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

Hey there, future physics master! Let's break down compound DC circuits and make sure you're totally ready for the exam. This guide is designed to be your go-to resource the night before the test, so let's make every minute count!

#Compound Circuits: Series, Parallel, and Beyond

#Equivalent Resistance in Circuits

#Series and Parallel Connections

• Series Connection: Think of it like a single lane road. All the charge (current) has to go through each resistor one after the other. There are no alternate paths.

  • Current (I) is the same through all components.
  • Total voltage is divided among the resistors.

• Parallel Connection: Imagine a multi-lane highway. Charge has multiple paths to flow.

  • Voltage (ΔV) is the same across all components.
  • Total current is divided among the branches.

#Equivalent Resistance Calculations

• Equivalent Resistance (R_eq): We simplify a group of resistors into one effective resistor. This makes circuit analysis much easier.

• Series Resistors: Add them up directly. It's that easy!

  $$R_{\text{eq, s}} = \sum_{i} R_{i} = R_1 + R_2 + R_3 + ...$$

• Parallel Resistors: Use the reciprocal method. It's a bit trickier, but you've got this!

  $$\frac{1}{R_{\text{eq, p}}} = \sum_{i} \frac{1}{R_{i}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$$

• Key Insight: Adding resistors in parallel decreases the overall equivalent resistance. More paths for current mean less resistance overall. 💡

Practice Question

Multiple Choice Questions

1. Two resistors, one with a resistance of 10 Ω and the other with a resistance of 20 Ω, are connected in parallel. What is the equivalent resistance of this combination?

   (A) 30 Ω (B) 15 Ω (C) 6.67 Ω (D) 0.15 Ω

2. Three identical resistors are connected in series. If the equivalent resistance of the combination is 30 Ω, what is the resistance of each individual resistor?

   (A) 10 Ω (B) 30 Ω (C) 60 Ω (D) 90 Ω

Free Response Question

A circuit consists of a 12 V battery with an internal resistance of 1 Ω connected to two resistors, R1 = 4 Ω and R2 = 6 Ω, in series.

(a) Draw a schematic diagram of the circuit, including the internal resistance of the battery.

(b) Calculate the total equivalent resistance of the circuit.

(c) Determine the current flowing through the circuit.

(d) Calculate the voltage drop across each resistor, R1 and R2. (e) Calculate the terminal voltage of the battery.

Answer Key

Multiple Choice Questions

1. (C) 6.67 Ω
2. (A) 10 Ω

Free Response Question

(a) Diagram should show a battery with an internal resistance in series with two resistors R1 and R2. (b) R_eq = 1 Ω + 4 Ω + 6 Ω = 11 Ω (2 points)

(c) I = V / R_eq = 12 V / 11 Ω = 1.09 A (2 points)

(d) V1 = I * R1 = 1.09 A * 4 Ω = 4.36 V, V2 = I * R2 = 1.09 A * 6 Ω = 6.54 V (2 points)

(e) V_terminal = V - Ir = 12 V - (1.09 A * 1 Ω) = 10.91 V (2 points)

#Circuits with Resistive Wires and Batteries

#Ideal vs. Nonideal Components

• Ideal Components: In an ideal world, wires have zero resistance, and batteries have no internal resistance. This makes calculations easier but isn't realistic.

• Real-World Components: Real wires have some resistance (though it's usually negligible), and real batteries have internal resistance that affects their performance.

• Electromotive Force (ε): This is the potential difference a battery provides when no current is flowing. It's the battery's ideal voltage.

#Internal Resistance Effects

• Nonideal Battery: A nonideal battery acts like an ideal battery in series with a small resistor (its internal resistance, r). 🔋

• Terminal Voltage: When current flows, the actual voltage you measure across the battery's terminals is less than its emf due to the voltage drop across its internal resistance.

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

  • ε is the electromotive force (ideal voltage).
  • I is the current flowing through the battery.
  • r is the internal resistance of the battery.

#Terminal Voltage Equation

• This equation shows how the battery's terminal voltage decreases as current increases due to internal resistance. It’s a key concept to understand how real batteries behave.

Practice Question

Multiple Choice Questions

1. A battery with an EMF of 10 V and an internal resistance of 0.5 Ω is connected to a 4.5 Ω resistor. What is the current flowing in the circuit?

   (A) 1 A (B) 2 A (C) 2.2 A (D) 2.5 A

2. A battery has an EMF of 12 V and a terminal voltage of 11 V when delivering a current of 2 A. What is the internal resistance of the battery?

   (A) 0.5 Ω (B) 1 Ω (C) 2 Ω (D) 6 Ω

Free Response Question

A 9 V battery with an internal resistance of 0.8 Ω is connected to a 2 Ω resistor.

(a) Draw a circuit diagram showing the battery, its internal resistance, and the external resistor.

(b) Calculate the current flowing through the circuit.

(c) Calculate the terminal voltage of the battery.

(d) Calculate the power dissipated in the external resistor.

(e) Calculate the power dissipated in the internal resistance of the battery.

Answer Key

Multiple Choice Questions

1. (B) 2 A
2. (A) 0.5 Ω

Free Response Question

(a) Diagram should show a battery with an internal resistance in series with an external resistor.

(b) I = V / (R + r) = 9 V / (2 Ω + 0.8 Ω) = 3.21 A (2 points)

(c) V_terminal = V - Ir = 9 V - (3.21 A * 0.8 Ω) = 6.43 V (2 points)

(d) P_resistor = I^2 * R = (3.21 A)^2 * 2 Ω = 20.67 W (2 points)

(e) P_internal = I^2 * r = (3.21 A)^2 * 0.8 Ω = 8.27 W (2 points)

#Current and Voltage Measurement

#Ammeter Usage and Placement

• Ammeter: Measures current (the flow of charge) in a circuit.

• Placement: Always connect an ammeter in series with the component you want to measure. Think of it as a toll booth on the road – all the current has to go through it.

• Ideal Ammeter: Has zero resistance so it doesn't affect the circuit's current. In reality, it has very low resistance.

#Voltmeter Usage and Placement

• Voltmeter: Measures the potential difference (voltage) between two points in a circuit. 📏

• Placement: Connect a voltmeter in parallel with the component you want to measure. It's like measuring the height difference between two points – you measure it alongside, not in the path.

• Ideal Voltmeter: Has infinite resistance so it doesn't draw any current from the circuit. In reality, it has very high resistance.

#Effects of Nonideal Measuring Devices

• Nonideal Ammeter: Its resistance increases the total series resistance, potentially reducing the current in the circuit.

• Nonideal Voltmeter: Its finite resistance can draw some current, altering the voltage reading, especially in circuits with high resistance.

Practice Question

Multiple Choice Questions

1. An ammeter is connected in a circuit to measure the current through a resistor. How should the ammeter be connected in relation to the resistor?

   (A) In parallel (B) In series (C) Neither series nor parallel (D) It doesn't matter

2. A voltmeter is connected in a circuit to measure the voltage across a resistor. How should the voltmeter be connected in relation to the resistor?

   (A) In series (B) In parallel (C) Neither series nor parallel (D) It doesn't matter

Free Response Question

A circuit consists of a 10 V battery and two resistors, R1 = 5 Ω and R2 = 10 Ω, connected in series. An ammeter and a voltmeter are used to measure the current through and voltage across R2 respectively.

(a) Draw a circuit diagram showing the battery, resistors, ammeter, and voltmeter in their correct positions.

(b) Calculate the ideal current through R2. (c) Calculate the ideal voltage across R2. (d) Explain how a non-ideal ammeter with a small resistance would affect the current measurement.

(e) Explain how a non-ideal voltmeter with a finite resistance would affect the voltage measurement.

Answer Key

Multiple Choice Questions

1. (B) In series
2. (B) In parallel

Free Response Question

(a) Diagram should show a battery in series with R1 and R2. Ammeter should be in series with R2, and voltmeter should be in parallel with R2. (2 points)

(b) I = V / (R1 + R2) = 10 V / (5 Ω + 10 Ω) = 0.67 A (2 points)

(c) V2 = I * R2 = 0.67 A * 10 Ω = 6.67 V (2 points)

(d) A non-ideal ammeter with small resistance would slightly decrease the total resistance of the circuit, leading to a slightly higher current reading than the ideal current. (2 points)

(e) A non-ideal voltmeter with finite resistance would draw some current, decreasing the measured voltage across R2. (2 points)

#Final Exam Focus

#High-Priority Topics

• Equivalent Resistance: Master series and parallel combinations. This is fundamental!
• Internal Resistance: Understand how it affects terminal voltage. It's often a source of error if not handled carefully.
• Measurement Devices: Know how to use ammeters and voltmeters correctly. Their placement is crucial.

#Common Question Types

• Circuit Analysis: Expect questions that require you to calculate equivalent resistance, current, voltage, and power in compound circuits.
• Conceptual Questions: Be ready to explain how nonideal components affect circuit behavior and measurements.
• Diagram Interpretation: Practice drawing and interpreting circuit diagrams, including the placement of measuring devices.

#Last-Minute Tips

• Time Management: Don't spend too long on any one question. Move on and come back if you have time.
• Units: Always include units in your calculations and answers. It's an easy way to avoid losing points.
• Diagrams: Draw clear, labeled circuit diagrams. They can help you visualize the problem and earn you partial credit even if you get the final answer wrong.
• Stay Calm: Take deep breaths and trust your preparation. You've got this!

Good luck, and go ace that exam! You're going to do great! 🎉