AP Physics C: Electricity & Magnetism • Electric Circuits

Compound Direct Current Circuits

Mastering the analysis of circuits that combine series and parallel elements into complex resistive networks.

A magnet with electric energy and an atomic field.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The story of circuit analysis is inseparable from the broader development of electrical science in the nineteenth century. When Alessandro Volta constructed the first reliable chemical battery in 1800, experimenters gained a steady source of current for the first time, but they lacked the mathematical language to predict how that current would distribute itself through branching conductors. Compound direct current circuits—networks that contain both series and parallel combinations of resistive elements—became the central puzzle of early electrical engineering, driving the formulation of the laws and techniques that still anchor circuit analysis today.

1800 • Volta's Pile

Alessandro Volta announces the first electrochemical battery, providing a continuous source of direct current and enabling systematic study of resistive circuits.

1827 • Ohm's Law Published

Georg Simon Ohm publishes his treatise relating voltage, current, and resistance (V = IR), giving physicists the first quantitative tool for analyzing simple circuits.

1845 • Kirchhoff's Circuit Laws

Gustav Kirchhoff formulates the junction rule (conservation of charge) and the loop rule (conservation of energy), making it possible to solve arbitrary compound circuits systematically.

1883 • Edison's DC Distribution Network

Thomas Edison's Pearl Street Station powers lower Manhattan using a compound DC network of series and parallel loads, demonstrating the practical necessity of compound circuit design.

The fundamental question that compound circuits pose is both simple and profound: given a network of resistors connected in an arbitrary mixture of series and parallel arrangements to one or more voltage sources, how do we determine the current through, and voltage across, every element? Answering this question requires the systematic application of Ohm's law and Kirchhoff's laws—the same tools that Kirchhoff devised nearly two centuries ago—together with the strategy of reducing complex topologies to equivalent resistances.

SECTION 2

Core Principles & Definitions

Before tackling a compound circuit, it is essential to internalize the rules that govern current flow and energy transfer in any DC network. A compound circuit (sometimes called a combination circuit) is one that cannot be classified as purely series or purely parallel; instead, it contains sub-groups of resistors in series nested within parallel branches, or vice versa. Analyzing such a circuit proceeds by identifying these sub-groups, reducing them to equivalent resistances, and then applying the fundamental laws.

1

Series Connection

Elements share the same current. Voltages add: Vtotal = V₁ + V₂ + ⋯. Equivalent resistance: Req = R₁ + R₂ + ⋯.
2

Parallel Connection

Elements share the same voltage. Currents add: Itotal = I₁ + I₂ + ⋯. Equivalent resistance: 1/Req = 1/R₁ + 1/R₂ + ⋯.
3

Kirchhoff's Junction Rule (KCL)

At any node, the algebraic sum of currents equals zero: ΣIin = ΣIout. This is conservation of charge.
4

Kirchhoff's Loop Rule (KVL)

Around any closed loop, the algebraic sum of potential differences is zero: ΣΔV = 0. This is conservation of energy per unit charge.
5

Equivalent Resistance Strategy

Systematically reduce series and parallel sub-groups to single equivalent resistors until the entire network is one resistor. Then work backward ("expand") to find individual voltages and currents.
✦

KEY TAKEAWAY

Think of a compound circuit like a highway system. A series segment is a single-lane road where every car (charge) must pass through: the traffic flow (current) is the same everywhere, but each toll booth (resistor) drops the speed (voltage). A parallel segment is a fork offering multiple lanes: the total traffic splits among the lanes, but each lane connects the same two interchanges, so the "height drop" (voltage) is the same across every lane. A compound circuit simply mixes single-lane stretches with multi-lane forks.

SECTION 3

Visual Explanation — A Compound Circuit Diagram

Compound DC Circuit — Series-Parallel Networkε24VR₁ = 4ΩAR₂ = 6ΩR₃ = 3ΩR₄ = 9ΩBIR₂ is in parallel with the series combination (R₃ + R₄). That parallel group is in series with R₁.
A battery (ε = 24 V) drives current through R₁ in series, then the current splits at junction A into two parallel branches: one containing R₂ alone and another containing R₃ + R₄ in series. The branches rejoin at junction B and return to the battery.

The diagram above illustrates the essential topology of a compound circuit. Notice that R₁ is in series with the rest of the circuit because all of the current from the battery must pass through it before reaching junction A. At junction A the current splits: some flows through R₂, and the remainder flows through the series pair R₃ + R₄. Because R₂ and the (R₃ + R₄) branch connect between the same two nodes (A and B), they are in parallel. This hierarchical nesting—series within parallel, or parallel within series—is the hallmark of compound circuits and the key to simplifying them.

SECTION 4

Mathematical Framework

The mathematical analysis of compound circuits rests on three pillars: Ohm's law, the series resistance formula, and the parallel resistance formula. For circuits that resist simplification by series-parallel reduction alone, Kirchhoff's laws provide a system of linear equations that can always be solved.

OHM'S LAW

V=IRV = IR

V = potential difference across the element (V), I = current through the element (A), R = resistance of the element (Ω). Applies to each resistor individually and to any equivalent resistance.

SERIES EQUIVALENT RESISTANCE

Req=R1+R2+R3+⋯R_eq = R₁ + R₂ + R₃ + ⋯

For resistors in series, the total resistance is the arithmetic sum. The same current flows through each resistor, and the total voltage drop equals the sum of individual drops.

PARALLEL EQUIVALENT RESISTANCE

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

For resistors in parallel, reciprocals add. All resistors share the same voltage, and the total current is the sum of the branch currents. For two resistors: Req = R₁R₂ / (R₁ + R₂).

KIRCHHOFF'S VOLTAGE LAW (KVL)

ΣΔVi=0(aroundanyclosedloop)Σ ΔV_i = 0 (around any closed loop)

Traversing a closed loop, the sum of EMFs and IR drops equals zero. By convention, moving through a resistor in the direction of assumed current gives −IR, and moving from − to + through a battery gives +ε.

💡

When Series-Parallel Reduction Fails

Some compound circuits (e.g., a Wheatstone bridge with an unbalanced galvanometer branch) cannot be reduced by series and parallel rules alone. In these cases, assign current variables to each branch, write KCL equations at each independent junction and KVL equations around independent loops, and solve the resulting system of linear equations. The number of independent equations always equals the number of unknown currents.

SECTION 5

Step-by-Step Reduction Strategy

The most common technique for solving compound circuits on the AP exam is series-parallel reduction followed by back-substitution. The procedure has two phases: a forward phase ("collapse") in which you simplify the network to a single equivalent resistance, and a reverse phase ("expand") in which you unpack each reduction to recover individual voltages and currents.

Reduction Flowchart — Collapse & ExpandSTEP 1Identify innermostseries or parallel groupSTEP 2Replace group withR_eq using formulasSTEP 3Repeat until oneequivalent R remainsSTEP 4Find total I fromI = ε / R_totalSTEP 5 — EXPANDReverse each reduction:V = IR or I = V/RSTEP 6 — VERIFYCheck: ΣV = ε aroundeach loop; ΣI at nodesExample: 24V Battery, R₁=4Ω in series with (R₂=6Ω ∥ [R₃=3Ω + R₄=9Ω])Collapse: R₃₄ = 3 + 9 = 12ΩR₂₃₄ = (6 × 12)/(6 + 12) = 72/18 = 4ΩR_total = 4 + 4 = 8ΩI_total = 24V / 8Ω = 3AV_parallel = 3A × 4Ω = 12V → I₂ = 12/6 = 2A, I₃₄ = 12/12 = 1A ✓
The six-step collapse-and-expand strategy. The bottom box shows the numerical reduction for the circuit from Section 3, confirming that the total current is 3 A and the branch currents through R₂ and R₃₄ are 2 A and 1 A respectively.

A critical insight is that during the expand phase, you exploit the fact that series elements share current and parallel elements share voltage. When you un-collapse a series pair, you already know the current (it equals the current through the equivalent), so you use V = IR to find each voltage. When you un-collapse a parallel pair, you already know the voltage (it equals the voltage across the equivalent), so you use I = V/R to find each branch current. This alternation of "same current" and "same voltage" reasoning is the backbone of compound circuit analysis.

SECTION 6

Worked Example — Full Compound Circuit Analysis

Consider the circuit shown in Section 3: a 24 V ideal battery in series with R₁ = 4 Ω, which then connects to a parallel pair consisting of R₂ = 6 Ω and the series combination R₃ + R₄ = 3 Ω + 9 Ω. We wish to find the total current, the voltage across and current through every resistor, and the power dissipated by each.

Compound Circuit — Collapse & Expand

  1. 1

    Step 1 — Identify the Innermost Group

    R₃ and R₄ are in series (they share the same branch and carry the same current). Combine them first: R₃₄ = R₃ + R₄ = 3 Ω + 9 Ω.
    R₃₄ = 12 Ω
  2. 2

    Step 2 — Reduce the Parallel Combination

    R₂ (6 Ω) is in parallel with R₃₄ (12 Ω). Using the two-resistor shortcut: R₂₃₄ = (R₂ × R₃₄) / (R₂ + R₃₄) = (6 × 12) / (6 + 12) = 72 / 18.
    R₂₃₄ = 4 Ω
  3. 3

    Step 3 — Find Total Resistance

    R₁ is in series with R₂₃₄: R_total = R₁ + R₂₃₄ = 4 Ω + 4 Ω.
    R_total = 8 Ω
  4. 4

    Step 4 — Total Current from the Battery

    Apply Ohm's law to the entire circuit: I_total = ε / R_total = 24 V / 8 Ω.
    I_total = 3 A
  5. 5

    Step 5 — Expand: Voltage Across R₁

    R₁ carries the full battery current (series). V₁ = I_total × R₁ = 3 A × 4 Ω = 12 V. The remaining voltage appears across the parallel section: V_parallel = ε − V₁ = 24 − 12.
    V₁ = 12 V; V_parallel = 12 V
  6. 6

    Step 6 — Expand: Branch Currents

    Both parallel branches see 12 V. I₂ = V_parallel / R₂ = 12 / 6 = 2 A. I₃₄ = V_parallel / R₃₄ = 12 / 12 = 1 A. Check: I₂ + I₃₄ = 2 + 1 = 3 A = I_total ✓.
    I₂ = 2 A; I₃₄ = 1 A
  7. 7

    Step 7 — Voltage Drops Across R₃ and R₄

    R₃ and R₄ share the branch current 1 A. V₃ = 1 × 3 = 3 V; V₄ = 1 × 9 = 9 V. Check: V₃ + V₄ = 3 + 9 = 12 V = V_parallel ✓.
    V₃ = 3 V; V₄ = 9 V
  8. 8

    Step 8 — Power Dissipated

    P = IV for each: P₁ = 3 × 12 = 36 W; P₂ = 2 × 12 = 24 W; P₃ = 1 × 3 = 3 W; P₄ = 1 × 9 = 9 W. Total = 36 + 24 + 3 + 9 = 72 W. Check: P_battery = εI = 24 × 3 = 72 W ✓.
    P_total = 72 W (consistent with battery output)

SECTION 7

Common Pitfalls & Comparisons

Students frequently lose points on the AP exam not from conceptual misunderstanding but from procedural errors during circuit reduction. The table below compares common mistakes with the correct approaches and highlights the reasoning behind each.

Common MistakeWhy It's WrongCorrect Approach
Adding all resistors in a compound circuit as if they were all in seriesParallel branches share voltage, not current; adding them directly overestimates total resistanceIdentify series and parallel sub-groups and reduce them separately using the appropriate formula
Forgetting to invert after summing reciprocals in the parallel formula1/R_eq = 1/R₁ + 1/R₂ gives the reciprocal of the answer; reporting 1/R_eq as R_eq yields a value that's far too smallAfter summing reciprocals, take one final reciprocal. For two resistors, use the product-over-sum shortcut: R_eq = R₁R₂/(R₁+R₂)
Assuming all resistors have the same voltageOnly resistors in parallel (connected between the same two nodes) share voltage; series resistors generally have different voltagesDetermine the topology first. Series → same I; Parallel → same V. Compute unknown quantities from Ohm's law
Applying KVL to a path that is not a closed loopKVL states ΣΔV = 0 only around a closed loop; an open path does not return to its starting potentialEnsure every KVL equation traces a complete closed path through the circuit back to the starting node
Not verifying the solution with conservation checksWithout checks, sign or arithmetic errors propagate; the exam graders expect self-consistent answersAlways verify: (1) currents sum at junctions, (2) voltages sum around loops, (3) total power dissipated = power delivered by the source

Common errors in compound circuit analysis and their corrections

✦

KEY TAKEAWAY

The most reliable exam strategy is to treat verification as an integral part of the solution, not an afterthought. Just as an engineer runs a finite-element simulation and then checks boundary conditions, you should always confirm that the sum of voltage drops around every loop equals the EMF and that currents balance at every junction. These quick checks catch arithmetic slips before they cost you points.

SECTION 8

Connection to Advanced Circuit Theory

The series-parallel reduction technique you have mastered is the foundation upon which more sophisticated methods build. On the AP Physics C exam, you may encounter circuits with multiple EMFs, internal resistances, or topologies (like the Wheatstone bridge) that resist simple reduction. The table below maps the concepts from this lesson to their advanced counterparts.

This Lesson (DC Compound Circuits)Advanced Extension
Series-parallel reduction to a single R_eqDelta-Wye (Δ-Y) transformations for non-reducible networks; Thévenin and Norton equivalent circuits for arbitrary two-terminal networks
Single-battery KVL loopsMulti-loop Kirchhoff analysis with simultaneous linear equations; mesh current method (matrix formulation)
KCL at single junctionsNode-voltage method: assign potentials at each independent node and write KCL in terms of voltages
Ideal batteries (no internal resistance)Real batteries with internal resistance r; terminal voltage VTV_T = ε − Ir; maximum power transfer theorem
Resistors only (steady-state DC)RC circuits: transient behavior, exponential charging/discharging with time constant τ = RC

Mapping compound DC circuit concepts to advanced topics

The ability to reduce a compound circuit is also a prerequisite for understanding RC transient circuits, which appear prominently on the AP exam. In an RC circuit, the effective resistance "seen" by the capacitor during charging or discharging often involves a compound network of resistors, and you must reduce that network to find the correct time constant τ. Mastering compound DC analysis now therefore pays dividends across multiple topics in the Electric Circuits unit.

SECTION 9

Practice Problems

Question 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. If the 20 Ω resistor were replaced with a wire (0 Ω), what would happen to the total current delivered by an ideal battery?
  • A. The total current would decrease because a wire has less resistance than the 20 Ω resistor it replaced.
  • B. The total current would increase because the parallel combination's equivalent resistance drops to zero, reducing the total circuit resistance to 10 Ω.
  • C. The total current would increase because the parallel combination's equivalent resistance drops to zero, reducing the total circuit resistance to just the 10 Ω series resistor, and all current bypasses the 30 Ω resistor.
  • D. The total current would remain the same because the battery always delivers the same current regardless of circuit changes.
Show answer (C)
Replacing the 20 Ω resistor with a wire (0 Ω) short-circuits the parallel combination. Any resistance in parallel with 0 Ω gives R_eq = 0 Ω, since 1/R_eq = 1/0 + 1/30 → ∞, so R_eq → 0. The 30 Ω resistor is effectively bypassed because all current takes the zero-resistance path. The total resistance becomes just R₁ = 10 Ω, so the current from the battery becomes I = ε/10, which is larger than ε/(10 + R_parallel) for any nonzero R_parallel. Choice A is wrong because less resistance means more current, not less. Choice B correctly states the current increases but is incomplete—it does not explain that the 30 Ω is bypassed. Choice D is wrong because an ideal battery maintains constant voltage, not constant current, so current depends on total resistance.

Question 2: BASIC CALCULATION

A 12 V battery is connected to a circuit in which a 6 Ω resistor (R₁) is in series with a parallel pair of 4 Ω (R₂) and 12 Ω (R₃). What is the current through R₃?
  • A. 0.33 A
  • B. 1.0 A
  • C. 1.5 A
  • D. 2.0 A
Show answer (A)
Parallel equivalent: R₂₃ = (4 × 12)/(4 + 12) = 48/16 = 3 Ω. Total resistance: R_total = 6 + 3 = 9 Ω. Total current: I = 12/9 = 4/3 A. Voltage across parallel section: V_parallel = (4/3) × 3 = 4 V. Current through R₃: I₃ = 4/12 = 1/3 A = 0.33 A. Choice A is correct. Choice B (1.0 A) is the current through R₂. Choice D (2.0 A) is the current through R₁ alone. Choice C (1.5 A) may result from other calculation errors such as incorrect equivalent resistance.

Question 3: INTERMEDIATE

Three resistors are connected to a 48 V ideal battery. R₁ = 8 Ω is in series with a parallel combination of R₂ = 24 Ω and R₃ = 12 Ω. Determine the total current from the battery, the voltage across the parallel combination, and the power dissipated by R₂.
  • A. I_total = 3.0 A, V_parallel = 24 V, P₂ = 24 W
  • B. I_total = 3.0 A, V_parallel = 24 V, P₂ = 48 W
  • C. I_total = 2.0 A, V_parallel = 16 V, P₂ = 10.67 W
  • D. I_total = 6.0 A, V_parallel = 48 V, P₂ = 96 W
Show answer (A)
Parallel equivalent: R₂₃ = (24 × 12)/(24 + 12) = 288/36 = 8 Ω. Total resistance: R_total = R₁ + R₂₃ = 8 + 8 = 16 Ω. Total current: I_total = 48/16 = 3.0 A. Voltage across R₁: V₁ = 3.0 × 8 = 24 V. Voltage across parallel section: V_parallel = 48 − 24 = 24 V. Current through R₂: I₂ = 24/24 = 1.0 A. Power dissipated by R₂: P₂ = I₂²R₂ = (1)²(24) = 24 W, or equivalently P₂ = V²/R = (24)²/24 = 24 W. Choice A is correct. Choice B incorrectly doubles the power (confusing P = IV with P = εI). Choice C uses an incorrect total resistance. Choice D ignores the series resistor entirely, treating only R₂ as the load.

Question 4: APPLIED

10 pts
An engineer designs a sensor circuit in which a 100 V DC power supply connects to three resistors. R₁ = 20 Ω is in series with the parallel combination of R₂ = 30 Ω and a branch containing R₃ = 10 Ω in series with R₄ = 20 Ω. The sensor element is R₃; it has a maximum safe current rating of 2.0 A. (a) Find the current through R₃. (b) Determine whether the sensor is within its safe operating limit. (c) Calculate the total power delivered by the power supply. (d) If R₂ were removed (open-circuited), find the new current through R₃ and state whether the sensor is still safe.
Show answer
RUBRIC (10 points): (a) Finding I through R₃ (4 points): • [1 pt] R₃₄ = R₃ + R₄ = 10 + 20 = 30 Ω • [1 pt] R_parallel = (R₂ × R₃₄)/(R₂ + R₃₄) = (30 × 30)/(30 + 30) = 900/60 = 15 Ω • [1 pt] R_total = R₁ + R_parallel = 20 + 15 = 35 Ω; I_total = 100/35 = 20/7 ≈ 2.86 A • [1 pt] V_parallel = I_total × 15 = (20/7)(15) = 300/7 ≈ 42.86 V; I₃ = V_parallel/R₃₄ = (300/7)/30 = 10/7 ≈ 1.43 A (b) Safety check (1 point): • [1 pt] I₃ ≈ 1.43 A < 2.0 A, so the sensor is within its safe operating limit. (c) Total power (2 points): • [1 pt] P_total = εI_total = 100 × (20/7) = 2000/7 ≈ 285.7 W • [1 pt] Equivalently, P = ε²/R_total = 10000/35 ≈ 285.7 W (consistency check) (d) R₂ removed (3 points): • [1 pt] With R₂ removed, only the R₃-R₄ branch remains in series with R₁: R_total_new = 20 + 10 + 20 = 50 Ω • [1 pt] I₃_new = 100/50 = 2.0 A • [1 pt] Since I₃_new = 2.0 A, which equals the maximum safe current rating, the sensor is at the threshold of its safe limit. Depending on interpretation, this is marginally safe but offers no safety margin, and any additional voltage fluctuation could exceed the rating.

Question 5: CRITICAL THINKING

7 pts
Consider a compound circuit with an ideal battery of EMF ε 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 equivalent resistance in terms of R. (b) Derive expressions for the current through each resistor and the voltage across each resistor. (c) Now suppose the battery has internal resistance r. Derive the new expression for the current through resistor A and show that the fraction of ε that appears across the external circuit is R_ext/(R_ext + r), where R_ext is the external equivalent resistance. (d) If one wished to maximize the power delivered to the external circuit, show that the condition is R_ext = r and find the maximum external power in terms of ε and r.
Show answer
RUBRIC (7 points): (a) Total equivalent resistance (1 point): • [1 pt] B ∥ C: R_BC = R×R/(R+R) = R/2. Total: R_ext = R + R/2 = 3R/2. (b) Currents and voltages (2 points): • [1 pt] I_total = ε/R_ext = ε/(3R/2) = 2ε/(3R). This is also IAI_A. VAV_A = IAI_A × R = 2ε/3. • [1 pt] V_parallel = ε − VAV_A = ε/3. By symmetry, IBI_B = ICI_C = (ε/3)/R = ε/(3R). Check: IBI_B + ICI_C = 2ε/(3R) = IAI_A ✓. (c) With internal resistance (2 points): • [1 pt] IAI_A = ε/(R_ext + r) = ε/(3R/2 + r). Terminal voltage: VTV_T = ε − IAI_A × r = ε × R_ext/(R_ext + r). • [1 pt] Therefore the fraction across the external circuit is VTV_T/ε = R_ext/(R_ext + r), QED. (d) Maximum power transfer (2 points): • [1 pt] P_ext = I²R_ext = [ε/(R_ext + r)]² × R_ext. To maximize, take dP_ext/dR_ext = 0: using the quotient rule, (R_ext + r)² − 2R_ext(R_ext + r) = 0 → R_ext + r − 2R_ext = 0 → R_ext = r. • [1 pt] P_max = [ε/(2r)]² × r = ε²/(4r). This is the maximum power transfer theorem.

SUMMARY

Compound Direct Current Circuits — Summary

Summary

A compound DC circuit contains resistors arranged in nested series and parallel sub-groups. The fundamental analysis strategy is collapse and expand: use the series formula (Req = ΣR) and the parallel formula (1/Req = Σ1/R) to reduce the network to a single equivalent resistance, find the total current via Ohm's law (V = IR), and then work backward—using the rule that series elements share current and parallel elements share voltage—to recover every individual current and voltage in the circuit.

Always verify your results using Kirchhoff's junction rule (ΣIin = ΣIout) and Kirchhoff's loop rule (ΣΔV = 0). Confirm that the total power dissipated by all resistors equals the power delivered by the source (P = εI). These conservation checks catch arithmetic errors and are expected on free-response answers. Mastery of compound circuit analysis also prepares you for RC transient analysis, multi-loop Kirchhoff problems, and the maximum power transfer theorem—topics that build directly on the skills developed here.

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