#AP Physics C: E&M - Unit 3: Circuits - The Night Before ⚡

Hey! Let's get you prepped for the exam with a high-impact review of circuits. This is your go-to guide for a confident test day. Let's make it click!

#3.0: Circuit Overview 🌐

Unit 3 focuses on connecting electrical devices, analyzing current flow, and understanding power and potential differences in circuits. It is a high-value topic, making up 17-23% of the AP exam. This unit is crucial for understanding how electrical components interact. Remember to review your AP Physics 1 notes on circuits if you need a refresher!

#3.1: Circuit Quantities 🔋

#Voltage, Current, and Resistance: The Big Three

Remember from Unit 1, voltage is work per unit charge. Now, let's add current and resistance to the mix. Think of it like water flowing through a hose:

Voltage (V): Water pressure 💧
Current (I): Amount of water flowing 💨
Resistance (R): Clogs or obstructions in the hose 🚧

Memory Aid

Use the water hose analogy to remember the relationship between voltage, current, and resistance. Voltage is the 'push', current is the 'flow', and resistance is what impedes the flow.

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Image from freeingenergy.com

#Current (I) 💨

Current is the rate of charge flow, measured in Amperes (A). The equation is:

$I = \frac{dQ}{dt}$

Where:

I is current (Amps)
Q is charge (Coulombs)
t is time (seconds)

Key Concept

Conventional Current: Always consider the direction of positive charge flow. Even though electrons are the mobile charge carriers, we use the positive direction for consistency with electric fields and potential differences.

#Microscopic View of Current

Current is also related to the drift velocity (v_d) of charge carriers. Imagine electrons slowly drifting through a wire, not zooming!

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Image from openstax.org

The equation linking current to drift velocity is:

$I = nAv_dq$

Where:

n is the number of charge carriers per unit volume
A is the cross-sectional area of the wire
v_d is the drift velocity
q is the charge of each carrier

#Current Density (J)

Current Density is a vector quantity that relates the current to the electric field (E) and material properties. It's given by:

$J = \frac{I}{A} = \sigma E$

Where:

J is the current density
σ is the conductivity (1/ρ)
E is the electric field

Quick Fact

Resistivity (ρ): A material property that describes how much it resists current flow. Higher temperature = higher resistivity (usually).

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Data from NYS Physics Reference Tables

#Resistance (R) 🛑

Resistance is the opposition to current flow in a circuit. It depends on the material's resistivity (ρ), length (L), and cross-sectional area (A):

$R = \rho \frac{L}{A}$

Quick Fact

Resistance is measured in Ohms (Ω). 1 Ω = 1 V/A

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#Ohm's Law 💡

Ohm's Law relates voltage, current, and resistance:

$V = IR$ or $I = \frac{V}{R}$ or $R = \frac{V}{I}$

It's all the same equation, just rearranged!

Memory Aid

Use the VIR triangle to easily remember Ohm's Law. Cover the variable you want to find, and the remaining two will show the operation (V=IR, I=V/R, R=V/I).

#Ohmic vs. Non-Ohmic Devices

Ohmic: Constant resistance, linear V-I graph.
Non-Ohmic: Resistance changes with voltage/current, non-linear V-I graph.

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Image from www.menihek.ca

#Power (P)

Electric Power is the rate at which electrical energy is used. The main equation is:

$P = IV$

Using Ohm's Law, we can also write:

$P = I^2R = \frac{V^2}{R}$

Quick Fact

Power is measured in Watts (W).

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#Circuit Symbols & Measuring Tools 🛠️

Here are some common circuit symbols. Don't worry about the ones not used in AP Physics C: E&M.

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Image from wikimedia.org

#Measuring Tools

Voltmeter: Measures potential difference (voltage) in parallel. High internal resistance.
Ammeter: Measures current in series. Low internal resistance.

Exam Tip

Always connect voltmeters in parallel and ammeters in series. Incorrect connections can cause short circuits or faulty readings.

#Series vs. Parallel Circuits

Series: Single path for current. Components are in a line.
Parallel: Multiple paths for current. Current divides between paths.

Common Mistake

Confusing series and parallel connections is a common error. Remember: series is a single path, parallel has multiple paths.

#Final Exam Focus 🎯

#High-Priority Topics:

Ohm's Law and its applications
Calculating resistance, current, and voltage in circuits
Understanding series and parallel connections
Power calculations
Interpreting V-I graphs for Ohmic and non-Ohmic devices

#Common Question Types:

Multiple Choice: Conceptual questions on current, resistance, and circuit analysis.
Free Response: Circuit analysis problems, calculating equivalent resistance, power dissipation, and interpreting graphs.

#Last-Minute Tips:

Time Management: Quickly identify the core concepts and equations needed for each problem.
Common Pitfalls: Be careful with units (amps, volts, ohms), and always double-check your calculations.
Strategies: Draw circuit diagrams clearly, label known and unknown values, and use the correct formulas.

#Practice Questions

Practice Question

#Multiple Choice Questions:

A cylindrical wire has a length L and a radius r. If both the length and radius are doubled, how does the resistance of the wire change?

(A) It is reduced to one-fourth. (B) It is reduced to one-half. (C) It remains the same. (D) It is doubled.
A circuit contains a battery and a resistor. If the voltage of the battery is doubled, what happens to the current through the resistor?

(A) It is reduced to one-fourth. (B) It is reduced to one-half. (C) It remains the same. (D) It is doubled.
A resistor is connected to a battery. If a second identical resistor is added in series, what happens to the total current through the circuit?

(A) It is reduced to one-fourth. (B) It is reduced to one-half. (C) It remains the same. (D) It is doubled.

#Free Response Question:

A circuit consists of a 12 V battery connected to three resistors: R1 = 10 Ω, R2 = 20 Ω, and R3 = 30 Ω. R1 and R2 are in series, and R3 is in parallel with the series combination of R1 and R2. (a) Draw a diagram of the circuit.

(b) Calculate the equivalent resistance of the circuit.

(d) Calculate the current through R3. (e) Calculate the power dissipated by R2. ### Answers:

#Multiple Choice Answers:

(B) The resistance of a wire is given by $R = \rho \frac{L}{A}$ . If both L and r are doubled, the new resistance is $R' = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4\pi r^2} = \frac{1}{2} \rho \frac{L}{\pi r^2} = \frac{1}{2}R$ . Thus, the resistance is reduced to one-half.
(D) According to Ohm's Law, $V = IR$ . If the voltage is doubled, the current through the resistor will also double, assuming the resistance remains constant.
(B) When an identical resistor is added in series, the total resistance doubles. According to Ohm's Law, the current is inversely proportional to the resistance, so the total current through the circuit is halved.

#Free Response Answer:

(a) Circuit Diagram:

   +----[R1]----+----[R2]----+
   |            |            |
12V |            +----[R3]----+
   |                           |
   +---------------------------+

(b) Equivalent Resistance (Req):

R1 and R2 are in series: R12 = R1 + R2 = 10 Ω + 20 Ω = 30 Ω
R12 and R3 are in parallel: $\frac{1}{R_{eq}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30}$
R_eq = 15 Ω

(c) Total Current (I_total):

Using Ohm's Law: $I_{total} = \frac{V}{R_{eq}} = \frac{12}{15} = 0.8 A$

(d) Current through R3 (I3):

The voltage across R3 is the same as the battery voltage: V3 = 12 V
Using Ohm's Law: $I_3 = \frac{V_3}{R_3} = \frac{12}{30} = 0.4 A$

(e) Power dissipated by R2 (P2):

The current through R1 and R2 (I12) is the total current minus the current through R3: I12 = I_total - I3 = 0.8A - 0.4A = 0.4 A
Power dissipated by R2: $P_2 = I_{12}^2 R_2 = (0.4)^2 * 20 = 3.2 W$