#AP Physics C: E&M - Capacitors: The Night Before ⚡

Hey! Let's get you prepped for the exam! Remember, you've got this. We're going to make sure you're feeling confident and ready to tackle those capacitor questions. Let’s dive in!

Remember, circuits with capacitors are very common on the FRQs. This is a high-value topic, so let's make sure you're solid on it!

#Capacitors in Series & Parallel 🔋

Just like resistors, capacitors can be combined in series and parallel, but their rules are flipped! Let's break it down:

#Parallel Capacitors

• In parallel, capacitors act like one big capacitor storing a large charge. Think of it as multiple lanes on a highway, all contributing to the total traffic flow.
• Total Charge: $Q_{total} = Q_1 + Q_2 + Q_3 + ...$
• Total Capacitance: $C_{total} = C_1 + C_2 + C_3 + ...$ (simply add them up!)

#Series Capacitors

• In series, the charge is the same on each capacitor, but the voltage is split. Think of it as a single-lane road with multiple toll booths.
• Total Charge: $Q_{total} = Q_1 = Q_2 = Q_3 = ...$
• Total Capacitance: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...$ (reciprocal addition, just like resistors in parallel!)

#Steady State Behavior 🎯

• Key Concept: In a DC circuit, a capacitor charges until its voltage equals the source voltage. Once charged, it acts like an open switch (no current flows through it).
• Initial State: When a capacitor is uncharged, current flows freely.
• Steady State: Once fully charged, no current flows through the capacitor. It's like a filled water tank – no more water can enter.

#RC Circuits 🏍️

RC circuits involve a resistor and capacitor in series. They're all about how quickly a capacitor charges or discharges, and this is controlled by the time constant.

#Charging a Capacitor

• Charge as a Function of Time: $Q(t) = Q_0(1 - e^{-t/RC})$
• Voltage as a Function of Time: $V(t) = V_0(1 - e^{-t/RC})$
• Current as a Function of Time: $I(t) = I_0e^{-t/RC}$

#Discharging a Capacitor

• Charge as a Function of Time: $Q(t) = Q_0e^{-t/RC}$
• Voltage as a Function of Time: $V(t) = V_0e^{-t/RC}$
• Current as a Function of Time: $I(t) = -I_0e^{-t/RC}$ (note the negative sign, indicating current is flowing in the opposite direction)

#Time Constant (τ)

• Definition: $\tau = RC$ (the time it takes for a capacitor to charge to about 63% of its max voltage or discharge to about 37% of its initial voltage).
• Importance: A larger time constant means slower charging/discharging. This is why RC circuits are used in filters and timing circuits.

#Final Exam Focus 🎯

• High-Priority Topics: Capacitor combinations (series/parallel), steady-state analysis, RC circuits (charging/discharging), time constant.
• Common Question Types:
  • Calculating equivalent capacitance.
  • Analyzing circuits at steady state.
  • Determining charge, voltage, and current as a function of time in RC circuits.
  • Graphing voltage and current in RC circuits.
• Time Management:
  • Quickly identify series and parallel combinations.
  • Focus on the initial and steady-state conditions first.
  • Use the time constant to estimate charging/discharging times.
• Common Pitfalls:
  • Mixing up series and parallel rules for capacitors.
  • Forgetting that capacitors block DC current at steady state.
  • Incorrectly applying the charging/discharging equations.

#Practice Questions

Practice Question

Multiple Choice Questions

1. A capacitor is charged to a voltage $V_0$ and then discharged through a resistor. Which of the following graphs best represents the voltage across the capacitor as a function of time? (A) A straight line with a negative slope (B) An exponential decay curve (C) A straight line with a positive slope (D) An exponential growth curve

2. Two capacitors, $C_1$ and $C_2$, are connected in series. If $C_1 = 2C_2$, what is the equivalent capacitance of the combination? (A) $\frac{2}{3}C_2$ (B) $\frac{3}{2}C_2$ (C) $2C_2$ (D) $3C_2$

3. In an RC circuit, if the resistance is doubled and the capacitance is halved, how does the time constant change? (A) It doubles (B) It halves (C) It remains the same (D) It quadruples

Free Response Question

Consider the following circuit:

A 20 V battery is connected to a circuit containing a 10 Ω resistor ($R_1$), a 20 Ω resistor ($R_2$), and a 5 μF capacitor ($C$). Initially, the capacitor is uncharged. The switch is closed at time t = 0. (a) What is the voltage across $R_1$ immediately after the switch is closed?

(b) What is the voltage across $R_2$ at steady state?

(c) What is the current through $R_2$ at steady state?

(d) Sketch a graph of the voltage across the capacitor as a function of time.

(e) Sketch a graph of the current through $R_1$ as a function of time.

(f) If the value of $R_2$ is decreased, how does the energy stored in the capacitor at steady state change? Explain your reasoning.

FRQ Scoring Rubric

(a) 2 points

• 1 point for stating that the voltage across the capacitor is zero at t=0
• 1 point for correctly calculating $V_{R1} = 20V$

(b) 2 points

• 1 point for recognizing that at steady state, the capacitor is fully charged and no current flows through it
• 1 point for correctly calculating $V_{R2} = 8V$

(c) 2 points

• 1 point for using Ohm's law to calculate the current
• 1 point for correctly calculating $I_{R2} = 0.4A$

(d) 2 points

• 1 point for showing an increasing curve
• 1 point for reaching a maximum voltage of 12V

(e) 2 points

• 1 point for showing a decreasing curve
• 1 point for starting at a maximum current and approaching zero

(f) 3 points

• 1 point for stating that the energy stored in the capacitor will increase
• 1 point for explaining that decreasing $R_2$ increases the current in the circuit
• 1 point for explaining that increased current increases voltage across $R_1$ and $C$, thus increasing the energy stored in the capacitor

Remember, you've got the knowledge and the skills to ace this exam. Stay calm, focused, and confident. You've got this!