Capacitors in Circuits

AP Physics C: E&M· AP Physics C: E&M CED — Electric Circuits· 14 min read

1. Equivalent Capacitance for Series and Parallel Combinations★★☆☆☆⏱ 4 min

When multiple capacitors are connected in a circuit network, we can simplify the network to a single equivalent capacitance $C_{e q}$ using rules based on connection type. Importantly, these rules are reversed from the equivalent resistance rules for resistors, which is a common point of confusion.

For parallel connections: All capacitors share the same potential difference across their plates, equal to the potential difference across the equivalent capacitor. Total charge stored by the combination is the sum of charge on each individual capacitor, per conservation of charge:

Q_{\text{total}} = Q_1 + Q_2 + ... + Q_n

Substituting $Q = C V$ and canceling the common potential difference gives the parallel equivalent capacitance rule:

C_{eq,\text{parallel}} = \sum_{i=1}^n C_i

For series connections: All capacitors in series carry the same charge $Q$ on their plates (charge is conserved through the series branch, so no net charge can accumulate between capacitor plates). The total potential difference across the combination is the sum of potential differences across each individual capacitor. Substituting $V = Q / C$ and canceling the common charge gives the series rule:

\frac{1}{C_{eq,\text{series}}} = \sum_{i=1}^n \frac{1}{C_i}

Intuition for the reversed rules comes from the parallel-plate capacitor formula $C = ε_{0} A / d$ : parallel capacitors add effective plate area $A$ , so capacitance adds, while series capacitors add effective plate separation $d$ , so reciprocals of capacitance add.

📐 Worked Example

Find the equivalent capacitance between points A and B for the following network: a 1 μF capacitor and a 3 μF capacitor are connected in series, and this combination is connected in parallel with a 4 μF capacitor.

1
First simplify the series combination of 1 μF and 3 μF using the series rule:
$\frac{1}{C_{12}} = \frac{1}{1\ \mu\text{F}} + \frac{1}{3\ \mu\text{F}} = \frac{4}{3\ \mu\text{F}} \implies C_{12} = 0.75\ \mu\text{F}$
2
This $C_{12}$ is in parallel with 4 μF, so add the capacitances:
$C_{eq} = 0.75\ \mu\text{F} + 4\ \mu\text{F} = 4.75\ \mu\text{F}$
3
Check for consistency: The series equivalent is smaller than the smallest individual capacitor, and the total parallel equivalent is larger than the largest individual capacitor, which matches the rules for capacitor combinations.

Exam tip:

If your result for equivalent capacitance of a series combination is larger than any individual capacitor in the series, you have swapped the series and parallel rules—reverse them immediately to get a correct result.

2. Charging RC Circuits★★★☆☆⏱ 5 min

A resistor-capacitor (RC) circuit is a circuit with one or more resistors and one or more capacitors that exhibits time-dependent behavior when connected to a voltage source. When an initially uncharged capacitor is connected in series with a battery and a resistor, it does not charge instantaneously. Charge builds up exponentially, with the rate of charging determined by the product of resistance and capacitance, called the time constant $τ$ .

🔬 Derivation

Goal:

Derive the time dependence of charge for a charging RC circuit

Starting from:

Kirchhoff's loop rule for a series circuit with emf $ε$ , resistance $R$ , and capacitance $C$

1
Apply loop rule to the closed circuit:
2
$\varepsilon - IR - V_C = 0$
3
Substitute $V_{C} = Q / C$ and $I = d Q / d t$ (current is the rate of change of charge on the capacitor):
4
$\varepsilon - R \frac{dQ}{dt} - \frac{Q}{C} = 0$

Result:

Separating variables and integrating from initial condition $Q (0) = 0$ (uncharged at $t = 0$ ) gives the solution: $Q (t) = Q_{max} (1 - e^{- t / τ})$ , $V_{C} (t) = ε (1 - e^{- t / τ})$ , $I (t) = \frac{ε}{R} e^{- t / τ}$ , where $τ = R C$ is the time constant. After one time constant, the capacitor reaches ~63% of its maximum charge; after 5 time constants, it is effectively fully charged.

📐 Worked Example

A 20 V battery, 500 Ω resistor, and 100 μF capacitor are connected in series to form a charging RC circuit. The capacitor is initially uncharged. What is the voltage across the capacitor 0.06 seconds after the circuit is connected?

1
Calculate the time constant:
$\tau = RC = (500\ \Omega)(100 \times 10^{-6}\ \text{F}) = 0.05\ \text{s}$
2
Use the charging voltage relation for an initially uncharged capacitor:
$V_C(t) = \varepsilon (1 - e^{-t/\tau})$
3
Substitute values $t = 0.06$ s:
$t/\tau = \frac{0.06}{0.05} = 1.2 \implies V_C = 20 (1 - e^{-1.2}) \approx 14.0\ \text{V}$
4
Check: After 1.2 time constants, we expect voltage greater than 63% of 20 V (12.6 V), which matches our result.

Exam tip:

Always confirm your initial conditions before writing exponential expressions. For an initially uncharged capacitor at $t = 0$ , $V_{C} (0) = 0$ and $I (0) = ε / R$ —this is a quick check to confirm you have the correct form of the exponential function.

3. Discharging RC Circuits★★★☆☆⏱ 5 min

When a fully charged capacitor is disconnected from its charging battery and connected to a resistor, it discharges, releasing its stored charge through the resistor, again with exponential time dependence. We use the same Kirchhoff's loop approach for discharging, with no battery emf in the circuit.

🔬 Derivation

Goal:

Derive the time dependence of charge for a discharging RC circuit

Starting from:

A capacitor with initial charge $Q_{0} = C V_{0}$ connected to a resistor $R$ at $t = 0$

1
Apply Kirchhoff's loop rule to the closed circuit:
2
$V_C - IR = 0$
3
Substitute $V_{C} = Q / C$ . For discharge, charge decreases over time, so $I = - d Q / d t$ (the negative sign accounts for decreasing $Q$ ):
4
$\frac{Q}{C} + R \frac{dQ}{dt} = 0$

Result:

Separating variables and integrating from initial condition $Q (0) = Q_{0}$ gives the solution: $Q (t) = Q_{0} e^{- t / τ}$ , $V_{C} (t) = V_{0} e^{- t / τ}$ , $I (t) = - \frac{V _{0}}{R} e^{- t / τ}$ , where $τ = R C$ is identical to the charging time constant. The negative sign indicates current direction is opposite to charging. After one time constant, ~37% of initial charge remains; after 5 time constants, the capacitor is effectively fully discharged.

📐 Worked Example

A 20 μF capacitor is charged to 100 V, then connected to a 50 kΩ resistor to discharge. What is the charge on the capacitor after 1 second?

1
Calculate initial charge on the capacitor:
$Q_0 = C V_0 = (20 \times 10^{-6}\ \text{F})(100\ \text{V}) = 0.002\ \text{C}$
2
Calculate the time constant:
$\tau = RC = (50 \times 10^3\ \Omega)(20 \times 10^{-6}\ \text{F}) = 1.0\ \text{s}$
3
Use the discharge charge relation:
$Q(t) = Q_0 e^{-t/\tau} = 0.002 e^{-1} \approx 7.36 \times 10^{-4}\ \text{C} = 0.736\ \text{mC}$
4
Check: After one time constant, we expect ~37% of the initial charge, which matches our result.

Exam tip:

In steady-state DC (after many time constants), a capacitor acts as an open circuit with zero current. If you are asked for steady-state current through a capacitor branch, the answer is always zero.

4. AP-Style Practice Worked Examples★★★★☆⏱ 5 min

📐 Worked Example

An uncharged 6 μF capacitor and a 500 Ω resistor are connected in series to a 12 V battery. What is the current in the circuit after two time constants have elapsed? Options: A) 3.3 mA B) 16.2 mA C) 0.33 mA D) 24 mA

1
For a charging RC circuit, calculate initial current:
$I_0 = \frac{\varepsilon}{R} = \frac{12\ \text{V}}{500\ \Omega} = 0.024\ \text{A} = 24\ \text{mA}$
2
Current decays exponentially as $I (t) = I_{0} e^{- t / τ}$ . After two time constants, $t / τ = 2$ :
$I = 24 e^{-2} \approx 24(0.135) \approx 3.3\ \text{mA}$
3
The correct answer is A. Option D is the initial current at $t = 0$ , B and C are calculation errors.

📐 Worked Example

A circuit consists of a 12 V battery connected to an open switch, a 40 kΩ resistor $R_{1}$ , and a series combination of a 5 μF capacitor $C_{1}$ and a 10 μF capacitor $C_{2}$ . Both capacitors are initially uncharged. At $t = 0$ , the switch is closed. (a) Calculate equivalent capacitance. (b) Calculate the time constant. (c) Find $V_{C 2} (t)$ and its value at $t = 0.3$ s.

1
Part (a): Capacitors are in series, so apply the series rule:
$\frac{1}{C_{eq}} = \frac{1}{5\ \mu\text{F}} + \frac{1}{10\ \mu\text{F}} = \frac{3}{10\ \mu\text{F}} \implies C_{eq} \approx 3.33\ \mu\text{F}$
2
Part (b): Calculate time constant:
$\tau = R_1 C_{eq} = (4 \times 10^4\ \Omega)(3.33 \times 10^{-6}\ \text{F}) = 0.133\ \text{s}$
3
Part (c): For series capacitors, charge on $C_{2}$ equals total charge $Q (t) = C_{e q} ε (1 - e^{- t / τ})$ . Voltage across $C_{2}$ is $V_{C 2} = Q / C_{2}$ :
$V_{C2}(t) = 4\left(1 - e^{-t/0.133}\right) \text{ V}$
4
Substitute $t = 0.3$ s:
$t/\tau \approx 2.25 \implies V_{C2} \approx 3.58\ \text{V}$

📐 Worked Example

A heart defibrillator uses a 50 μF capacitor charged to 3000 V. After charging, the capacitor is discharged through a patient's chest with effective resistance 50 Ω. The defibrillator must deliver 95% of its stored charge to be effective. How long does discharge take?

1
Start with the discharge relation, set charge delivered equal to 95% of initial charge:
$Q_0 - Q(t) = 0.95 Q_0 \implies e^{-t/\tau} = 0.05$
2
Calculate the time constant:
$\tau = RC = (50\ \Omega)(50 \times 10^{-6}\ \text{F}) = 0.0025\ \text{s}$
3
Solve for $t$ by taking the natural logarithm of both sides:
$-t/\tau = \ln(0.05) \approx -2.996 \implies t \approx 7.5\ \text{ms}$

5. Common Pitfalls

Wrong move:

Swapping series and parallel rules, calculating $C_{e q}$ for series as the sum of individual capacitances

Why:

Students memorize resistor rules first and confuse the reversed rules for capacitors

Correct move:

If you forget the rules, rederive from first principles: parallel capacitors share voltage, charge adds so capacitance adds; series capacitors share charge, voltage adds so reciprocals of capacitance add

Wrong move:

Forgetting the negative sign for current during discharge, leading to wrong current direction

Why:

Students use the same positive sign convention as charging, where charge is increasing

Correct move:

Explicitly define current direction, and note that charge on the capacitor decreases during discharge so $d Q / d t$ is negative, leading to negative current relative to the charging direction

Wrong move:

Using $τ = 1/ (R C)$ instead of $τ = R C$ for the time constant

Why:

Students confuse the decay constant (units 1/s) with the time constant (units of seconds)

Correct move:

Check units first: $R \times C$ has units of seconds, which matches the requirement for a time constant

Wrong move:

Only using the resistor labeled near the capacitor for time constant calculation, forgetting other series resistors

Why:

Students overlook resistors elsewhere in the series branch with the capacitor

Correct move:

Always find the total equivalent resistance seen by the capacitor when calculating the time constant, including all resistors in its series path

Wrong move:

Writing a positive potential change when crossing a capacitor from positive to negative plate in a Kirchhoff loop

Why:

Students mix up capacitor sign conventions with resistor or battery conventions

Correct move:

Follow the same rule as batteries: crossing from negative to positive plate gives a positive potential change, crossing from positive to negative gives a negative potential drop

6. Quick Reference Cheatsheet

Category	Formula	Notes
Parallel Capacitor Equivalent	$C_{e q} = C_{1} + C_{2} + ... + C_{n}$	All capacitors share the same potential difference
Series Capacitor Equivalent	$\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + ... + \frac{1}{C _{n}}$	All capacitors carry the same charge
RC Time Constant	$τ = R C$	Same for charging and discharging, units = seconds
Charging: Charge	$Q (t) = C ε (1 - e^{- t / τ})$	Initial condition: $Q (0) = 0$ (uncharged)
Charging: Voltage/Current	$V_{C} (t) = ε (1 - e^{- t / τ}), I (t) = \frac{ε}{R} e^{- t / τ}$	Current decays from maximum to zero
Discharging: Charge/Voltage	$Q (t) = Q_{0} e^{- t / τ}, V_{C} (t) = V_{0} e^{- t / τ}$	Initial condition: $Q (0) = Q_{0} = C V_{0}$ (fully charged)
Discharging: Current	$I (t) = - \frac{V _{0}}{R} e^{- t / τ}$	Negative sign = current direction opposite to charging
Steady-State DC Behavior	$I = 0$	Fully charged capacitor acts as an open circuit

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Equivalent capacitance calculation
2022 · FRQ
Discharging RC circuit analysis

Going deeper

unit overviewAP Physics C: E&M Unit 3 Overview

What's Next

Mastery of capacitors in circuits is a required prerequisite for the remaining topics in Unit 3 of AP Physics C: E&M, including energy storage in capacitors and analysis of multi-loop capacitive circuits. Without understanding how to calculate equivalent capacitance and model time-dependent exponential behavior, you will not be able to solve common free-response questions involving energy transfer or complex RC networks, which frequently appear on the AP exam. This topic also builds the core foundation for AC circuits later in the course, where capacitors act as frequency-dependent impedance elements, drawing directly on your understanding of capacitive behavior in DC circuits. Practice applying these rules regularly to build fluency for exam day.