Compound Direct Current (DC) Circuits
Identify genuine series and parallel groups and place meters where their quantities are measured.
AP Physics 2 Topic 11.5 · College Board CED printed p. 67, 68, 69 / PDF p. 74, 75, 76 · See this topic in the course outline
Reduce a network only after identifying which nodes each component connects. Then expand it again to recover branch currents and voltage drops.
For a supporting original reading, see Resistors in Circuits. Identify genuine series and parallel groups before reducing them; ideal meters must be connected to the quantity they measure.
Before you start
- Circuit nodes and schematics
- Ohm’s law and power
Topology gives equal currents or equal voltages
Elements are in series if charge passing through one must pass through each of the others with no alternative path between them. Their steady currents are equal. The total voltage drop is the sum of drops, so .
Elements are in parallel if each connects across the same two nodes. Their voltage differences are equal and their branch currents add. Thus . For two resistors, . A parallel equivalent is smaller than the smallest individual resistance: adding a path increases total current for the same voltage, so reduces equivalent resistance.
| Connection | Equivalent | Shared quantity | How to expand after solving total |
|---|---|---|---|
| Series | 9 Ω | Current | Each voltage is IR |
| Parallel | 2 Ω | Voltage | Each current is ΔV/R |
A complicated network can be replaced by a single equivalent resistor as seen by the source, then expanded using those shared quantities. Some networks, such as a bridge with current through a cross-connection, do not reduce to series and parallel groups; use Kirchhoff equations there.
Internal resistance and resistive wires
Ideal wires have negligible resistance and no voltage drop. This is a good approximation when their resistance is much smaller than the other resistances in the circuit. It is not appropriate to erase all resistance from a battery short. Ideal batteries have negligible internal resistance, so their terminal voltage equals their emf ε independent of current in the model.
A real discharging battery is modeled as an ideal source ε in series with internal resistance r. With external equivalent resistance R, . Additional resistive wire segments add in series where the same current flows, with their own I Rwire drops and heating.
Emf equals open-circuit terminal voltage when no current flows through the battery. Under load, terminal voltage is lower because energy per charge is transferred internally. More current produces a larger Ir loss. The ε − Ir expression here describes discharge; charging a battery involves a different current direction and sign convention.
Put the meter where its quantity is defined
An ammeter measures current at a point in a branch. Insert it in series so the branch’s charge passes through it. An ideal ammeter has zero resistance and introduces no additional voltage drop; connecting one directly across a battery would create a short rather than measure useful branch current.
A voltmeter compares potential between two points. Connect it in parallel across the chosen element or endpoints. An ideal voltmeter has infinite resistance, so it draws no current and does not alter the circuit. A voltmeter inserted in series would block the ideal conducting path.
A nonideal ammeter has some resistance, so it can reduce the branch current it measures. A nonideal voltmeter has finite resistance, so it draws some extra current and changes the network being measured; with a source or feeding resistance, it may lower the loaded voltage. The size and direction of every affected reading depend on topology. AP requires qualitative effects, not numerical meter-loading calculations.
Unless specified, all batteries, wires, and meters are ideal. Unequal-potential batteries directly in parallel are outside assessed scope. For experiments, inspect polarity, range, and connections with the supply off before taking readings.
Reduce, solve, expand
A 12 V ideal battery supplies a 2 Ω resistor in series with parallel 6 Ω and 3 Ω branches. Find total current, all resistor voltages and currents, and total power.
Label the feed-to-branch node B and source-negative node D. The 6 Ω and 3 Ω resistors share B–D, so they are parallel. The 2 Ω feed has no alternative path before B and carries total current. An ideal ammeter can be inserted in that feed; a voltmeter across B–D measures either parallel branch voltage.
Why this step? Topology establishes the equal voltage and current conditions before arithmetic.
The parallel group has Rp = (6 × 3)/(6 + 3) = 2 Ω. Total Req = 2 + 2 = 4 Ω.
Why this step? The two branch resistors share both nodes; their group is in series with the first resistor.
Total current is 12/4 = 3 A. The series 2 Ω drop is 6 V, leaving 6 V across each branch. Branch currents are 6/6 = 1 A and 6/3 = 2 A.
Why this step? Expand using same series current and same parallel voltage.
Psource = 12 × 3 = 36 W. Resistor powers are 18 W, 6 W, and 12 W, which sum to 36 W.
Why this step? Power conservation checks both current division and voltage drops.
Answer and check
Itotal = 3 A; voltages are 6 V across each resistor; currents 3 A, 1 A, 2 A respectively; total power 36 W.
The indicated ideal ammeter reads 3 A and the B–D voltmeter reads 6 V.
Open-circuit voltage is not loaded voltage
A battery has emf 9.0 V and internal resistance 1.0 Ω. It drives a 5.0 Ω external resistor through wire resistance 0.50 Ω. Find current and battery terminal voltage; describe ideal meter placement.
All resistances are in one loop: Rtotal = 1.0 + 5.0 + 0.50 = 6.50 Ω. I = 9.0/6.50 = 1.38 A.
Why this step? The battery’s internal resistor and wires carry the same current as the load.
Vterminal = 9.0 − (1.38)(1.0) = 7.62 V. The 5 Ω load gets 6.92 V and the wire drops 0.692 V.
Why this step? Terminal voltage includes the whole external circuit, while the load has only its own drop.
Place an ideal ammeter in series in the loop and an ideal voltmeter across the battery terminals. The ammeter adds no resistance; the voltmeter draws no current.
Why this step? Each ideal meter preserves the property it measures through the appropriate connection.
The load-only voltmeter reads I(5 Ω) = 6.92 V. The terminal meter reads the load plus wire drop, 7.62 V; the series ammeter reads 1.38 A. Do not connect the ammeter between P and N, which would add a bypass path.
Why this step? Each voltmeter compares its own two labeled nodes; terminal and load voltage differ because the wire is resistive.
Answer and check
I = 1.38 A and Vterminal = 7.62 V. With the circuit open, current is zero and the terminal voltmeter reads 9.0 V.
Ideal removes two resistance drops, not the load
An ideal battery has negligible internal resistance. Its terminal potential difference equals its emf independently of current in the ideal model. For a real battery, the emf is the terminal potential difference measured with no current; internal voltage loss is then zero. An ideal wire has negligible resistance, so its endpoints have essentially the same potential even while finite current flows. A finite load can still oppose charge flow, carry current and convert energy thermally.
For a discharging real battery, retain a series internal resistor r and any total series wire resistance R_wire: I = ε/(R_load + r + R_wire). Battery terminal voltage is ε − Ir; load voltage is IR_load. The wire drop separates those two external voltage measurements. The ideal limit sets both r and R_wire negligible relative to the finite load.
Check the size of omitted terms. If r + R_wire is a small fraction of a finite load, the ideal model may be accurate enough. If the load is removed and the only remaining path is a short, the wire and internal resistance supply the physical current limit; deleting them too gives incompatible nonzero-source and zero-wire voltage constraints. That failed model has no sensible finite steady solution. It is not a prediction of a measurable infinite current.
Recover the ideal result as resistances shrink
A 6.0 V emf drives a 6.0 Ω load. Compare (i) an ideal battery and ideal wires and (ii) internal resistance 0.50 Ω with total wire resistance 0.50 Ω. The battery is discharging.
- Ideal: I = 6.0/6.0 = 1.0 A. Both omitted resistances are negligible, so terminal and load voltages are 6.0 V; wire drop and internal heating vanish in the model.
- Nonideal: I = 6.0/(6.0 + 0.50 + 0.50) = 0.857 A. Terminal voltage = 6.0 − 0.857(0.50) = 5.57 V; load voltage = 0.857(6.0) = 5.14 V; total wire drop = 0.429 V.
- Source power is εI = 5.14 W. Load power is I²(6.0) = 4.41 W, and the internal and wire resistors each dissipate I²(0.50) = 0.367 W. Their sum is 5.14 W.
- As r and R_wire tend to zero with the finite load retained, current tends to 1.0 A and both voltage losses tend to zero.
Answer and check
Ideal: 1.0 A, terminal/load 6.0 V. Nonideal: 0.857 A, terminal 5.57 V, load 5.14 V; finite resistance explains the smaller values.
Make the model respond
Focused Physics 2 model
Two-resistor DC circuit
Predict first. Predict how total current changes when the same resistors are rewired in parallel.
Rₑq = R₁ + R₂; I₁ = I₂ = I; Vbattery = V₁ + V₂; P = I²R
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
Series: I₁ = 0.2 A, I₂ = 0.2 A, and total current = 0.2 A.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Battery voltage | 12 | V | |
| Equivalent resistance | 60 | Ω | |
| Battery current | 0.2 | A | |
| R₁ current | 0.2 | A | |
| R₂ current | 0.2 | A | |
| R₁ voltage drop | 4 | V | |
| R₂ voltage drop | 8 | V | |
| R₁ power | 0.8 | W | |
| R₂ power | 1.6 | W | |
| Battery power | 2.4 | W | |
| Power residual Pbattery − P₁ − P₂ | -2.220 × 10^−16 | W |
In series, every resistor carries the same current, and the voltage drops sum to the battery voltage. In parallel, every branch has the full battery voltage, and the branch currents sum at the junction. In either topology, the ideal battery power equals the sum of resistor powers. The battery moves charge around a closed circuit; resistors transfer electric energy to thermal energy.
Assumptions and limits. Ideal constant-voltage battery, ideal wires, two ohmic resistors with fixed resistance and no temperature dependence. Steady DC state; conventional-current direction from the positive terminal. No internal battery resistance or transient behavior.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
Add a resistor in parallel across an existing ideal-battery load. What happens to equivalent resistance and total source current? Does the original branch’s voltage change?
Worked answer and reason
Equivalent resistance decreases and total source current increases. The original branch voltage stays at the ideal battery’s fixed voltage, so its current is unchanged if its resistance is fixed.
Additional paths increase total conductance; an ideal source fixes the common node difference.
Question 2
Why does a nonideal ammeter in series alter current, and a nonideal voltmeter across a resistor alter the network? Which ideal limits avoid those effects?
Worked answer and reason
The ammeter adds series resistance; a finite-resistance voltmeter adds a parallel current path. Zero ammeter resistance and infinite voltmeter resistance avoid those changes.
A meter is itself a circuit element; qualitative loading follows from its topology.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- All parallel branches carry the same current.
- They share voltage; branch current depends on each resistance.
- Battery emf always equals terminal voltage.
- A discharging nonideal battery loses Ir internally; emf is its open-circuit voltage.
- A voltmeter belongs in the same series path as the resistor.
- It connects across the resistor’s endpoints; an ideal series voltmeter would interrupt current.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 9 · network reduction and branch expansion
- Problem 10 · nonideal source and measurement design
- Problem 17 · resistance experiment, uncertainty, and power
- Problem 18 · nonreducible bridge and independent equations
- Problem 19 · branching RC endpoints and empirical time scale
Explain and apply the model
Try these problems before opening the solutions. Show the physical model, your calculation or sketch, and the reason your conclusion follows.
What does ideal mean in the circuit model?
A battery of emf 12.0 V supplies a 10.0 Ω load through an unbranched loop. First use an ideal battery and ideal wires. Then compare a discharging nonideal battery with r = 1.0 Ω and total series wire resistance R_wire = 1.0 Ω. Load resistance remains constant. All stated voltage measurements are by ideal voltmeters.
(a) State the resistance assumptions for both an ideal battery and ideal wires. Draw the ideal circuit and calculate current, terminal voltage, load voltage and wire voltage drop. Explain how finite current is compatible with zero ideal-wire voltage drop.
(b) Draw the nonideal series model with the internal resistor inside the battery boundary and identify the two terminal-measurement nodes and the two load-measurement nodes. Find current and both measured voltages. Write a closed-loop voltage check and verify source power equals the sum of all three resistive powers.
(c) In another trial keep the 10.0 Ω load but reduce r and total wire resistance to 0.010 Ω each. Calculate the fractional current overestimate from using the ideal model, relative to the true nonideal current. Then explain why the same neglect argument fails if the finite load is removed and the battery is connected directly across its wires.
Written working workspace
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For the teacher
- Check endpoints before reduction.
- Have students recover branch values after finding equivalent resistance.
- Ask why neglecting wire resistance fails in a nearly shorted loop.
Source objectives and scope
- 11.5.A — Describe the equivalent resistance of multiple resistors connected in a circuit.
- 11.5.B — Describe a circuit with resistive wires and a battery with internal resistance.
- 11.5.C — Describe the measurement of current and potential difference in a circuit.
AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements. Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- College Board AP Physics 2 Course and Exam Description, Unit 11 (PDF pages 68–81) — Official topic order, learning objectives, all essential-knowledge clauses and boundaries. Equation pages 68,71,73,75,77–80 and relevant continuation pages visually inspected; independent reviewed-source-equation overlay consulted.
- OpenStax University Physics Volume 2, 9.3 Resistivity and Resistance — Independent verification of real material temperature dependence and the fixed-temperature qualification for linear metallic response. Teaching and assessment prose are original.
- OpenStax University Physics Volume 2, 10.3 Kirchhoff’s Rules — Independent verification of conservation-based branch and loop analysis; no calculus is needed in these authored tasks.
- Corrected original reading: Current and Resistance — Follow charge-flow rate and conventional-current direction; keep electron drift opposite the arrows.Supporting context; original AP teaching and practice above remain independently authored.
- Corrected original reading: Introduction to Circuits — Practice node connectivity, switches and common symbols. Check that any short-circuit idealization retains a finite current-limiting resistance.Supporting context; original AP teaching and practice above remain independently authored.
- Corrected original reading: Resistors in Circuits — Connect resistor voltage and current to energy transfer per second. Brightness comparisons require a stated bulb model.Supporting context; original AP teaching and practice above remain independently authored.
- Corrected original reading: Kirchhoff’s Laws — Trace actual closed paths and keep signed voltage changes. Use independent equations rather than combining disconnected branches into one loop.Supporting context; original AP teaching and practice above remain independently authored.
- Corrected original reading: Capacitors in Circuits — Compare equal-charge series and equal-voltage parallel capacitor arrangements.Supporting context; original AP teaching and practice above remain independently authored.
- Corrected original reading: RC Circuits — Use initial and long-time states, τ and qualitative curves. Source exponential and logarithmic transient calculations extend beyond required AP Physics 2 modeling.Supporting context; original AP teaching and practice above remain independently authored.
Glossary and deeper reading · 11 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Electric current
Electric current is the net rate of charge flow through a specified cross section. It is measured in amperes, or coulombs per second, with direction defined by positive-charge motion. For example, the same steady current passes through each element of a simple unbranched series loop.
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Potential difference
Potential difference is the change in electric potential between two locations. It represents electric potential-energy change per unit charge and is measured in volts. For example, the potential difference across a resistor describes the energy transferred per coulomb passing through it.
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Circuit node
A circuit node is a connected conducting region treated as having one potential in the ideal-wire model. Branches meet at nodes, so topology rather than the drawing’s shape determines how elements are connected. For example, moving a wire on a schematic without changing its connections leaves the circuit nodes unchanged.
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Ohm’s law
Ohm’s law describes an element whose current is proportional to its potential difference under the stated conditions. Such an element has constant resistance over the modeled range, giving a straight-line current–voltage relationship through the origin. For example, a fixed-temperature ideal resistor follows this model, whereas a warming filament bulb need not.
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Electric charge
Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.
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Electrical resistance
Electrical resistance relates the potential difference across an element to the current through it. It is measured in ohms and depends on the element’s material, geometry, and conditions. For example, a longer uniform wire has greater resistance when cross-sectional area and material remain fixed.
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Internal resistance
Internal resistance models energy dissipation inside a real electrical source. When the source delivers current, its terminal voltage can be smaller than its emf because of this internal voltage drop. For example, connecting a heavy load to a battery can lower the voltage measured at its terminals.
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Electromotive force
Electromotive force is energy supplied per unit charge by a source or induction process. Despite its name it is measured in volts, rather than newtons, and can exist even when no current flows. For example, an ideal battery supplies an emf while an induced emf can be produced by changing magnetic flux.
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Ammeter
An ammeter measures current through a branch. It is connected in series with that branch and ideally has negligible resistance so it changes the current as little as possible. For example, inserting an ammeter into the wire leading to a resistor measures the current through that resistor.
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Voltmeter
A voltmeter measures potential difference between two nodes. It is connected in parallel and ideally has very high resistance so it draws negligible current. For example, placing its probes on opposite ends of a resistor measures that resistor’s voltage drop.
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Power
Power is the rate at which energy is transferred or converted. It is measured in watts, with one watt equal to one joule per second. For example, two motors can lift the same load through the same height but have different average powers if they take different times.