AP Physics C: E&M Unit 3 — Learning DC Circuit Analysis

Series and Parallel Resistors

When you build a DC circuit, you are really building paths for charge to move. Resistors are elements that oppose current and convert electrical energy into other forms (often thermal energy). In AP Physics C, you’ll constantly translate between a messy physical circuit and a clean mathematical model. The first major simplification tool is replacing groups of resistors with a single equivalent resistance.

What resistance means (and why equivalent resistance is useful)

Resistance connects the microscopic idea “charges bump into atoms” to the macroscopic relationship between voltage and current. For an ohmic resistor, the potential difference across it is proportional to the current through it:

$V = IR$

Here $V$ is the potential difference across the resistor, $I$ is the current through it, and $R$ is the resistance.

Equivalent resistance matters because many circuit questions boil down to: “Given a battery (or known voltages), what are the currents and voltages everywhere?” If you can replace part of a circuit with an equivalent resistor, you reduce the number of unknowns—often turning a multi-step algebra problem into one line of Ohm’s law.

A helpful mental model is a “water pipe” analogy: voltage is like pressure difference, current is like flow rate, and resistance is like how narrow/rough the pipe is. Equivalent resistance answers: “How hard is it, overall, to push charge through this network?”

Series resistors

Series means components are connected end-to-end so there is only one path for current through them. Because charge cannot “choose” a different route, the current is the same through each series resistor.

What changes, instead, is the voltage: each resistor takes a share of the total potential drop.

How the equivalent resistance in series works

Suppose resistors $R_1$ and $R_2$ are in series and the same current $I$ flows through both. The total potential difference across the pair is

$V_{total} = V_1 + V_2$

Using Ohm’s law on each resistor gives

$V_{total} = IR_1 + IR_2 = I(R_1 + R_2)$

So the equivalent resistance is

$R_{eq} = R_1 + R_2$

This generalizes to any number of resistors in series: you simply add them.

Example (series): finding current and voltage drops

A $12\ \text{V}$ ideal battery is connected to $R_1 = 2\ \Omega$ and $R_2 = 4\ \Omega$ in series.

1) Equivalent resistance:

$R_{eq} = 2 + 4 = 6\ \Omega$

2) Circuit current:

$I = \frac{V}{R_{eq}} = \frac{12}{6} = 2\ \text{A}$

3) Voltage drops:

$V_1 = IR_1 = (2)(2) = 4\ \text{V}$

$V_2 = IR_2 = (2)(4) = 8\ \text{V}$

Check: $4\ \text{V} + 8\ \text{V} = 12\ \text{V}$, consistent.

Common pitfall: Students sometimes add voltages across series resistors incorrectly by assuming they are equal. Voltages are only equal if resistances are equal (since the current is the same).

Parallel resistors

Parallel means components share the same two nodes, creating multiple paths for charge. Because both ends of each resistor connect to the same two nodes, the voltage across each parallel branch is the same.

What changes is the current: current splits among the branches depending on their resistances. A lower resistance branch draws more current.

How the equivalent resistance in parallel works

For two resistors $R_1$ and $R_2$ in parallel, each has the same voltage $V$. The currents are

$I_1 = \frac{V}{R_1}$

$I_2 = \frac{V}{R_2}$

Total current is the sum:

$I_{total} = I_1 + I_2 = V\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$

Define $R_{eq}$ by $I_{total} = \frac{V}{R_{eq}}$. Then

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

For many resistors in parallel, you add reciprocals.

Two key qualitative consequences you should internalize:

• $R_{eq}$ for parallel is always less than the smallest branch resistance (because adding another path makes it easier for current to flow).
• If you add an additional parallel branch, the total current drawn from the source increases (for a fixed source voltage).

Example (parallel): equivalent resistance and branch currents

A $12\ \text{V}$ ideal battery is connected to two resistors in parallel: $R_1 = 6\ \Omega$ and $R_2 = 3\ \Omega$.

1) Equivalent resistance:

$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$

$R_{eq} = 2\ \Omega$

2) Total current:

$I_{total} = \frac{V}{R_{eq}} = \frac{12}{2} = 6\ \text{A}$

3) Branch currents (same voltage across each branch):

$I_1 = \frac{12}{6} = 2\ \text{A}$

$I_2 = \frac{12}{3} = 4\ \text{A}$

Check: $2\ \text{A} + 4\ \text{A} = 6\ \text{A}$.

Common pitfall: Treating parallel resistors as if they share the same current. In parallel, they share the same voltage; the current splits.

Mixed series-parallel networks (strategy, not memorization)

Real AP problems often have combinations. The key skill is recognizing “chunks” that are purely series or purely parallel based on node connections.

A practical process:

1) Identify nodes (junction points) and see which components share the same two nodes (parallel) or share a single path (series).
2) Replace the obvious series/parallel group with its equivalent resistance.
3) Repeat until you have a simpler circuit.
4) Once you find a total current or voltage, work backward (undo the replacements) to find branch currents and individual drops.

Notation reference (common symbols)

Exam Focus

• Typical question patterns:
  • Compute $R_{eq}$ for a network, then use $V = IR$ to find the battery current.
  • Find the current/voltage on each resistor after simplifying a circuit and then “expanding” back out.
  • Qualitative reasoning: what happens to total current or brightness of bulbs when a branch is added/removed.
• Common mistakes:
  • Adding resistances in parallel as $R_1 + R_2$ instead of using reciprocals.
  • Forgetting that series means same current, parallel means same voltage.
  • Losing track of which elements truly share nodes (misidentifying series/parallel because of how the circuit is drawn rather than how it is connected).

Kirchhoff’s Rules

Series/parallel reductions are powerful, but not every circuit can be simplified that way—especially when there are multiple batteries, “bridge” connections, or complex junctions. Kirchhoff’s Rules give you a universal method to analyze any lumped DC circuit using conservation laws.

The big idea: conservation laws in circuit form

Kirchhoff’s rules are not new physics—they are conservation principles translated into circuit language:

• Kirchhoff’s Junction Rule (KJR) comes from conservation of charge: charge doesn’t pile up at an ideal junction in steady-state DC.
• Kirchhoff’s Loop Rule (KLR) comes from conservation of energy: going around a closed loop, the net change in electric potential is zero.

These rules are essential because AP Physics C expects you to set up systems of equations, solve for unknown currents, then compute voltages and power.

Kirchhoff’s Junction Rule (current law)

At any junction (node), the sum of currents entering equals the sum leaving.

$\sum I_{in} = \sum I_{out}$

A sign-convention way to write it is “algebraic sum of currents at a node equals zero,” but what matters is consistency: pick entering as positive (or leaving as positive) and stick with it.

Why it works: In steady DC, charge density at a node is not changing. If more current entered than left, charge would accumulate and the electric field would change until currents adjusted.

Common misconception: Thinking current is “used up” by a resistor. Resistors dissipate energy, not charge; current is the same on both sides of a resistor in a single branch.

Kirchhoff’s Loop Rule (voltage law)

For any closed loop you traverse, the algebraic sum of potential changes is zero.

$\sum \Delta V = 0$

Why it works: Electric potential is tied to energy per charge. After a charge goes around a closed loop and returns to the start, it must have the same energy per charge it began with (otherwise you’d get perpetual motion).

Sign conventions for loop equations (what usually trips people)

You choose a direction to walk around the loop (clockwise or counterclockwise). As you cross elements, you add potential rises and drops based on your traversal direction and assumed current directions.

• Across an ideal resistor: if you move in the direction of the assumed current, potential drops by $IR$.

$\Delta V_{resistor} = -IR$

If you move opposite the assumed current direction, potential rises:

$\Delta V_{resistor} = +IR$

• Across an ideal battery (emf source): going from the negative terminal to the positive terminal is a rise of $\mathcal{E}$.

$\Delta V_{battery} = +\mathcal{E}$

Going from positive to negative is a drop:

$\Delta V_{battery} = -\mathcal{E}$

Here $\mathcal{E}$ is the emf (electromotive force), measured in volts.

Very important: Your assumed current directions can be wrong. That’s okay—if the algebra gives a negative value for a current, it means the actual direction is opposite your assumption.

A consistent method for solving Kirchhoff problems

1) Label currents in each branch. Don’t overthink direction; pick something.
2) Apply the junction rule at independent nodes (each gives one equation).
3) Choose independent loops and apply the loop rule to each (each gives one equation).
4) Use Ohm’s law relationships $V = IR$ in loop equations automatically through the resistor voltage terms.
5) Solve the resulting linear system.

A practical tip: the number of independent equations you need equals the number of unknown currents (or unknowns you define). Picking good loop directions and current labels can make the algebra far simpler.

Worked problem 1: one junction, two loops

Circuit description (text-based): A battery of emf $\mathcal{E} = 12\ \text{V}$ feeds a resistor $R_1 = 2\ \Omega$, then the circuit splits into two parallel branches with resistors $R_2 = 4\ \Omega$ (top branch) and $R_3 = 6\ \Omega$ (bottom branch), then recombines and returns to the battery.

Define currents:

• $I_1$ through $R_1$ (before the split)
• $I_2$ through $R_2$
• $I_3$ through $R_3$

Because the split is a junction:

$I_1 = I_2 + I_3$

Now write loop equations. Choose both loops to go from the battery through $R_1$, then through one branch resistor, then back.

Loop through $R_2$ branch (take battery as a rise):

$\mathcal{E} - I_1 R_1 - I_2 R_2 = 0$

Loop through $R_3$ branch:

$\mathcal{E} - I_1 R_1 - I_3 R_3 = 0$

Substitute numbers:

$12 - 2I_1 - 4I_2 = 0$

$12 - 2I_1 - 6I_3 = 0$

Along with

$I_1 - I_2 - I_3 = 0$

Solve efficiently by expressing $I_2$ and $I_3$ in terms of $I_1$:

From first loop:

$4I_2 = 12 - 2I_1$

$I_2 = 3 - 0.5I_1$

From second loop:

$6I_3 = 12 - 2I_1$

$I_3 = 2 - \frac{1}{3}I_1$

Plug into junction equation:

$I_1 = (3 - 0.5I_1) + \left(2 - \frac{1}{3}I_1\right)$

$I_1 = 5 - \left(\frac{1}{2} + \frac{1}{3}\right)I_1$

$I_1 = 5 - \frac{5}{6}I_1$

$\frac{11}{6}I_1 = 5$

$I_1 = \frac{30}{11}\ \text{A} \approx 2.73\ \text{A}$

Then

$I_2 = 3 - 0.5\left(\frac{30}{11}\right) = \frac{18}{11}\ \text{A} \approx 1.64\ \text{A}$

$I_3 = 2 - \frac{1}{3}\left(\frac{30}{11}\right) = \frac{12}{11}\ \text{A} \approx 1.09\ \text{A}$

Check: $\frac{18}{11} + \frac{12}{11} = \frac{30}{11}$, consistent.

What you should learn from this: Kirchhoff’s method naturally handles the parallel split even if you don’t first compute equivalent resistances.

Worked problem 2: a “negative current” is not failure

Suppose you have two opposing batteries in one loop with resistors. If you guess the current direction incorrectly, your loop equation will yield $I < 0$. That simply means the net emf drives current the other way.

For a single loop with batteries $\mathcal{E}_1$ and $\mathcal{E}_2$ opposing and total resistance $R_{tot}$, if you traverse so that one battery is a rise and the other a drop, you might get

$\mathcal{E}_1 - \mathcal{E}_2 - IR_{tot} = 0$

So

$I = \frac{\mathcal{E}_1 - \mathcal{E}_2}{R_{tot}}$

If $\mathcal{E}_2 > \mathcal{E}_1$, then $I$ is negative relative to your assumed direction.

What goes wrong most often with Kirchhoff’s Rules

• Students mix sign conventions mid-loop (for example, writing a resistor as $+IR$ even though they are traversing with the current).
• Students forget that junction equations use actual branch currents, not “currents after a resistor.” In a single branch, current is the same everywhere.
• Students try to force series/parallel logic onto a circuit that isn’t reducible, leading to incorrect simplifications.

Exam Focus

• Typical question patterns:
  • Set up and solve 2–3 simultaneous equations for branch currents using KJR and KLR.
  • Determine potential differences between two labeled nodes after solving for currents.
  • Circuits with multiple sources: find current direction (including interpreting negative results).
• Common mistakes:
  • Sign errors in loop equations (especially across resistors and batteries).
  • Writing redundant loop equations (not independent) and then being unable to solve uniquely.
  • Assuming current splits equally at a junction without considering resistances.

Power in Circuits

So far, you’ve mainly used circuits to find currents and voltages. But circuits are often built to deliver energy—lighting a bulb, running a motor, heating a resistor, charging a device. Power is the rate at which electrical energy is transferred or converted.

What electrical power means

Power is energy per unit time:

$P = \frac{dE}{dt}$

In circuits, when a charge $q$ moves through a potential difference $\Delta V$, the change in electric potential energy is

$\Delta U = q\Delta V$

If charge flows at a rate $I = \frac{dq}{dt}$, then the rate of energy transfer is

$P = I\Delta V$

In practice, across a circuit element with voltage $V$ and current $I$ (with a consistent sign convention), you use

$P = IV$

Power formulas for resistors (the big three)

For a resistor, $V = IR$, so you can rewrite $P = IV$ in two extremely useful ways:

$P = I^2 R$

$P = \frac{V^2}{R}$

These are not “new” formulas; they are the same relationship expressed with different known quantities.

• Use $P = I^2 R$ when you know current through the resistor.
• Use $P = \frac{V^2}{R}$ when you know the voltage across it.

Why this matters: Many AP questions ask about brightness (bulbs), heating, or energy usage. Brightness is typically modeled as proportional to power dissipated in the bulb’s resistance.

Power supplied vs power dissipated (energy accounting)

In an ideal battery, chemical energy is converted into electrical energy. The battery “raises” potential by emf $\mathcal{E}$. If current $I$ leaves the battery’s positive terminal, the power delivered by the battery to the circuit is often treated as

$P_{source} = I\mathcal{E}$

Resistors then dissipate power (convert electrical energy to thermal energy), given by the formulas above.

A powerful consistency check in DC steady state: total power supplied equals total power dissipated (for ideal wires and ideal meters). If your computed powers don’t balance, you likely have a sign error or a wrong current.

Worked problem: power in series and parallel (and what changes)

Consider a $12\ \text{V}$ ideal battery and two resistors $R_1 = 6\ \Omega$ and $R_2 = 6\ \Omega$.

Case A: series

Equivalent resistance:

$R_{eq} = 6 + 6 = 12\ \Omega$

Current:

$I = \frac{12}{12} = 1\ \text{A}$

Power in each resistor:

$P_1 = I^2 R_1 = (1)^2(6) = 6\ \text{W}$

$P_2 = 6\ \text{W}$

Total dissipated:

$P_{tot} = 12\ \text{W}$

Battery power:

$P_{battery} = IV = (1)(12) = 12\ \text{W}$

So energy accounting works.

Case B: parallel

Equivalent resistance:

$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

$R_{eq} = 3\ \Omega$

Total current:

$I_{tot} = \frac{12}{3} = 4\ \text{A}$

Each branch has the full $12\ \text{V}$, so branch current is

$I_{branch} = \frac{12}{6} = 2\ \text{A}$

Power in each resistor:

$P_{each} = \frac{V^2}{R} = \frac{12^2}{6} = 24\ \text{W}$

Total dissipated: $48\ \text{W}$. Battery supplies

$P_{battery} = IV = (4)(12) = 48\ \text{W}$

Conceptual takeaway: Putting the same resistors in parallel dramatically increases the current drawn from a fixed-voltage source, so the source delivers much more power. This is why household wiring uses parallel branches (each appliance gets full voltage), and why adding too many high-power devices can trip breakers.

Energy over time (kilowatt-hours logic)

Once you know power, energy over a time interval $t$ (constant power) is

$E = Pt$

AP problems may ask how much energy is converted to heat in a resistor over some time, or they may combine circuit analysis with thermodynamics ideas (like temperature rise) in more advanced contexts.

Common misconceptions with power

• Confusing voltage with power: high voltage does not automatically mean high power. Power depends on both voltage and current.
• Using the wrong power form: for a resistor, $P = \frac{V^2}{R}$ uses the voltage across that resistor, not necessarily the battery voltage (unless it’s directly across the battery).
• Assuming “current is the same everywhere” even in parallel networks, then computing power incorrectly.

Exam Focus

• Typical question patterns:
  • Compute power dissipated by each resistor after finding currents/voltages (often tied to bulb brightness).
  • Compare power before and after a circuit change (adding a parallel branch, changing one resistor value).
  • Use energy $E = Pt$ to connect circuits to real-world energy consumption.
• Common mistakes:
  • Plugging the battery voltage into $P = \frac{V^2}{R}$ for a resistor that does not have the full battery voltage across it.
  • Forgetting that in series, current is the same but voltage divides; in parallel, voltage is the same but current divides—this flips which power formula is most convenient.
  • Reporting negative power without interpreting sign conventions: in many AP contexts, you report magnitudes for dissipated power and interpret negative sign (if used) as power delivered vs absorbed.