AP PHYSICS C: ELECTRICITY AND MAGNETISM • ELECTRIC CIRCUITS

Simple Circuits

How voltage, current, and resistance combine to govern energy flow in series and parallel networks.

Historical Context & Motivation

The study of electric circuits grew from a centuries-long effort to understand and harness the flow of charge. Before the late eighteenth century, static electricity was a laboratory curiosity—dramatic but impractical. The breakthrough came when experimenters discovered how to maintain a steady current, transforming electricity from a parlor trick into the backbone of modern technology. Each milestone below marked a conceptual leap that brought the physics of circuits closer to the rigorous, quantitative framework you will use on the AP exam.

1800
Volta's Pile
Alessandro Volta constructed the first true battery by stacking zinc and copper discs separated by brine-soaked cardboard, providing the first source of continuous EMF and enabling systematic study of current.
1827
Ohm's Law Published
Georg Ohm established the linear relationship V = IR between voltage, current, and resistance, initially met with skepticism but ultimately forming the cornerstone of circuit analysis.
1845
Kirchhoff's Circuit Laws
Gustav Kirchhoff formulated the junction rule (conservation of charge) and the loop rule (conservation of energy), enabling the systematic solution of complex multi-loop networks.
1841–1842
Joule Heating
James Prescott Joule quantified the heat dissipated in a resistor as P = I²R, linking electrical energy to thermal energy and underscoring the role of power in circuit design.

These discoveries frame a central question that the rest of this lesson addresses: given a network of resistors and an ideal battery, how do we predict the current through—and voltage across—every element in the circuit? Answering that question with precision is the goal of simple circuit analysis.

Core Principles & Definitions

Before solving any circuit, you need a firm grip on four foundational ideas. These concepts connect the macroscopic quantities you measure in a lab—voltage readings on a multimeter, current readings on an ammeter—to the underlying physics of charge flow and energy transfer.

1

Electromotive Force (EMF)

The EMF (ε) of a battery is the energy per unit charge supplied by the source. It equals the open-circuit voltage and drives current through the external circuit.
2

Ohm's Law

For an ohmic resistor, V = IR. The voltage drop across the element is directly proportional to the current through it, with resistance R as the constant of proportionality.
3

Kirchhoff's Junction Rule (KJR)

At any node, the sum of currents entering equals the sum leaving: ΣI_in = ΣI_out. This is conservation of charge in steady state.
4

Kirchhoff's Loop Rule (KLR)

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

Power Dissipation

The rate of energy dissipation in a resistor is P = IV = I²R = V²/R. Power is always positive for a resistor, reflecting energy converted to heat.
KEY TAKEAWAY
KEY TAKEAWAY

Visual Explanation — Series vs. Parallel

Left: In a series circuit the same current I passes through every element, and voltages add. Right: In a parallel circuit every branch shares the same voltage, and currents add. The equivalent resistance formulas are shown below each.

The diagram above captures the two fundamental topologies you will encounter in simple circuits. In the series configuration on the left, charge has only one path from the high-potential terminal of the battery back to the low-potential terminal, so the current through R₁ and R₂ is identical. Applying Kirchhoff's loop rule around the single loop yields ε = IR₁ + IR₂ = I(R₁ + R₂), which immediately gives the series equivalent resistance Req = R₁ + R₂. In the parallel configuration on the right, both resistors connect directly across the battery, so each experiences the full EMF. The junction rule at the top node requires the total current to split: I = I₁ + I₂ = ε/R₁ + ε/R₂, leading to 1/Req = 1/R₁ + 1/R₂. Notice that adding resistors in series always increases the equivalent resistance, while adding resistors in parallel always decreases it.

Mathematical Framework

The quantitative backbone of simple circuit analysis rests on Ohm's law and Kirchhoff's two rules. Combining these allows you to reduce any resistor network to a single equivalent resistance, find the total current, and then work backwards to determine individual branch currents and voltage drops.

OHM'S LAW
V = IR
V is the potential difference across the resistor (V), I is the current through it (A), and R is the resistance (Ω). This holds for ohmic materials where R is constant.
SERIES EQUIVALENT RESISTANCE
R_eq = R₁ + R₂ + R₃ + ⋯ + Rₙ
Derived by applying the loop rule to a single-loop circuit: ε = I(R₁ + R₂ + ⋯), so the sum acts as a single resistor carrying the same current.
PARALLEL EQUIVALENT RESISTANCE
1/R_eq = 1/R₁ + 1/R₂ + ⋯ + 1/Rₙ
Derived by applying the junction rule (currents add) and noting each branch shares voltage V: I = V/R₁ + V/R₂ + ⋯ = V(1/R₁ + 1/R₂ + ⋯). For two resistors, the shortcut is R_eq = R₁R₂/(R₁ + R₂).
POWER DISSIPATED
P = IV = I²R = V²/R
Three equivalent forms obtained by substituting Ohm's law. Choose the form that uses the two quantities you already know to avoid propagating rounding errors.
Internal Resistance

Reducing Combination Circuits

Most AP problems present a combination circuit—a mixture of series and parallel connections. The strategy is to identify the innermost purely series or purely parallel group, replace it with its equivalent resistance, and repeat until the entire network collapses to a single resistor. After finding the total current from ε and Req, you "un-collapse" step by step to recover individual currents and voltages.

A combination circuit with R₁ in series with a parallel pair (R₂ ∥ R₃) is reduced in two steps. The calculation box shows the full numerical solution for ε = 12 V.

The reduction strategy illustrated above works for any network that is reducible—one where every pair of elements can be classified as purely series or purely parallel. Certain topologies (like a Wheatstone bridge with an unbalanced galvanometer branch) are not reducible and require full Kirchhoff loop/junction analysis. On the AP exam, however, most simple-circuit FRQs involve reducible networks, so mastering this collapse-and-expand technique is essential.

Worked Example

Consider a circuit with a 24 V battery (internal resistance r = 1 Ω) connected to three external resistors: R₁ = 5 Ω in series with the parallel combination of R₂ = 12 Ω and R₃ = 4 Ω. Find the total current, the terminal voltage, the power dissipated in R₃, and the voltage across R₂.

1
Step 1 — Find the parallel equivalent of R₂ and R₃1/Rp = 1/R₂ + 1/R₃ = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3. Therefore Rp = 3 Ω.
Rp = 3 Ω
2
Step 2 — Find the total equivalent resistanceThe external resistance is Rext = R₁ + Rp = 5 + 3 = 8 Ω. Including internal resistance: Rtotal = Rext + r = 8 + 1 = 9 Ω.
Rtotal = 9 Ω
3
Step 3 — Total current from ε and R_totalI = ε / Rtotal = 24 V / 9 Ω = 8/3 A ≈ 2.67 A.
I ≈ 2.67 A
4
Step 4 — Terminal voltageVterm = ε − Ir = 24 − (8/3)(1) = 24 − 8/3 = 64/3 V ≈ 21.3 V. This is the voltage available to the external circuit.
Vterm ≈ 21.3 V
5
Step 5 — Voltage across the parallel pair and power in R₃Voltage across R₁: V₁ = IR₁ = (8/3)(5) = 40/3 V ≈ 13.3 V. Voltage across the parallel pair: Vp = IRp = (8/3)(3) = 8 V. Both R₂ and R₃ share this voltage. Power in R₃: P₃ = Vp²/R₃ = 64/4 = 16 W. Voltage across R₂ is also 8 V (same node connections).
Vp = 8 V; P₃ = 16 W

Series vs. Parallel — Comparison

Key distinctions between series and parallel resistor configurations.
PropertySeriesParallel
CurrentSame through all elements: I₁ = I₂ = ISplits: I = I₁ + I₂ + ⋯
VoltageDivides: ε = V₁ + V₂ + ⋯Same across all branches: V₁ = V₂ = ε
R_eqR_eq = ΣRᵢ (always larger than any single R)1/R_eq = Σ(1/Rᵢ) (always smaller than the smallest R)
Effect of adding a resistorTotal R increases → total I decreasesTotal R decreases → total I increases
If one resistor burns outEntire circuit breaks (open circuit)Other branches still carry current
KEY TAKEAWAY
DESIGN INSIGHT

Connection to Advanced Theory

Simple DC resistor circuits are the gateway to a much richer landscape of circuit analysis. In the AP Physics C curriculum and in later coursework, you will encounter capacitors and inductors that store energy rather than dissipate it, leading to time-dependent (transient) behavior governed by differential equations. The table below maps the simple-circuit concepts you have learned to their more general counterparts.

How simple DC circuit concepts generalize to AC and transient analysis.
Simple Circuit ConceptAdvanced Extension
Ohm's law: V = IRGeneralized impedance: V = IZ (Z complex, includes R, jωL, 1/jωC)
Kirchhoff's loop & junction rulesMesh analysis & nodal analysis — systematic matrix methods for arbitrary networks
Series/parallel reductionY-Δ (star-delta) transforms for non-reducible topologies
DC steady-state (constant I)RC and RL transients: I(t) = (ε/R)e^(−t/τ) with time constant τ
Power: P = I²RAC power: P_avg = I_rms² R, power factor cos φ

On the AP Physics C exam, RC circuits are explicitly tested and form the next natural topic after mastering simple circuits. The key insight is that the resistor equations you have learned still apply at each instant—Kirchhoff's loop rule simply yields a first-order ODE when a capacitor is present, because the capacitor voltage depends on the integral of the current. A solid command of Ohm's law and Kirchhoff's rules is therefore prerequisite infrastructure for every circuit topic that follows.

Practice Problems

1
Two identical resistors are first connected in series across a battery of EMF ε, then reconnected in parallel across the same battery. How does the total power delivered by the battery change when switching from series to parallel? A) Power is halved B) Power stays the same C) Power is doubled D) Power is quadrupled
2
A 9.0 V battery is connected to a 6.0 Ω resistor in series with a 3.0 Ω resistor. What is the voltage across the 3.0 Ω resistor? A) 1.5 V B) 3.0 V C) 6.0 V D) 9.0 V
3
Three resistors R₁ = 10 Ω, R₂ = 15 Ω, and R₃ = 30 Ω are connected in parallel across a 30 V source. What is the total current drawn from the source? A) 1.0 A B) 3.0 A C) 6.0 A D) 9.0 A
PROBLEM 4APPLIED
A battery of EMF ε = 20 V and internal resistance r = 2.0 Ω is connected to an external circuit consisting of R₁ = 6.0 Ω in series with the parallel combination of R₂ = 8.0 Ω and R₃ = 24 Ω. (a) Determine the equivalent external resistance. (b) Calculate the current drawn from the battery. (c) Find the terminal voltage of the battery. (d) Calculate the power dissipated in R₂. (e) If R₃ is replaced by a wire (short-circuited), determine the new current drawn from the battery and explain qualitatively what happens to the power dissipated in R₁ compared to the original circuit.
PROBLEM 5CRITICAL THINKING
A student has a black-box circuit containing a battery and an unknown resistor network. Using only a voltmeter and an ammeter, the student measures a terminal voltage of 11.4 V when the current is 0.60 A, and 10.5 V when the current is 1.50 A. (a) Determine the EMF and internal resistance of the battery. (b) Predict the short-circuit current (maximum current the battery can supply). (c) Determine the external resistance in each trial. (d) Explain why the terminal voltage decreases as the current increases.
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