#AP Physics C: E&M - RC Circuits Study Guide ⚡

Hey there! Let's get you prepped for the AP exam with a focused review of RC circuits. This guide is designed to be your go-to resource, especially the night before the test. We'll break down the key concepts, highlight important formulas, and give you some exam-busting tips. Let's do this!

#Equivalent Capacitance

#Capacitor Combinations 🔌

Capacitors in circuits can be simplified into a single equivalent capacitance ( $C_{eq}$ ). This helps in analyzing complex circuits.
Series Capacitors:
- The reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances: $\frac{1}{C_{\text{eq, s}}} = \sum_{i} \frac{1}{C_i}$

Quick Fact

The equivalent capacitance of series capacitors is always smaller than the smallest individual capacitance.

Parallel Capacitors:
- The equivalent capacitance is the sum of individual capacitances: $C_{\text{eq, p}} = \sum_{i} C_i$

Key Concept

Charge Conservation: In a series connection, each capacitor holds the same magnitude of charge.

Practice Question

Multiple Choice:

Two capacitors, with capacitances of 2 $\mu$ F and 4 $\mu$ F, are connected in series. What is the equivalent capacitance of this combination? (A) 6 $\mu$ F (B) 3 $\mu$ F (C) 4/3 $\mu$ F (D) 8/3 $\mu$ F (E) 1/6 $\mu$ F
Three capacitors, each with a capacitance of 3 $\mu$ F, are connected in parallel. What is the equivalent capacitance of this combination? (A) 1 $\mu$ F (B) 3 $\mu$ F (C) 6 $\mu$ F (D) 9 $\mu$ F (E) 27 $\mu$ F

Free Response:

Consider the circuit below, where C1 = 2 $\mu$ F, C2 = 4 $\mu$ F, and C3 = 6 $\mu$ F. The voltage source provides 12V.

Circuit Diagram

(a) Calculate the equivalent capacitance of the circuit. (b) Determine the total charge stored in the circuit. (c) Find the voltage across capacitor C1. (d) Find the charge on capacitor C2. Answer Key:

(a) Equivalent Capacitance

C2 and C3 are in parallel, so their equivalent capacitance is: $C_{23} = C_2 + C_3 = 4 \mu F + 6 \mu F = 10 \mu F$
C1 and C23 are in series, so the total equivalent capacitance is: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{2 \mu F} + \frac{1}{10 \mu F} = \frac{6}{10 \mu F}$ $C_{eq} = \frac{10}{6} \mu F = \frac{5}{3} \mu F \approx 1.67 \mu F$

(b) Total Charge

The total charge stored in the circuit is: $Q = C_{eq}V = (\frac{5}{3} \times 10^{-6} F)(12 V) = 20 \times 10^{-6} C = 20 \mu C$

(c) Voltage Across C1

Since C1 is in series with the equivalent capacitance of C2 and C3, it carries the same charge as the total charge, i.e., 20 $\mu$ C
The voltage across C1 is: $V_1 = \frac{Q}{C_1} = \frac{20 \mu C}{2 \mu F} = 10 V$

(d) Charge on C2

The voltage across the parallel combination of C2 and C3 is: $V_{23} = V - V_1 = 12 V - 10 V = 2 V$
The charge on C2 is: $Q_2 = C_2V_{23} = (4 \mu F)(2 V) = 8 \mu C$

#RC Circuit Behavior

#Fundamental Differential Equation

Kirchhoff's loop rule gives us the fundamental equation for an RC circuit: $\mathcal{E} = \frac{dq}{dt}R + \frac{q}{C}$
- $\mathcal{E}$ is the electromotive force (voltage source).
- $R$ is the resistance.
- $C$ is the capacitance.
- $q$ is the charge on the capacitor.
- $\frac{dq}{dt}$ is the current through the resistor.

#Time Constant 🕰️

The time constant ( $\tau$ ) determines how quickly a capacitor charges or discharges.
It's calculated as: $\tau = R_{\text{eq}}C_{\text{eq}}$

Quick Fact

For a charging capacitor, $\tau$ is the time to reach about 63% of its final charge. For discharging, it's the time to drop to about 37% of its initial charge.

#Capacitor Charging

Initially, an uncharged capacitor acts like a wire, allowing easy charge flow.
As the capacitor charges:
- The charge on its plates increases.
- The current in the circuit decreases.
- The stored electric potential energy increases.

Key Concept

These values approach steady-state asymptotically. A fully charged capacitor has maximum voltage and zero current.

#Capacitor Discharging 📉

When discharging:
- Charge and stored energy decrease.
- Current decreases.

Exam Tip

After a time much greater than $\tau$ , the circuit reaches a steady state. You can often simplify the analysis at this point.

Practice Question

Multiple Choice:

An RC circuit has a resistance of 100 $\Omega$ and a capacitance of 10 $\mu$ F. What is the time constant of the circuit? (A) 1 ms (B) 10 ms (C) 100 ms (D) 1 s (E) 10 s
In an RC discharging circuit, if the initial charge on the capacitor is Q0, what is the charge on the capacitor after one time constant ( $\tau$ )? (A) 0 (B) 0.37 Q0 (C) 0.5 Q0 (D) 0.63 Q0 (E) Q0

Free Response:

A 100 $\mu$ F capacitor is charged to a potential of 50 V. It is then connected to a 200 $\Omega$ resistor, and the capacitor discharges through the resistor.

Discharging RC Circuit

(a) Calculate the time constant of the circuit. (b) Determine the initial charge on the capacitor. (c) Find the current in the circuit at t = 0. (d) Calculate the charge on the capacitor after one time constant. (e) Calculate the energy dissipated by the resistor after one time constant.

Answer Key:

(a) Time Constant

The time constant is: $\tau = RC = (200 \Omega)(100 \times 10^{-6} F) = 0.02 s$

(b) Initial Charge

The initial charge on the capacitor is: $Q_0 = CV_0 = (100 \times 10^{-6} F)(50 V) = 5 \times 10^{-3} C = 5 mC$

(c) Initial Current

The initial current is: $I_0 = \frac{V_0}{R} = \frac{50 V}{200 \Omega} = 0.25 A$

(d) Charge After One Time Constant

The charge after one time constant is: $Q(\tau) = Q_0e^{-1} = (5 \times 10^{-3} C)(e^{-1}) \approx 1.84 \times 10^{-3} C = 1.84 mC$

(e) Energy Dissipated After One Time Constant

The energy initially stored in the capacitor is: $U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(100 \times 10^{-6} F)(50 V)^2 = 0.125 J$
The energy stored in the capacitor after one time constant is: $U(\tau) = \frac{1}{2}\frac{Q(\tau)^2}{C} = \frac{1}{2}\frac{(1.84 \times 10^{-3} C)^2}{100 \times 10^{-6} F} \approx 0.0169 J$
The energy dissipated by the resistor is: $\Delta U = U_0 - U(\tau) = 0.125 J - 0.0169 J \approx 0.108 J$

#Final Exam Focus

High-Priority Topics:
- Equivalent capacitance (series and parallel combinations)
- Time constant ( $\tau = RC$ )
- Charging and discharging behavior (understanding the exponential nature)
- Applying Kirchhoff's loop rule to RC circuits
Common Question Types:
- Calculating equivalent capacitance in complex circuits.
- Determining the time constant and its significance.
- Analyzing charge, current, and voltage changes over time in RC circuits.
- Solving differential equations for RC circuits.
- Graphing charge, current, and voltage as functions of time.
Last-Minute Tips:
- Time Management: Quickly identify the type of problem and apply the relevant formulas. Don't get bogged down in lengthy calculations.
- Common Pitfalls: Watch out for unit conversions (especially $\mu$ F to F). Double-check your series and parallel calculations.
- Strategies for Challenging Questions: Draw circuit diagrams and label all known and unknown quantities. Break down complex problems into smaller, manageable steps.

Memory Aid

Remember "Q-C-V" and "I-R-V" relationships. Charge (Q) is related to Capacitance (C) and Voltage (V), and Current (I) is related to Resistance (R) and Voltage (V). Use these to navigate through the problems.

Good luck, you've got this! 💪

#AP Physics C: E&M - RC Circuits Study Guide ⚡

#Equivalent Capacitance

#Capacitor Combinations 🔌

Capacitors in circuits can be simplified into a single equivalent capacitance ( $C_{eq}$ ). This helps in analyzing complex circuits.
Series Capacitors:
- The reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances: $\frac{1}{C_{\text{eq, s}}} = \sum_{i} \frac{1}{C_i}$

Quick Fact

The equivalent capacitance of series capacitors is always smaller than the smallest individual capacitance.

Parallel Capacitors:
- The equivalent capacitance is the sum of individual capacitances: $C_{\text{eq, p}} = \sum_{i} C_i$

Key Concept

Charge Conservation: In a series connection, each capacitor holds the same magnitude of charge.

Practice Question

Multiple Choice:

Two capacitors, with capacitances of 2 $\mu$ F and 4 $\mu$ F, are connected in series. What is the equivalent capacitance of this combination? (A) 6 $\mu$ F (B) 3 $\mu$ F (C) 4/3 $\mu$ F (D) 8/3 $\mu$ F (E) 1/6 $\mu$ F
Three capacitors, each with a capacitance of 3 $\mu$ F, are connected in parallel. What is the equivalent capacitance of this combination? (A) 1 $\mu$ F (B) 3 $\mu$ F (C) 6 $\mu$ F (D) 9 $\mu$ F (E) 27 $\mu$ F

Free Response:

Consider the circuit below, where C1 = 2 $\mu$ F, C2 = 4 $\mu$ F, and C3 = 6 $\mu$ F. The voltage source provides 12V.

Circuit Diagram

(a) Equivalent Capacitance

C2 and C3 are in parallel, so their equivalent capacitance is: $C_{23} = C_2 + C_3 = 4 \mu F + 6 \mu F = 10 \mu F$
C1 and C23 are in series, so the total equivalent capacitance is: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{2 \mu F} + \frac{1}{10 \mu F} = \frac{6}{10 \mu F}$ $C_{eq} = \frac{10}{6} \mu F = \frac{5}{3} \mu F \approx 1.67 \mu F$

(b) Total Charge

The total charge stored in the circuit is: $Q = C_{eq}V = (\frac{5}{3} \times 10^{-6} F)(12 V) = 20 \times 10^{-6} C = 20 \mu C$

(c) Voltage Across C1

Since C1 is in series with the equivalent capacitance of C2 and C3, it carries the same charge as the total charge, i.e., 20 $\mu$ C
The voltage across C1 is: $V_1 = \frac{Q}{C_1} = \frac{20 \mu C}{2 \mu F} = 10 V$

(d) Charge on C2

The voltage across the parallel combination of C2 and C3 is: $V_{23} = V - V_1 = 12 V - 10 V = 2 V$
The charge on C2 is: $Q_2 = C_2V_{23} = (4 \mu F)(2 V) = 8 \mu C$

#RC Circuit Behavior

#Fundamental Differential Equation

Kirchhoff's loop rule gives us the fundamental equation for an RC circuit: $\mathcal{E} = \frac{dq}{dt}R + \frac{q}{C}$
- $\mathcal{E}$ is the electromotive force (voltage source).
- $R$ is the resistance.
- $C$ is the capacitance.
- $q$ is the charge on the capacitor.
- $\frac{dq}{dt}$ is the current through the resistor.

#Time Constant 🕰️

The time constant ( $\tau$ ) determines how quickly a capacitor charges or discharges.
It's calculated as: $\tau = R_{\text{eq}}C_{\text{eq}}$

Quick Fact

For a charging capacitor, $\tau$ is the time to reach about 63% of its final charge. For discharging, it's the time to drop to about 37% of its initial charge.

#Capacitor Charging

Initially, an uncharged capacitor acts like a wire, allowing easy charge flow.
As the capacitor charges:
- The charge on its plates increases.
- The current in the circuit decreases.
- The stored electric potential energy increases.

Key Concept

These values approach steady-state asymptotically. A fully charged capacitor has maximum voltage and zero current.

#Capacitor Discharging 📉

When discharging:
- Charge and stored energy decrease.
- Current decreases.

Exam Tip

After a time much greater than $\tau$ , the circuit reaches a steady state. You can often simplify the analysis at this point.

Practice Question

Multiple Choice:

An RC circuit has a resistance of 100 $\Omega$ and a capacitance of 10 $\mu$ F. What is the time constant of the circuit? (A) 1 ms (B) 10 ms (C) 100 ms (D) 1 s (E) 10 s
In an RC discharging circuit, if the initial charge on the capacitor is Q0, what is the charge on the capacitor after one time constant ( $\tau$ )? (A) 0 (B) 0.37 Q0 (C) 0.5 Q0 (D) 0.63 Q0 (E) Q0

Free Response:

A 100 $\mu$ F capacitor is charged to a potential of 50 V. It is then connected to a 200 $\Omega$ resistor, and the capacitor discharges through the resistor.

Discharging RC Circuit

Answer Key:

(a) Time Constant

The time constant is: $\tau = RC = (200 \Omega)(100 \times 10^{-6} F) = 0.02 s$

(b) Initial Charge

The initial charge on the capacitor is: $Q_0 = CV_0 = (100 \times 10^{-6} F)(50 V) = 5 \times 10^{-3} C = 5 mC$

(c) Initial Current

The initial current is: $I_0 = \frac{V_0}{R} = \frac{50 V}{200 \Omega} = 0.25 A$

(d) Charge After One Time Constant

The charge after one time constant is: $Q(\tau) = Q_0e^{-1} = (5 \times 10^{-3} C)(e^{-1}) \approx 1.84 \times 10^{-3} C = 1.84 mC$

(e) Energy Dissipated After One Time Constant

The energy initially stored in the capacitor is: $U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(100 \times 10^{-6} F)(50 V)^2 = 0.125 J$
The energy stored in the capacitor after one time constant is: $U(\tau) = \frac{1}{2}\frac{Q(\tau)^2}{C} = \frac{1}{2}\frac{(1.84 \times 10^{-3} C)^2}{100 \times 10^{-6} F} \approx 0.0169 J$
The energy dissipated by the resistor is: $\Delta U = U_0 - U(\tau) = 0.125 J - 0.0169 J \approx 0.108 J$

#Final Exam Focus

High-Priority Topics:
- Equivalent capacitance (series and parallel combinations)
- Time constant ( $\tau = RC$ )
- Charging and discharging behavior (understanding the exponential nature)
- Applying Kirchhoff's loop rule to RC circuits
Common Question Types:
- Calculating equivalent capacitance in complex circuits.
- Determining the time constant and its significance.
- Analyzing charge, current, and voltage changes over time in RC circuits.
- Solving differential equations for RC circuits.
- Graphing charge, current, and voltage as functions of time.
Last-Minute Tips:
- Time Management: Quickly identify the type of problem and apply the relevant formulas. Don't get bogged down in lengthy calculations.
- Common Pitfalls: Watch out for unit conversions (especially $\mu$ F to F). Double-check your series and parallel calculations.
- Strategies for Challenging Questions: Draw circuit diagrams and label all known and unknown quantities. Break down complex problems into smaller, manageable steps.

Memory Aid

Good luck, you've got this! 💪

Resistor-Capacitor (RC) Circuits

Mia Gonzalez

#AP Physics C: E&M - RC Circuits Study Guide ⚡

#Equivalent Capacitance

#Capacitor Combinations 🔌

#RC Circuit Behavior

#Fundamental Differential Equation

#Time Constant 🕰️

#Capacitor Charging

#Capacitor Discharging 📉

#Final Exam Focus

Explore more resources

How are we doing?

Resistor-Capacitor (RC) Circuits

Mia Gonzalez

#AP Physics C: E&M - RC Circuits Study Guide ⚡

#Equivalent Capacitance

#Capacitor Combinations 🔌

#RC Circuit Behavior

#Fundamental Differential Equation

#Time Constant 🕰️

#Capacitor Charging

#Capacitor Discharging 📉

#Final Exam Focus

Explore more resources

How are we doing?