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  2. AP Physics 2
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Resistivity and Resistance

Elijah Ramirez

Elijah Ramirez

9 min read

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Study Guide Overview

This study guide covers current, resistance, and Ohm's Law, including current density, resistivity, resistors in series and parallel, and Ohmic vs. non-Ohmic devices. It also explains power and energy in circuits, focusing on calculating electrical power. Finally, it discusses non-ideal batteries and EMF, emphasizing the relationship between EMF, terminal voltage, and internal resistance. The guide includes practice questions and a final exam focus section with high-priority topics, common question types, and last-minute tips.

#AP Physics 2: Electricity & Circuits - Night Before Review ⚡

Hey! Let's get you prepped for the exam. We're going to break down the key concepts, hit the high-yield stuff, and make sure you're feeling confident. Let's do this!

#Current, Resistance, and Ohm's Law

#Current Density

Key Concept

Current density (J) is like a microscopic view of current, showing how much charge is flowing through a specific area. It's a vector, meaning it has both magnitude and direction. It's related to the electric field (E) by the equation: J=EρJ = \frac{E}{\rho}J=ρE​ where ρ is the resistivity of the material. Think of resistivity as how much a material fights against the flow of current.

Quick Fact
  • Current Density (J): Current per unit area (A/m²)
    • Resistivity (ρ): Material's opposition to current (Ω⋅m)

Current Density

Image: Relationship between current density (J) and electric field (E)

#Resistance

Key Concept

Resistance (R) is the overall opposition to current flow in a specific object or component, taking into account its material (resistivity), length (L), and cross-sectional area (A). It's calculated as: R=ρLAR = \frac{\rho L}{A}R=AρL​ Think of it like this: a longer wire (larger L) is harder for current to flow through (higher resistance), and a thicker wire (larger A) is easier (lower resistance).

Quick Fact
  • Resistance (R): Opposition to current (Ω)
    • Longer wire = Higher R
    • Thicker wire = Lower R

Resistance

Image: Resistance depends on material, length, and area

Exam Tip

Remember: Resistivity (ρ) is a material property, while Resistance (R) is a property of a specific object. Don't mix them up!

#Resistors in Series and Parallel

Memory Aid

Series: Think of a single lane of traffic – all cars (current) must go through each resistor. Total resistance adds up. Parallel: Think of multiple lanes – current can split up. Total resistance goes down.

  • Series Resistors (Rₛ): Rs=R1+R2+R3+...R_s = R_1 + R_2 + R_3 + ...Rs​=R1​+R2​+R3​+...

  • Parallel Resistors (Rₚ): 1Rp=1R1+1R2+1R3+...\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...Rp​1​=R1​1​+R2​1​+R3​1​+...

Resistors

Image: Series and parallel resistor configurations

#Ohm's Law 💡

Key Concept

Ohm's Law is the cornerstone of circuit analysis, relating voltage (V), current (I), and resistance (R): V=IRV = IRV=IR It can be rearranged to solve for current (I=V/RI = V/RI=V/R) or resistance (R=V/IR = V/IR=V/I).

Quick Fact
  • Ohm's Law: V = IR
    • Ohmic Device: Linear V-I graph
    • Non-Ohmic Device: Non-linear V-I graph

Ohms Law

Image: Different forms of Ohm's law

Ohmic vs Non-Ohmic

Image: Graphical representation of Ohmic and non-Ohmic behavior

Exam Tip

Be ready to identify Ohmic vs. non-Ohmic behavior from V-I graphs. Ohmic materials have a straight line relationship.

Practice Question

  1. A wire has a length L and a cross-sectional area A. If the length of the wire is doubled and the cross-sectional area is halved, what happens to the resistance of the wire?

    (A) The resistance is unchanged. (B) The resistance is doubled. (C) The resistance is quadrupled. (D) The resistance is halved.

    Answer: (C) The resistance is quadrupled. Because resistance is proportional to length and inversely proportional to area, doubling the length doubles the resistance, and halving the area doubles it again.

#Power and Energy in Circuits

#Electrical Power

Key Concept

Power (P) is the rate at which electrical energy is converted into other forms. The most common equation is: P=IVP = IVP=IV Using Ohm's Law, we can also express power as: P=I2R=V2RP = I^2R = \frac{V^2}{R}P=I2R=RV2​ Choose the equation that uses the information given in the problem.

Quick Fact
  • Power (P): Rate of energy conversion (Watts)
    • P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}P=IV=I2R=RV2​

Power Equations

Image: Power equations in circuits

Power Equations

Image: Alternative power equations based on Ohm's law

Practice Question

  1. A hair dryer is rated at 1200W when connected to 120V. What is the resistance of the dryer?

    Answer: P=V2RP = \frac{V^2}{R}P=RV2​ R=V2P=(120V)21200W=12ΩR = \frac{V^2}{P} = \frac{(120V)^2}{1200W} = 12 \OmegaR=PV2​=1200W(120V)2​=12Ω

#Non-Ideal Batteries and EMF

#Electromotive Force (EMF)

Key Concept

Electromotive force (EMF or ε) is the total energy a battery can provide per unit charge. It's the voltage before any current flows. Real batteries have internal resistance (r), which causes the terminal voltage (Vₜ) to be less than the EMF when current flows. The relationship is: ϵ=VT+Ir\epsilon = V_T + Irϵ=VT​+Ir

Quick Fact
  • EMF (ε): Total voltage provided by a battery
    • Internal Resistance (r): Resistance within the battery
    • Terminal Voltage (Vₜ): Actual voltage available in a circuit
    • ϵ=VT+Ir\epsilon = V_T + Irϵ=VT​+Ir

EMF

Image: EMF and internal resistance of a battery

Common Mistake

Don't confuse EMF with terminal voltage. EMF is the ideal voltage, while terminal voltage is the actual voltage available when current is flowing. The difference is due to the internal resistance of the battery.

#Final Exam Focus

#High-Priority Topics

  • Ohm's Law and Circuit Analysis: Be able to apply Ohm's law in various circuit configurations (series, parallel, and combinations).
  • Power Calculations: Know how to calculate power using different forms of the power equation. Be able to calculate energy dissipated in a resistor.
  • Resistors in Series and Parallel: Master the rules for combining resistors. This is a must for circuit problems.
  • EMF and Internal Resistance: Understand how internal resistance affects the terminal voltage of a battery.
  • Conceptual Understanding: Focus on understanding the relationship between current, voltage, resistance, and power. Be able to explain these concepts.

#Common Question Types

  • Circuit Analysis: Solving for current, voltage, and resistance in complex circuits.
  • Graphical Analysis: Interpreting V-I graphs to determine if a device is Ohmic or non-Ohmic.
  • Conceptual Questions: Explaining the physical meaning of resistivity, resistance, and EMF.
  • Multi-Concept Problems: Combining concepts from different units (e.g., circuits and energy).

#Last-Minute Tips

  • Time Management: Don't spend too long on a single question. Move on and come back if time permits.
  • Units: Always include units in your calculations and answers.
  • Diagrams: Draw circuit diagrams to visualize the problem. It helps a lot!
  • Practice: Review practice problems from the textbook and past AP exams. This is the best way to prepare.
  • Stay Calm: Take a deep breath. You've got this!
Practice Question

Free Response Question

Consider the circuit below, consisting of a battery with an EMF of 12V and an internal resistance of 1Ω, connected to two resistors, R1 = 5Ω and R2 = 10Ω, in parallel.

Circuit Diagram

(a) Calculate the equivalent resistance of the parallel combination of R1 and R2. (b) Calculate the total current flowing through the battery.

(c) Calculate the terminal voltage of the battery.

(d) Calculate the power dissipated in resistor R1. Answer:

(a) Equivalent resistance of parallel resistors: 1Req=1R1+1R2=15Ω+110Ω=310Ω\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5\Omega} + \frac{1}{10\Omega} = \frac{3}{10\Omega}Req​1​=R1​1​+R2​1​=5Ω1​+10Ω1​=10Ω3​ Req=103Ω≈3.33ΩR_{eq} = \frac{10}{3} \Omega \approx 3.33 \OmegaReq​=310​Ω≈3.33Ω

(b) Total current flowing through the battery: Total resistance in the circuit is the equivalent resistance of the parallel resistors plus the internal resistance of the battery. Rtotal=Req+r=3.33Ω+1Ω=4.33ΩR_{total} = R_{eq} + r = 3.33\Omega + 1\Omega = 4.33\OmegaRtotal​=Req​+r=3.33Ω+1Ω=4.33Ω I=ϵRtotal=12V4.33Ω≈2.77AI = \frac{\epsilon}{R_{total}} = \frac{12V}{4.33\Omega} \approx 2.77AI=Rtotal​ϵ​=4.33Ω12V​≈2.77A

(c) Terminal voltage of the battery: VT=ϵ−Ir=12V−(2.77A)(1Ω)=9.23VV_T = \epsilon - Ir = 12V - (2.77A)(1\Omega) = 9.23VVT​=ϵ−Ir=12V−(2.77A)(1Ω)=9.23V

(d) Power dissipated in resistor R1: First, find the voltage across the parallel resistors. This is the same as the terminal voltage of the battery, since the internal resistance is in series with the rest of the circuit. VR1=VT=9.23VV_{R1} = V_T = 9.23VVR1​=VT​=9.23V PR1=VR12R1=(9.23V)25Ω≈17.0WP_{R1} = \frac{V_{R1}^2}{R_1} = \frac{(9.23V)^2}{5\Omega} \approx 17.0 WPR1​=R1​VR12​​=5Ω(9.23V)2​≈17.0W

Scoring Rubric:

(a) 2 points

  • 1 point for correct formula for parallel resistance
  • 1 point for correct calculation

(b) 3 points

  • 1 point for calculating total resistance
  • 1 point for using Ohm's law
  • 1 point for correct calculation

(c) 2 points

  • 1 point for correct formula for terminal voltage
  • 1 point for correct calculation

(d) 3 points

  • 1 point for recognizing the voltage across parallel resistors
  • 1 point for using the correct power equation
  • 1 point for correct calculation

Good luck! You've got this! 💪

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Question 1 of 13

What are the units of current density (J)? ⚡

Ω⋅m

A/m²

V/m

Ω