AP Physics 2 Unit 3 Electric Circuits — Conceptual Foundations, Problem-Solving, and RC Transients

Electric Current: What Flows (and What Doesn’t)

When you hear “current,” it’s tempting to picture a substance being used up as it moves through a circuit. That mental model causes many of the most common mistakes in circuit problems. Electric current is not “electricity being consumed”—it is a rate of charge flow.

Defining current in a useful way

Electric current I is defined as the net amount of charge \Delta Q that passes a cross-section of a conductor per time interval \Delta t:

I = \frac{\Delta Q}{\Delta t}

• I is measured in amperes (A), where 1\ \text{A} = 1\ \text{C/s}.
• The definition is about how much charge crosses a location, not about how much charge is “inside” a component.

In most AP Physics 2 circuit situations, you will treat current as steady (constant in time) in DC circuits after they reach equilibrium—except during capacitor charging/discharging, where current changes with time.

Conventional current vs electron flow

A historical convention defines current direction as the direction positive charge would move. This is called conventional current. In a metal wire, the actual mobile charges are electrons, so electron drift is opposite the conventional current direction.

Why this matters: circuit rules (Ohm’s law, Kirchhoff’s rules) are written using conventional current direction. You can still think about electrons if you want, but keep the sign conventions consistent.

Where current comes from: the “push” and the path

Current requires two ingredients:

1. A potential difference (energy per charge) that can drive charge motion.
2. A closed conducting path so charges can continuously circulate.

A battery can maintain a potential difference, but if the circuit is open (a break in the path), charges may redistribute slightly, but there is no sustained current.

A microscopic picture (conceptual, not math-heavy)

In a metal, there are many free electrons moving randomly. When an electric field is established in the wire, those electrons acquire a small net drift velocity superimposed on their random motion. The drift speed is typically very small, yet the circuit responds quickly when a switch closes because the electric field is established throughout the circuit at a substantial fraction of the speed of light.

This helps resolve a common confusion: “If electrons drift slowly, why does the bulb light immediately?” The energy transfer is mediated by the electric field and the electromagnetic interactions in the circuit, not by one electron traveling from the battery to the bulb in real time.

Worked example: interpreting current as charge flow

A wire carries a steady current of 2.0\ \text{A}. How much charge passes a point in 30\ \text{s}?

Use the definition:

I = \frac{\Delta Q}{\Delta t}

Solve for \Delta Q:

\Delta Q = I\Delta t

Substitute:

\Delta Q = (2.0\ \text{C/s})(30\ \text{s}) = 60\ \text{C}

Interpretation: “60 coulombs of charge crosses that point,” not “60 coulombs are used up.”

Exam Focus

• Typical question patterns:
  • Convert between current and charge transfer over time using I = \Delta Q/\Delta t.
  • Concept questions about why bulbs light quickly, or what happens to current in different parts of a circuit.
  • Determine current directions in branches (especially before applying Kirchhoff’s junction rule).
• Common mistakes:
  • Saying current is “consumed” by circuit elements; instead, energy is transferred while charge continues to circulate.
  • Mixing electron flow direction with conventional current direction mid-solution.
  • Treating current as automatically the same everywhere even in circuits with branches (it’s the same only along a single series path).

Electric Potential Difference, EMF, and Energy Transfer in Circuits

Circuits are fundamentally about energy per charge. Charges move because electric potential differences provide a way to convert stored or supplied energy into other forms (thermal, light, mechanical, chemical).

Electric potential difference (voltage)

Electric potential difference \Delta V between two points is defined as the change in electric potential energy \Delta U per unit charge q:

\Delta V = \frac{\Delta U}{q}

A volt (V) is a joule per coulomb: 1\ \text{V} = 1\ \text{J/C}.

A very productive way to think is:

• Voltage tells you how much energy each coulomb of charge gains or loses moving between two points.
• Components like resistors cause a drop in electric potential associated with converting electrical energy into thermal energy.

EMF: why batteries are not “just a voltage drop”

A battery (or any source) does work on charges to move them from lower to higher electric potential. The parameter that describes this is electromotive force (emf) \mathcal{E}.

Despite the name, emf is not a force; it is an energy-per-charge quantity measured in volts. In an ideal battery with no internal resistance, the potential rise across the battery equals \mathcal{E}.

In a real battery, chemical processes push charges “uphill” in potential, but internal resistance can cause some energy to be dissipated inside the battery itself, reducing the terminal voltage under load.

Power and energy in circuits

Circuits frequently ask: “How fast is electrical energy being transferred?” That rate is electric power P.

If a charge q loses potential energy \Delta U = q\Delta V moving through a potential difference \Delta V, then per unit time:

P = I\Delta V

For a resistor, using Ohm’s law (developed later), you also get equivalent forms:

P = I^2R

P = \frac{(\Delta V)^2}{R}

These are all the same relationship written in different variables. Which one is easiest depends on what you know.

Worked example: connecting voltage to energy per charge

A device has a potential difference of 12\ \text{V} across it. How much energy is transferred per coulomb of charge passing through?

Use \Delta V = \Delta U/q:

\Delta U = q\Delta V

For q = 1\ \text{C}:

\Delta U = (1\ \text{C})(12\ \text{J/C}) = 12\ \text{J}

Interpretation: each coulomb transfers 12\ \text{J} of energy.

Worked example: power from current and voltage

A resistor has 6.0\ \text{V} across it and carries 0.50\ \text{A}. Find the power dissipated.

P = I\Delta V

P = (0.50)(6.0) = 3.0\ \text{W}

Exam Focus

• Typical question patterns:
  • Use \Delta V = \Delta U/q to interpret voltage as energy per charge.
  • Compute power with P = IV or the resistor forms P = I^2R and P = (\Delta V)^2/R.
  • Explain qualitatively where energy is transferred in a circuit (battery supplies, resistors dissipate).
• Common mistakes:
  • Treating emf \mathcal{E} and terminal voltage as automatically identical even when internal resistance is present.
  • Confusing “voltage at a point” with “voltage across a component” (voltage is a difference between two points).
  • Using the wrong power formula for given variables (for example, using P = I^2R when only voltage is known and current is not).

Resistance, Resistivity, and Ohmic vs Non-Ohmic Behavior

To understand circuits, you need a model for how materials oppose current. That opposition is resistance, but resistance is not just a property of the material—it also depends on geometry.

What resistance means physically

Resistance R measures how much a component resists the flow of current for a given potential difference. The defining relationship for a resistor is Ohm’s law (for ohmic materials under constant conditions):

\Delta V = IR

• \Delta V is the potential difference across the component.
• I is the current through it.
• R is measured in ohms (Ω), where 1\ \Omega = 1\ \text{V/A}.

It’s crucial to hear the subtlety: Ohm’s law is not a universal law of nature that every object obeys. It is an empirical relationship that holds for ohmic conductors over a certain operating range (often metals at constant temperature).

Ohmic vs non-ohmic: reading I-vs-V behavior

An ohmic device has a linear relationship between current and voltage (straight line through the origin on an I vs V graph). Its resistance is approximately constant.

A non-ohmic device has a nonlinear I-vs-V curve. Common non-ohmic examples:

• Diodes (current flows primarily one direction after a threshold)
• Incandescent bulbs (filament heats up; resistance increases with temperature)

AP questions may present a graph and ask you to interpret whether the device is ohmic and to compute resistance from the slope.

Resistance from material and geometry: resistivity

A material property called resistivity \rho links the material to resistance. For a uniform cylindrical wire of length L and cross-sectional area A:

R = \rho\frac{L}{A}

• Longer wires have more resistance (more collisions/energy dissipation along the path).
• Thicker wires have less resistance (more “lanes” for charge to flow).

This relationship supports real design decisions: power lines are thick to reduce resistance and thus reduce power loss.

Temperature effects (conceptual)

For many metals, increasing temperature increases resistivity (and thus resistance) because lattice vibrations increase, producing more scattering of charge carriers. For some semiconductors, behavior can differ. On the AP level, you’re typically expected to reason qualitatively: “hotter filament → higher resistance → different current.”

Worked example: using resistivity and geometry

A wire has resistivity \rho = 1.7\times 10^{-8}\ \Omega\cdot\text{m}, length L = 2.0\ \text{m}, and area A = 1.0\times 10^{-6}\ \text{m}^2. Find its resistance.

Use:

R = \rho\frac{L}{A}

Substitute:

R = (1.7\times 10^{-8})\frac{2.0}{1.0\times 10^{-6}} = 3.4\times 10^{-2}\ \Omega

Interpretation: a thick copper wire has very small resistance, which is why connecting wires are often treated as ideal (zero resistance) in many circuit models.

Exam Focus

• Typical question patterns:
  • Apply \Delta V = IR to relate changes in voltage, current, and resistance.
  • Use R = \rho L/A to compare wires with different lengths and diameters.
  • Interpret I-vs-V graphs (ohmic vs non-ohmic) and extract resistance.
• Common mistakes:
  • Assuming every device obeys Ohm’s law (it’s not generally true for bulbs/diodes).
  • Confusing resistivity \rho (material property) with resistance R (depends on geometry too).
  • Mixing up how area affects resistance: larger A means smaller R.

Ideal Circuit Elements, Potential Changes, and Measuring Current/Voltage

Before you combine resistors and sources, it helps to be precise about what different elements do to potential and how measurements work.

Ideal wires and nodes: why many points share the same potential

In circuit diagrams, connecting wires are usually treated as having negligible resistance. That means there is essentially no potential drop along a wire segment. Points connected by ideal wire are at the same potential and form a node.

This matters because “voltage at a point” is shorthand for the potential relative to some chosen reference node (often called ground). In most AP problems, you can avoid absolute potentials and focus on potential differences across components.

Batteries, sources, and sign conventions

An ideal battery raises the electric potential of charges by \mathcal{E} when moving from its negative terminal to its positive terminal. In loop problems, you decide a direction to traverse the loop; then:

• Going from negative to positive across an ideal source is a potential rise of +\mathcal{E}.
• Going from positive to negative is a potential change of -\mathcal{E}.

For resistors, if you traverse in the direction of current, potential decreases by IR (a drop). If you traverse opposite the current direction, potential increases by IR.

The most important idea: Kirchhoff’s loop rule is bookkeeping for energy per charge. A charge that goes around a complete loop returns to the same energy state, so total rises and drops sum to zero.

Ammeters and voltmeters (and why they are built the way they are)

To measure current through a component, you place an ammeter in series with it. An ideal ammeter has negligible resistance so it doesn’t change the circuit’s current.

To measure potential difference across a component, you place a voltmeter in parallel with it. An ideal voltmeter has extremely large resistance so it draws negligible current.

These design choices are not arbitrary; they are required so the measuring device doesn’t significantly disturb what it measures.

Common “short circuit” reasoning

A short circuit is a low-resistance path that bypasses a component. If a wire directly connects two nodes, the potential difference between those nodes is essentially zero, so any component connected across those same two nodes has approximately zero voltage across it.

A typical AP conceptual question: “What happens to the bulb’s brightness if you add a wire in parallel with it?” The bulb is bypassed, so its voltage drops toward zero; current prefers the low-resistance wire, and the bulb goes out (in the ideal model).

Worked example: diagnosing a measurement setup

You want to measure the current through a resistor. Should the meter go across it or in the path?

• If you place the ammeter across the resistor (parallel), you create a very low-resistance branch, effectively a short. That can drastically increase current and may “blow” the meter.
• Correct: put the ammeter in series so the same current that goes through the resistor goes through the ammeter.

Exam Focus

• Typical question patterns:
  • Identify node potentials and recognize where voltage drops do and do not occur (ideal wires).
  • Determine correct placement of ammeters and voltmeters.
  • Predict effects of adding/removing wires (shorts) or opening/closing switches.
• Common mistakes:
  • Putting an ammeter in parallel or a voltmeter in series in diagrams.
  • Thinking current “chooses the path of least resistance” as if other paths get zero current (in reality, current splits among paths according to resistance).
  • Forgetting that a wire in parallel with an element forces the same potential on both ends, often reducing the element’s voltage.

Series and Parallel Resistor Circuits: Building Equivalent Resistance Intuition

Most circuit problems become manageable when you can replace part of a circuit with an equivalent resistance R_{\text{eq}}. The key is understanding what series and parallel connections really mean in terms of shared current and shared voltage.

Series: same current, voltage divides

Resistors are in series if the same current must pass through each one (there is no branching between them). In a series chain:

• Current is the same through each: I_1 = I_2 = I_3 = I.
• The total potential difference across the chain is the sum of individual drops.

Equivalent resistance for series resistors:

R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots

Why it makes sense: adding more resistors in series makes it harder for current to flow, so resistance increases.

Voltage division (conceptual): If a fixed current flows, the drop across each resistor is \Delta V_i = IR_i, so bigger resistors take a larger share of the total voltage.

Parallel: same voltage, current divides

Resistors are in parallel if they share the same two nodes—meaning each resistor has the same potential difference across it.

• Voltage is the same across each branch: \Delta V_1 = \Delta V_2 = \Delta V.
• Current splits among branches, and branch currents add to the total.

Equivalent resistance for parallel resistors:

\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots

Why it makes sense: adding a parallel branch provides an additional path for charge to flow, so overall resistance decreases.

A useful reality check: for resistors in parallel, R_{\text{eq}} must be smaller than the smallest individual resistor.

How to decide series vs parallel (a reliable method)

Students often memorize pictures rather than using definitions, and then get confused in “messy” diagrams. A robust approach:

• Series test: Do the two components share a single connection point that nothing else connects to (no branch at their junction)? If yes, they are in series.
• Parallel test: Are both ends of one component connected to the same two nodes as the other component? If yes, they are in parallel.

Power and brightness reasoning for identical bulbs/resistors

Brightness of a bulb (in simplified AP modeling) is linked to power dissipation. For a resistor-like bulb:

P = I\Delta V

If bulbs are identical, you can compare power in each bulb to compare brightness.

• In series, identical bulbs share the same current, so each has the same power, but the current is smaller than if a single bulb were connected to the battery—so each is dimmer.
• In parallel, identical bulbs each get the full battery voltage, so each dissipates about the same power as a single bulb alone—so each has similar brightness to the single-bulb case (assuming an ideal battery).

Worked example: series-parallel reduction

Two resistors R_1 = 4\ \Omega and R_2 = 6\ \Omega are in series. This series combination is in parallel with R_3 = 3\ \Omega. Find R_{\text{eq}}.

Step 1: series first:

R_{12} = R_1 + R_2 = 4 + 6 = 10\ \Omega

Step 2: parallel with R_3:

\frac{1}{R_{\text{eq}}} = \frac{1}{10} + \frac{1}{3} = \frac{3}{30} + \frac{10}{30} = \frac{13}{30}

So:

R_{\text{eq}} = \frac{30}{13}\ \Omega \approx 2.31\ \Omega

Sanity check: parallel with a 3\ \Omega resistor should give less than 3\ \Omega—it does.

Exam Focus

• Typical question patterns:
  • Reduce a network to R_{\text{eq}} and then find total current and/or branch currents.
  • Brightness or power comparisons for bulbs in series vs parallel.
  • Determine whether resistors are truly in series/parallel in non-obvious diagrams.
• Common mistakes:
  • Assuming resistors “next to each other” are in series without checking for branching.
  • Forgetting that parallel elements share the same two nodes (same voltage), not “same side of the page.”
  • Claiming current is the same in parallel branches; actually voltage is the same and current divides.

Kirchhoff’s Junction and Loop Rules: Solving Real Multi-Loop Circuits

Series-parallel reduction is powerful, but many AP circuits cannot be simplified that way—especially with multiple sources or bridge-like connections. Kirchhoff’s rules let you write equations based on conservation laws.

Junction rule (current conservation)

At a junction (node), charge cannot accumulate in steady state, so current in equals current out:

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

This is conservation of charge applied to circuits.

Loop rule (energy conservation)

Around any closed loop, the algebraic sum of potential changes is zero:

\sum \Delta V = 0

This is conservation of energy per unit charge.

To apply it, you must choose:

1. A direction to traverse the loop.
2. A direction for each current (you can guess; negative answers just mean the real direction is opposite).

Then assign signs consistently:

• Resistor: in direction of current, \Delta V = -IR; opposite direction, \Delta V = +IR.
• Ideal source: from negative to positive, +\mathcal{E}; from positive to negative, -\mathcal{E}.

A systematic solving strategy

For a circuit with multiple unknown currents:

1. Label currents in each branch (use arrows).
2. Write junction equations at key nodes.
3. Write loop equations for independent loops.
4. Solve the resulting system of linear equations.

A big mindset shift: you are not “finding the one right equation.” You are building a solvable system from conservation laws.

Including internal resistance (real batteries)

Real sources can be modeled as an ideal emf \mathcal{E} in series with an **internal resistance** r.

If the circuit draws current I, the terminal voltage across the battery’s external terminals is:

V_{\text{terminal}} = \mathcal{E} - Ir

This matches the energy picture: some energy per charge is dissipated inside the battery.

If the battery is being charged (current forced into its positive terminal in the appropriate direction), the sign situation can lead to terminal voltage larger than \mathcal{E}. On AP problems, you’ll usually be guided by the loop-rule signs.

Worked example: two-loop Kirchhoff system

A circuit has two loops that share a resistor. Left loop: battery \mathcal{E}_1 = 12\ \text{V} in series with resistor R_1 = 4\ \Omega. Right loop: battery \mathcal{E}_2 = 6\ \text{V} in series with resistor R_2 = 3\ \Omega. The loops share a middle resistor R_3 = 2\ \Omega between the top node and bottom node. Let mesh currents I_1 (left loop) and I_2 (right loop) both be clockwise.

In the shared resistor, currents oppose: actual current through R_3 (top to bottom for left loop direction) is I_1 - I_2.

Loop 1 (clockwise): rise across battery then drops across resistors:

\mathcal{E}_1 - I_1R_1 - (I_1 - I_2)R_3 = 0

Substitute values:

12 - 4I_1 - 2(I_1 - I_2) = 0

Simplify:

12 - 4I_1 - 2I_1 + 2I_2 = 0

12 - 6I_1 + 2I_2 = 0

Loop 2 (clockwise):

\mathcal{E}_2 - I_2R_2 - (I_2 - I_1)R_3 = 0

Note: for loop 2, the shared resistor drop uses current I_2 - I_1.

Substitute values:

6 - 3I_2 - 2(I_2 - I_1) = 0

Simplify:

6 - 3I_2 - 2I_2 + 2I_1 = 0

6 + 2I_1 - 5I_2 = 0

Now solve the system:

-6I_1 + 2I_2 = -12

2I_1 - 5I_2 = -6

Multiply the second equation by 3 to eliminate I_1:

6I_1 - 15I_2 = -18

Add to the first equation:

(-6I_1 + 2I_2) + (6I_1 - 15I_2) = -12 - 18

-13I_2 = -30

I_2 = \frac{30}{13}\ \text{A} \approx 2.31\ \text{A}

Substitute back into 6 + 2I_1 - 5I_2 = 0:

2I_1 = 5I_2 - 6

I_1 = \frac{5(30/13) - 6}{2} = \frac{150/13 - 78/13}{2} = \frac{72/13}{2} = \frac{36}{13}\ \text{A} \approx 2.77\ \text{A}

Then current through shared resistor is I_1 - I_2 = (36/13 - 30/13) = 6/13\ \text{A}.

Interpretation: both loop currents are clockwise as assumed (positive), and the shared resistor carries a modest net current.

Exam Focus

• Typical question patterns:
  • Write and solve Kirchhoff junction + loop equations for multi-loop circuits.
  • Include internal resistance and find terminal voltage or current under load.
  • Determine direction of current in a branch (often revealed by a negative solution).
• Common mistakes:
  • Sign errors in loop equations (especially across shared resistors or when traversing opposite current direction).
  • Writing too few independent equations for the number of unknown currents.
  • Treating emf sources as always potential rises regardless of traversal direction.

Capacitors: Charge Storage, Energy, and Combining Capacitors

Resistors dissipate energy; capacitors store energy in an electric field. In AP Physics 2, capacitors are central because they create time-dependent behavior in circuits.

What a capacitor is

A capacitor is a device that stores separated charge: one plate becomes positively charged, the other negatively charged. The amount of charge stored per potential difference is the capacitance C:

C = \frac{Q}{\Delta V}

• C is measured in farads (F), where 1\ \text{F} = 1\ \text{C/V}.
• Q is the magnitude of charge on either plate (plates have equal and opposite charge).

A key conceptual point: capacitance depends on geometry and dielectric material, not on how much charge you currently put on it.

Energy stored in a capacitor

Charging a capacitor requires doing work to move charge onto a plate against an increasing potential difference. The stored energy is:

U = \frac{1}{2}C(\Delta V)^2

Equivalent forms (same relationship rewritten):

U = \frac{1}{2}Q\Delta V

U = \frac{Q^2}{2C}

In circuits, that energy can later be released (for example, during discharge through a resistor).

Dielectrics (conceptual emphasis)

A dielectric is an insulating material placed between capacitor plates. In many cases, inserting a dielectric increases capacitance because the material polarizes and reduces the effective electric field for a given free charge, allowing more charge to be stored at the same voltage.

On AP problems, you are often expected to reason qualitatively: “With the battery still connected (fixed voltage), increasing C increases Q = C\Delta V.” If the capacitor is isolated (fixed charge), increasing C decreases \Delta V = Q/C.

Combining capacitors in series and parallel

Capacitors combine oppositely to resistors.

Parallel capacitors share the same voltage (same two nodes). Charges add:

C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots

Why: at the same \Delta V, total charge stored is the sum of charges on each capacitor.

Series capacitors share the same charge magnitude (in steady state, the same charge must appear on each because the intermediate conductor cannot accumulate net charge). Voltages add:

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots

A useful check: series combination yields a capacitance smaller than the smallest individual capacitor.

Worked example: series vs parallel capacitor combination

Two capacitors C_1 = 4\ \mu\text{F} and C_2 = 6\ \mu\text{F}.

Parallel:

C_{\text{eq}} = 4 + 6 = 10\ \mu\text{F}

Series:

\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}

So:

C_{\text{eq}} = \frac{12}{5}\ \mu\text{F} = 2.4\ \mu\text{F}

Exam Focus

• Typical question patterns:
  • Combine capacitors to find C_{\text{eq}} and then find charge/voltage/energy.
  • Compare what changes when a dielectric is inserted with the battery connected vs disconnected.
  • Use energy expressions to compare stored energy before and after configuration changes.
• Common mistakes:
  • Using resistor combination rules for capacitors (they are reversed: series uses reciprocals, parallel adds).
  • Forgetting that series capacitors have the same charge magnitude (in steady state).
  • Confusing “charge on the capacitor” with “net charge on the whole device” (net is zero, but separated charges are nonzero).

RC Circuits: Transient Charging and Discharging (The Time Constant)

Pure resistor circuits usually reach steady current essentially instantly in the ideal model. Capacitors change that: when a capacitor charges or discharges through a resistor, the current and voltage change over time in a predictable exponential way. This is a major emphasis in AP Physics 2 Unit 3.

What happens physically during charging

Consider a series circuit with a battery \mathcal{E}, a resistor R, and a capacitor C.

Right after the switch is closed:

• The capacitor is initially uncharged, so \Delta V_C \approx 0.
• The full battery voltage appears across the resistor, so the initial current is largest.

As time passes:

• Charge accumulates on the capacitor plates.
• The capacitor’s voltage \Delta V_C increases.
• That reduces the voltage left for the resistor, so current decreases.

Eventually (long time later):

• The capacitor voltage approaches the battery emf.
• The current approaches zero.

A key steady-state rule: In a DC circuit at long times, a capacitor behaves like an open circuit (no current through that branch).

The time constant

The characteristic time scale is the time constant \tau:

\tau = RC

• Larger R means slower charging (current is more limited).
• Larger C means more charge must be accumulated for a given voltage change, also slower.

A common interpretation: after one time constant during charging, the capacitor voltage reaches about 63% of its final value; during discharging, it drops to about 37% of its initial value. (You are often expected to use the qualitative “one time constant” idea, sometimes with approximate percentages.)

Charging equations

For charging from 0 toward \mathcal{E}:

\Delta V_C(t) = \mathcal{E}\left(1 - e^{-t/(RC)}\right)

Charge on the capacitor:

Q(t) = C\mathcal{E}\left(1 - e^{-t/(RC)}\right)

Current in the circuit:

I(t) = \frac{\mathcal{E}}{R}e^{-t/(RC)}

Notice the structure:

• Voltage and charge rise toward a maximum.
• Current decays toward zero.

Discharging equations

For a capacitor initially at voltage \Delta V_0 discharging through a resistor R:

\Delta V_C(t) = \Delta V_0 e^{-t/(RC)}

Charge:

Q(t) = Q_0 e^{-t/(RC)}

Current magnitude:

I(t) = \frac{\Delta V_0}{R}e^{-t/(RC)}

The sign of current depends on the direction you defined as positive; the exponential gives the magnitude behavior.

Why exponential behavior appears (conceptual reasoning)

At any moment during charging, Kirchhoff’s loop rule gives:

• Battery rise equals resistor drop plus capacitor voltage.

As the capacitor voltage grows, the resistor’s share of the battery voltage shrinks. Because current through the resistor depends on that resistor voltage, the rate of change decreases as you approach the final state. That “rate proportional to how far you are from the final value” structure leads mathematically to exponentials.

You don’t need calculus on the AP exam to use these results, but you should understand the story they tell.

Graph shapes you should recognize

• I(t) decays exponentially from \mathcal{E}/R to 0 during charging.
• \Delta V_C(t) rises exponentially from 0 to \mathcal{E} during charging.
• During discharge, both \Delta V_C(t) and I(t) decay exponentially toward 0.

Worked example: using the time constant

A circuit has R = 200\ \text{k}\Omega and C = 5.0\ \mu\text{F}.

Compute the time constant:

\tau = RC = (200\times 10^3)(5.0\times 10^{-6}) = 1.0\ \text{s}

Interpretation: after about 1\ \text{s} of charging, \Delta V_C is about 63% of the battery voltage; after about 5\ \text{s} (five time constants), it’s very close to fully charged.

Worked example: current at a given time during charging

A battery \mathcal{E} = 12\ \text{V} charges a capacitor through R = 3.0\ \text{k}\Omega and C = 200\ \mu\text{F}. Find the current at t = 0.60\ \text{s}.

Step 1: compute \tau:

\tau = RC = (3.0\times 10^3)(200\times 10^{-6}) = 0.60\ \text{s}

Step 2: use current function:

I(t) = \frac{\mathcal{E}}{R}e^{-t/(RC)}

Initial current \mathcal{E}/R:

\frac{\mathcal{E}}{R} = \frac{12}{3.0\times 10^3} = 4.0\times 10^{-3}\ \text{A}

Now at t = \tau, the factor is e^{-1}:

I(0.60\ \text{s}) = (4.0\times 10^{-3})e^{-1} \approx 1.47\times 10^{-3}\ \text{A}

So the current is about 1.5\ \text{mA}.

Exam Focus

• Typical question patterns:
  • Determine whether a capacitor acts like an open circuit (long time DC) or like a wire initially (at the instant just after switching, for an uncharged capacitor).
  • Compute \tau = RC and interpret capacitor voltage/current after some number of time constants.
  • Use exponential functions for I(t), Q(t), and \Delta V_C(t) in charging/discharging.
• Common mistakes:
  • Thinking current is constant during charging (it is not; it decays).
  • Treating a capacitor as a resistor in steady state; in DC steady state it blocks current.
  • Plugging incorrect “final values” into the charging equation (for charging from 0 to \mathcal{E}, the final capacitor voltage is \mathcal{E}, not something else).

Multi-Element Circuits with Capacitors: Steady-State, Switching, and Reasoning Shortcuts

AP problems often combine several resistors and capacitors with switches. You usually do not need to solve a full differential equation for every configuration; instead, you use two powerful ideas:

1. What capacitors do at t = 0^+ (just after a switch action).
2. What capacitors do at t \to \infty (long after the switch action in a DC circuit).

The two “extreme time” models

Just after switching (immediately after a change):

• The capacitor’s voltage cannot change instantaneously.
• So if a capacitor initially has voltage \Delta V_C(0^-), then \Delta V_C(0^+) = \Delta V_C(0^-).

This often makes a charged capacitor behave like a voltage source (holding that initial voltage for the instant).

Long after switching in a DC circuit:

• Current through any capacitor goes to zero.
• The capacitor behaves like an open circuit.

These two models let you analyze the circuit at two times using only algebra and Kirchhoff/series-parallel rules.

Capacitor voltage continuity (why it can’t jump)

If capacitor voltage changed instantly, that would require an instantaneous change in charge Q since Q = C\Delta V. An instantaneous charge change implies infinite current I = \Delta Q/\Delta t, which is not physically possible in ordinary circuits. So capacitor voltage is continuous even though current can change abruptly.

Example scenario: switch changes the circuit topology

A common setup:

• A capacitor charges through a resistor from a battery for a long time.
• Then a switch disconnects the battery and connects the capacitor across a different resistor network to discharge.

To solve, you typically:

1. Find the capacitor’s voltage just before switching (often equals battery voltage if charged fully).
2. Use continuity to set initial voltage for the discharge phase.
3. Find the equivalent resistance “seen” by the capacitor in the discharge configuration.
4. Compute \tau = R_{\text{eq}}C for the discharge.
5. Write \Delta V_C(t) = \Delta V_0 e^{-t/\tau} or use qualitative time-constant reasoning.

“Resistance seen by the capacitor” (a practical method)

When finding \tau for a discharge or charge process, you often need the equivalent resistance that controls current to/from the capacitor.

A common AP method:

• Replace the capacitor by an open circuit (because you’re analyzing the resistor network it connects to), and replace ideal voltage sources with wires (set them to 0 V) when finding the equivalent resistance for time constant calculations.

Then compute R_{\text{eq}} between the two nodes where the capacitor connects.

This works because the time constant depends on the Thevenin-equivalent resistance seen by the capacitor.

Worked example: capacitor discharging through a network

A capacitor C = 10\ \mu\text{F} is initially charged to \Delta V_0 = 20\ \text{V}. At t = 0 it is connected across two resistors in parallel: R_1 = 4\ \text{k}\Omega and R_2 = 6\ \text{k}\Omega.

Step 1: find equivalent discharge resistance:

\frac{1}{R_{\text{eq}}} = \frac{1}{4\times 10^3} + \frac{1}{6\times 10^3} = \frac{3}{12\times 10^3} + \frac{2}{12\times 10^3} = \frac{5}{12\times 10^3}

So:

R_{\text{eq}} = \frac{12\times 10^3}{5} = 2.4\times 10^3\ \Omega

Step 2: time constant:

\tau = R_{\text{eq}}C = (2.4\times 10^3)(10\times 10^{-6}) = 2.4\times 10^{-2}\ \text{s}

Step 3: voltage after t = 0.048\ \text{s} (two time constants):

\Delta V_C(t) = \Delta V_0 e^{-t/\tau} = 20e^{-0.048/0.024} = 20e^{-2}

\Delta V_C(t) \approx 20(0.135) \approx 2.7\ \text{V}

Interpretation: discharging through a relatively low resistance happens quickly.

Common conceptual pitfall: “the capacitor is a break, so nothing happens”

It’s true that at long times a capacitor blocks DC current, but during transients it absolutely allows current (because its voltage is changing). So if a question asks what happens right after a switch closes, you must consider whether the capacitor is initially charged or uncharged—because that determines the initial current.

Real-world applications (why RC matters)

RC behavior appears any time a system stores charge/field energy and loses energy through resistance:

• Camera flashes (capacitor discharge)
• Timing circuits and delays
• Filters in electronics (capacitors block DC but pass changing signals to varying degrees)
• Sensing (capacitive touch sensing involves changing capacitance and timing)

You do not need advanced electronics for AP, but recognizing that “capacitors create time dependence” is the big idea.

Exam Focus

• Typical question patterns:
  • Analyze a circuit at t = 0^+ and at t \to \infty after a switch changes connections.
  • Find R_{\text{eq}} seen by a capacitor and compute \tau = R_{\text{eq}}C.
  • Determine initial and final capacitor voltages/charges using continuity and steady-state open-circuit behavior.
• Common mistakes:
  • Letting capacitor voltage jump instantly at switching (it cannot).
  • Forgetting to recompute R_{\text{eq}} after the switch changes the circuit.
  • Treating capacitor branches as open circuits at all times rather than only at long-time DC steady state.