework

The analysis of compound circuits rests on three equations used in combination: Ohm's law, the series resistance formula, and the parallel resistance formula. You will apply these iteratively—first inward (reducing sub-networks) then outward (distributing current and voltage).

OHM'S LAW
V = I × R
V = potential difference (V), I = current (A), R = resistance (Ω). Applies to any single resistor or any equivalent resistance.
SERIES EQUIVALENT RESISTANCE
R_series = R₁ + R₂ + R₃ + …
Elements in series carry the same current. The total resistance is always greater than the largest individual resistance.
PARALLEL EQUIVALENT RESISTANCE
1/R_parallel = 1/R₁ + 1/R₂ + 1/R₃ + …
Elements in parallel share the same voltage. The equivalent resistance is always less than the smallest individual resistance. For exactly two resistors: Rparallel = (R₁ × R₂) / (R₁ + R₂).
POWER DISSIPATED
P = I × V = I²R = V²/R
P = power (W). Any of these three forms may be convenient depending on which quantities are known for a given element.
AP EXAM TIP
SECTION 5

Step-by-Step Reduction Strategy

The general procedure for solving any compound circuit can be summarized in two phases: reduce inward (simplify sub-networks to find total resistance and total current) and expand outward (distribute voltages and currents back to individual elements). The diagram below visualizes this two-phase approach applied to a three-resistor compound circuit.

Two-Phase Reduction StrategyPHASE 1: REDUCE INWARDPHASE 2: EXPAND OUTWARDStep 1Identify series and parallel groups.R₂ ∥ R₃ → parallel pair between A & BStep 2Replace parallel group with R₂₃ = 4 Ω(6×12)/(6+12) = 72/18 = 4 ΩStep 3Add in series: R_total = 4 + 4 = 8 ΩI_total = ε / R_total = 12 / 8 = 1.5 AStep 4V across R₁ = I × R₁ = 1.5 × 4 = 6 VRemaining: 12 − 6 = 6 V across parallel pairStep 5Branch currents via Ohm's law:I₂ = 6/6 = 1.0 A I₃ = 6/12 = 0.5 AStep 6Verify: I₂ + I₃ = 1.0 + 0.5 = 1.5 A ✓Junction rule satisfied at node A.VERIFICATIONP_total = ε × I = 12 × 1.5 = 18 WP₁ = 9 W, P₂ = 6 W, P₃ = 3 W → Sum = 18 W ✓ (Energy conservation)
The left column shows the inward reduction (parallel → series → total), while the right column shows the outward expansion (total current → voltage drops → branch currents). The verification box confirms energy conservation.
  • Identify all series and parallel sub-groups by tracing current paths from the battery.
  • Replace each identified sub-group with its equivalent resistance, starting from the innermost group.
  • Repeat until a single equivalent resistance remains; apply Ohm's law to find total current.
  • Expand outward: use Ohm's law and Kirchhoff's rules to find voltage and current for each original element.
  • Verify that branch currents sum correctly at junctions and voltage drops sum to zero around every loop.
SECTION 6

Worked Example

Consider the circuit shown in Section 3: a 12 V battery connected to R₁ = 4 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. Find the equivalent resistance, total current, voltage across each resistor, current through each resistor, and power dissipated in each resistor.

Step 1 — Find the Parallel Equivalent of R₂ and R₃

R₂ and R₃ share nodes A and B, so they are in parallel. Using the two-resistor shortcut: R23 = (R₂ × R₃) / (R₂ + R₃) = (6 × 12) / (6 + 12) = 72 / 18.
R₂₃ = 4 Ω

Step 2 — Find the Total Equivalent Resistance

R₁ is in series with R₂₃. For series resistors: Rtotal = R₁ + R₂₃ = 4 + 4.
R_total = 8 Ω

Step 3 — Find the Total Current

Apply Ohm's law to the entire circuit: Itotal = ε / Rtotal = 12 / 8.
I_total = 1.5 A

Step 4 — Voltage Drop Across R₁

Since R₁ is in series, it carries the full 1.5 A. V₁ = Itotal × R₁ = 1.5 × 4.
V₁ = 6 V

Step 5 — Voltage Across the Parallel Pair

By Kirchhoff's loop rule, the remaining voltage appears across the parallel combination: V23 = ε − V₁ = 12 − 6. This is also V₂ = V₃ because parallel elements share the same voltage.
V₂ = V₃ = 6 V

Step 6 — Branch Currents in the Parallel Pair

Apply Ohm's law to each branch: I₂ = V₂ / R₂ = 6 / 6 = 1.0 A, and I₃ = V₃ / R₃ = 6 / 12 = 0.5 A. Check: I₂ + I₃ = 1.0 + 0.5 = 1.5 A = Itotal. ✓
I₂ = 1.0 A, I₃ = 0.5 A

Step 7 — Power Dissipated

P₁ = I₁² × R₁ = (1.5)² × 4 = 9 W. P₂ = I₂² × R₂ = (1.0)² × 6 = 6 W. P₃ = I₃² × R₃ = (0.5)² × 12 = 3 W. Total power = 9 + 6 + 3 = 18 W, which matches Pbattery = ε × I = 12 × 1.5 = 18 W. ✓
P₁ = 9 W, P₂ = 6 W, P₃ = 3 W
SECTION 7

Series vs. Parallel vs. Compound — A Comparison

Comparison of series, parallel, and compound circuit properties
PropertyPure SeriesPure ParallelCompound
CurrentSame through all elementsSplits at junctionsSame in series sections; splits in parallel sections
VoltageDivides proportional to RSame across all branchesDivides in series sections; shared in parallel sections
R_eqR₁ + R₂ + …(1/R₁ + 1/R₂ + …)⁻¹Combine sub-groups iteratively
Effect of removing one REntire circuit breaks (open)Other branches unaffectedDepends on location; may affect only a sub-branch or the whole circuit
Typical useVoltage dividers, daisy-chained holiday lightsHousehold outlets, car battery loadsMost real circuits: computers, power grids, lab setups
✦ KEY TAKEAWAY
KEY TAKEAWAY
SECTION 8

Connection to Advanced Circuit Theory

The reduction technique you have learned works beautifully for ladder networks—circuits that can be fully decomposed into nested series-parallel groups. However, some circuits (such as the Wheatstone bridge or more general mesh topologies) cannot be reduced by simple series-parallel substitution. For those, physicists and engineers turn to Kirchhoff's loop and junction equations written as a system of simultaneous linear equations, or to more advanced methods like mesh analysis and nodal analysis covered in college-level circuit theory.

AP-level compound circuits vs. college-level circuit theory
FeatureAP Physics 2 (This Lesson)College Circuit Theory
Circuit typesSeries, parallel, series-parallel compoundAny topology including bridge and mesh networks
Primary methodSeries-parallel reduction + Ohm's lawMesh/nodal analysis, Thévenin/Norton equivalents
ComponentsIdeal resistors, ideal battery (no internal resistance on most problems)Resistors, capacitors, inductors, dependent sources, non-ideal batteries
Math requiredAlgebra (fractions, reciprocals)Linear algebra, differential equations (for AC and transients)

Even so, the intuition you build here—recognizing series and parallel relationships, applying conservation of charge at junctions and conservation of energy around loops—carries directly into every future circuit course. Many advanced methods are simply efficient bookkeeping systems for the same Kirchhoff's laws you already know.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL
In a compound circuit, a 10 Ω resistor is in series with a parallel combination of a 20 Ω and a 30 Ω resistor. Which of the following correctly describes the current through and the voltage across the 10 Ω resistor compared with the 20 Ω resistor?
PROBLEM 2 — BASIC CALCULATION
A 24 V battery is connected to a 6 Ω resistor in series with two resistors (8 Ω and 24 Ω) in parallel. What is the total current drawn from the battery?
PROBLEM 3 — INTERMEDIATE
In a compound circuit, a 12 V battery drives current through R₁ = 4 Ω in series with a parallel pair of R₂ = 6 Ω and R₃ = 12 Ω. What is the power dissipated in R₂?
PROBLEM 4 — APPLIED
A student has a 12 V battery, an ammeter, a voltmeter, connecting wires, and three resistors of unknown value (labeled X, Y, and Z). The student suspects that when Y and Z are connected in parallel and then placed in series with X, the total current from the battery will be 2.0 A. Design an experiment to verify this hypothesis and determine the resistance of each resistor.
PROBLEM 5 — CRITICAL THINKING
A compound circuit consists of a battery of EMF ε with negligible internal resistance connected to three identical resistors, each of resistance R. Resistor A is in series with a parallel combination of resistors B and C. (a) Derive an expression for the total power dissipated in the circuit in terms of ε and R. (b) If resistor C is removed (open-circuited), determine the new total power and calculate the ratio P_new / P_original. (c) Explain physically why the total power changes when a parallel branch is removed.
SUMMARY

Summary

Varsity Tutors • AP Physics 2: Algebra-Based • Compound Direct Current (DC) Circuits