11.8 Resistor-Capacitor (RC) Circuits

RC circuits combine resistors and capacitors so that charge, current, and voltage change over time instead of staying fixed. The time constant $\tau = R_{\mathrm{eq}}C_{\mathrm{eq}}$ controls how fast a capacitor charges or discharges, and capacitor combinations follow rules that are the reverse of resistor combinations.

Why This Matters for the AP Physics C: E&M Exam

This topic pulls together Kirchhoff's loop rule, capacitors, and resistors into one model where quantities depend on time. You will be asked to write and interpret the differential equation from the loop rule, predict short-term and long-term behavior, and use the exponential charging and discharging functions. Because Unit 11 is one of the most heavily weighted units on the exam, you should be ready to handle RC circuits both as quick conceptual checks and as multi-step calculations.

The free-response section includes an Experimental Design and Analysis question, and RC circuits fit that style well. You may need to design a procedure to measure a time constant, linearize exponential data to find a slope, or justify claims using your data and the underlying physics.

Key Takeaways

• Capacitors in series add as reciprocals, so the equivalent capacitance is less than the smallest one; capacitors in parallel simply add.
• Series capacitors carry the same magnitude of charge on each plate because of conservation of charge.
• Apply Kirchhoff's loop rule to get the governing equation $\varepsilon = \frac{dq}{dt}R + \frac{q}{C}$.
• The time constant is $\tau = R_{\mathrm{eq}} C_{\mathrm{eq}}$: about 63% charged after one $\tau$, about 37% of charge left after one $\tau$ when discharging.
• At $t = 0$ an uncharged capacitor acts like a wire (maximum current); after $t \gg \tau$ a fully charged capacitor acts like an open branch (zero current).
• Charging and discharging follow exponential functions, so charge, voltage, current, and stored energy all approach steady state asymptotically.

Equivalent Capacitance

Capacitors can be arranged in different configurations within a circuit, and these arrangements affect how they store charge and energy.

Series Connection. When capacitors are connected end-to-end, they form a series combination.

• The equivalent capacitance is calculated using:

$\frac{1}{C_{\mathrm{eq}, \mathrm{s}}}=\sum_{i} \frac{1}{C_{i}}$

• For two capacitors in series, this simplifies to: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$
• The equivalent capacitance is always smaller than the smallest individual capacitance in the series.
• Each capacitor in series has the same magnitude of charge on its plates due to conservation of charge.

Parallel Connection. When capacitors are connected across the same two points, they form a parallel combination.

• The equivalent capacitance is simply the sum of individual capacitances:

$C_{\mathrm{eq}, p}=\sum_{i} C_{i}$

• For two capacitors in parallel: $C_{eq} = C_1 + C_2$
• Parallel capacitors all experience the same voltage across their terminals.
• The total charge is distributed among the capacitors proportionally to their capacitance values.

Notice the pattern is the opposite of resistors: series resistors add directly, but series capacitors add as reciprocals.

RC Circuit Behavior

Fundamental Differential Equation

When a resistor and capacitor are combined in a circuit with a voltage source, their behavior is governed by a differential equation derived from Kirchhoff's loop rule.

$\mathcal{E}=\frac{d q}{d t} R+\frac{q}{C}$

This equation describes how the charge on the capacitor changes over time, where:

• $\mathcal{E}$ is the electromotive force (voltage source)
• $R$ is the resistance
• $C$ is the capacitance
• $q$ is the charge on the capacitor
• $\frac{d q}{d t}$ represents the current flowing through the resistor

Solving this equation gives the time-dependent behavior of the circuit, showing how charge, current, and voltage evolve after a switch is opened or closed.

Time Constant

The time constant characterizes how quickly an RC circuit responds to changes.

$\tau=R_{\mathrm{eq}} C_{\mathrm{eq}}$

The time constant has several important interpretations:

• It represents the time required for a charging capacitor to reach approximately 63% of its final value.
• During discharge, it is the time needed for the charge to decrease to about 37% of its initial value.
• After about 5 time constants, the circuit is treated as having essentially reached steady state.
• The time constant has units of seconds and measures how fast or slow the circuit responds.

For complex circuits with multiple resistors and capacitors, find the equivalent resistance and equivalent capacitance first, then use them in $\tau$.

Capacitor Charging

When a capacitor is connected to a voltage source through a resistor, it begins to charge. This process follows an exponential pattern set by the time constant.

• Initial behavior: An uncharged capacitor initially acts like a wire, allowing maximum current to flow.
• Charge accumulation: As charge builds up on the plates, the voltage across the capacitor increases.
• Current reduction: The increasing capacitor voltage opposes the source voltage, gradually reducing the current.
• Mathematical description: The charge on the capacitor during charging follows: $q(t) = Q(1 - e^{-t/\tau})$ where $Q = C\mathcal{E}$ is the maximum charge.
• Voltage and current: $V_C(t) = \mathcal{E}(1 - e^{-t/\tau})$ $I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}$
• Energy storage: As the capacitor charges, the stored electric potential energy increases according to $U = \frac{1}{2}CV_C^2 = \frac{q^2}{2C}$ and approaches a maximum asymptotically.
• Steady-state behavior: After a long time ($t \gg \tau$), the capacitor is fully charged for modeling purposes. The potential difference across it reaches its maximum value equal to the battery emf, and the current in that branch becomes zero.

Capacitor Discharging

When a charged capacitor is connected across a resistor with no voltage source, it discharges through the resistor.

• Initial behavior: At the moment of connection, maximum current flows because of the full voltage across the capacitor.
• Immediate change in stored quantities: As soon as discharging begins, the charge on the capacitor and the energy stored in its electric field both start to decrease.
• Exponential decay: The charge, voltage, and current all decrease exponentially with time.
• Mathematical description: The charge during discharging follows: $q(t) = Q e^{-t/\tau}$ where $Q$ is the initial charge.
• Voltage and current: $V_C(t) = V_0 e^{-t/\tau}$ $I(t) = \frac{V_0}{R}e^{-t/\tau}$ where $V_0$ is the initial voltage.
• Energy change: As the capacitor discharges, the stored electric potential energy decreases with time and is dissipated as thermal energy in the resistor. The stored energy approaches zero asymptotically.
• Steady-state behavior: After a long time ($t \gg \tau$), the capacitor is effectively discharged, with charge, capacitor voltage, and current all approximately zero.

How to Use This on the AP Physics C: E&M Exam

Problem Solving

• Reduce capacitor networks to a single $C_{\mathrm{eq}}$ and resistor networks to a single $R_{\mathrm{eq}}$ before finding $\tau$.
• For series capacitors, remember the charge is shared equally, so start from the common charge to find each voltage.
• Always check the two limiting cases. At $t = 0$, treat an uncharged capacitor as a wire and find the initial current from the resistors. At $t \gg \tau$, treat the capacitor as an open branch with zero current.

Free Response

• If asked to set up the model, start from Kirchhoff's loop rule and write $\varepsilon = \frac{dq}{dt}R + \frac{q}{C}$, then state initial conditions like $q(0)$.
• When solving for behavior over time, use the charging or discharging exponential form and identify $\tau$, the maximum charge, and the initial values clearly.
• To find a time, set the exponential expression equal to the target value and take a natural log, as shown in the practice problems below.

Experimental Design and Analysis

• A common approach is to measure capacitor voltage or current versus time, then linearize. Taking the natural log of an exponential turns it into a straight line whose slope is related to $1/\tau$.
• From the slope and a known $R$ or $C$, you can solve for the unknown circuit value, and you can discuss sources of error such as nonideal meters or resistor heating.

Common Misconceptions

• Capacitors do not follow the same combination rules as resistors. Series capacitors add as reciprocals (smaller result), and parallel capacitors add directly.
• A capacitor does not "block" current immediately. When uncharged, it acts like a wire at $t = 0$ and allows the largest current; it blocks current only after it is fully charged.
• The capacitor does not reach its final charge in one time constant. One $\tau$ gives about 63% charged, not 100%.
• Current does not keep flowing through a fully charged capacitor branch. At steady state that branch carries zero current, so any series resistor in that branch has zero voltage drop.
• The time constant uses equivalent values. Use $R_{\mathrm{eq}}$ and $C_{\mathrm{eq}}$ for the relevant branch, not just one component, when the circuit has multiple elements.

Practice Problem 1: Equivalent Capacitance

Solution

a) For capacitors in series, we use the formula: $\frac{1}{C_{\mathrm{eq}, \mathrm{s}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

Substituting the values: $\frac{1}{C_{\mathrm{eq}, \mathrm{s}}}=\frac{1}{4.0 \mu F}+\frac{1}{6.0 \mu F}+\frac{1}{12.0 \mu F}$ $\frac{1}{C_{\mathrm{eq}, \mathrm{s}}}=\frac{3}{12.0 \mu F}+\frac{2}{12.0 \mu F}+\frac{1}{12.0 \mu F}=\frac{6}{12.0 \mu F}=\frac{1}{2.0 \mu F}$

Therefore, $C_{\mathrm{eq}, \mathrm{s}} = 2.0 \mu F$

b) For capacitors in parallel, we use the formula: $C_{\mathrm{eq}, p}=C_1+C_2+C_3$

Substituting the values: $C_{\mathrm{eq}, p}=4.0 \mu F+6.0 \mu F+12.0 \mu F=22.0 \mu F$

Practice Problem 2: RC Circuit Time Constant

Solution

a) The time constant of an RC circuit is given by: $\tau = RC$

Substituting the values: $\tau = (100 \Omega)(470 \times 10^{-6} F) = 0.047 \text{ seconds}$

b) To find the time needed to reach 99% of the maximum charge, we use the charging equation: $q(t) = Q(1 - e^{-t/\tau})$

We need to find t when $q(t) = 0.99Q$: $0.99Q = Q(1 - e^{-t/\tau})$

$0.99 = 1 - e^{-t/\tau}$

$e^{-t/\tau} = 0.01$ $-t/\tau = \ln(0.01)$ $t = -\tau \ln(0.01)$ $t = 0.047 \text{ s} \times 4.605 = 0.216 \text{ seconds}$

Therefore, it will take approximately 0.216 seconds for the capacitor to charge to 99% of its maximum value.

Practice Problem 3: Capacitor Discharge

Solution

a) The initial current can be calculated using Ohm's law: $I_0 = \frac{V_0}{R} = \frac{50 \text{ V}}{10 \text{ k}\Omega} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$

b) For a discharging capacitor, the voltage follows: $V(t) = V_0 e^{-t/\tau}$

First, we need to find the time constant: $\tau = RC = (10 \times 10^3 \Omega)(25 \times 10^{-6} \text{ F}) = 0.25 \text{ seconds}$

Now, we need to find t when $V(t) = 5 \text{ V}$: $5 \text{ V} = 50 \text{ V} \times e^{-t/0.25}$ $\frac{5}{50} = e^{-t/0.25}$ $0.1 = e^{-t/0.25}$ $\ln(0.1) = -t/0.25$ $t = -0.25 \times \ln(0.1) = 0.25 \times 2.303 = 0.576 \text{ seconds}$

Therefore, it will take approximately 0.576 seconds for the voltage to drop to 5 V.

Related AP Physics C: E&M Guides

• 11.1 Electric Current
• 11.2 Electric Circuits
• 11.3 Resistance, Resistivity, and Ohm's Law
• 11.5 Compound Direct Current Circuits
• 11.4 Electric Power
• 11.6 Kirchhoff's Loop Rule