# Current and Resistance

> AP Physics C: Electricity and Magnetism · Unit 3: Electric Circuits
> Source: https://www.owlsprep.com/study/ap-physics-c-em-u3-current-and-resistance/

This module covers core definitions of electric current, resistance, and resistivity, connects microscopic conductor properties to macroscopic circuit behavior, applies Ohm's law, temperature dependence, and calculates Joule heating power dissipation for AP Physics C: E&M.

**Prerequisites:** Electric potential and potential difference across conductors; Conservation of charge; Electric fields in conductive materials

## Learning objectives

- Define electric current, current density, resistance, and resistivity
- Relate macroscopic current to microscopic drift velocity of charge carriers
- Calculate resistance of a conductor from material resistivity and geometry
- Distinguish Ohm's law from the general definition of resistance
- Calculate temperature-dependent resistance of a conductor
- Calculate power dissipation in resistive materials via Joule heating

## Electric Current, Drift Velocity, and Current Density

Electric current is the foundational quantity for all circuit analysis, describing net charge flow through a cross-sectional surface. AP Physics C exclusively uses **conventional current** convention, where current direction matches the flow of positive charge, opposite to the direction of electron flow in metallic conductors.

**Electric Current** — The rate of net charge flow through a given cross-sectional surface

*Notation:* $I$

*Example:* A current of 1 ampere equals 1 coulomb of charge passing through the surface per second.

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

For uniform current flow through a conductor of constant cross-sectional area $A$, we define current density $\vec{J}$, a vector pointing in the direction of conventional current. Microscopically, when an electric field is applied to a conductor, free charges accelerate then scatter off the crystal lattice, resulting in a net average velocity called drift velocity $v_d$. For a material with $n$ charge carriers per unit volume, each of charge $q$, we derive the relation:

$$J = n q v_d \implies I = n q v_d A$$

**Worked example:** A copper wire with cross-sectional area $2.0 \times 10^{-6}\ \text{m}^2$ carries a current of $10\ \text{A}$. Copper has $n = 8.5 \times 10^{28}$ free electrons per cubic meter. Find the magnitude of drift velocity and the current density in the wire.

1. Calculate current density for uniform flow, which is current divided by cross-sectional area:

   $$J = \frac{I}{A} = \frac{10\ \text{A}}{2.0 \times 10^{-6}\ \text{m}^2} = 5.0 \times 10^6\ \text{A/m}^2$$
2. Use the microscopic relation $J = n e v_d$, where $e = 1.6 \times 10^{-19}\ \text{C}$ is electron charge. Rearrange to solve for $v_d$:

   $$v_d = \frac{J}{n e}$$
3. Substitute values to get the final drift velocity:

   $$v_d = \frac{5.0 \times 10^6}{(8.5 \times 10^{28})(1.6 \times 10^{-19})} \approx 3.7 \times 10^{-4}\ \text{m/s}$$

> **Exam tip:** If a question asks for the direction of current, always give the conventional direction (opposite to electron drift velocity). AP exam graders will deduct points for giving electron direction unless explicitly asked.

## Resistance, Resistivity, and Ohm's Law

Resistivity is an intrinsic material property that describes how strongly a material opposes current flow. Resistance is an extrinsic property that depends on both the material's resistivity and the size/shape of the conductor.

**Resistance and Resistivity** — Resistivity $\rho$: intrinsic material property with units $\Omega \cdot \text{m}$. Resistance $R$: total opposition of a conductor to current flow, with units ohms ($\Omega$).

*Notation:* $R$ (resistance), $\rho$ (resistivity)

*Example:* Copper has very low resistivity, making it ideal for manufacturing conducting wires.

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

Where $L$ is the length of the conductor along the direction of current flow, and $A$ is the cross-sectional area perpendicular to current flow. Ohm's law is an empirical law that only applies to ohmic materials, where potential difference is proportional to current. A common misconception is that $R = V/I$ is Ohm's law: this is just the definition of resistance at a given operating point, which holds even for non-ohmic materials like diodes.

$$V = I R$$

**Worked example:** A cylindrical carbon resistor has length $2.0\ \text{cm}$ and radius $0.5\ \text{cm}$. Carbon has resistivity $\rho = 3.5 \times 10^{-5}\ \Omega \cdot \text{m}$. Find the resistance of the resistor. If connected across a $12\ \text{V}$ battery, what current flows through the resistor (assume ohmic behavior)?

1. Convert all units to SI (meters):

   $$L = 0.02\ \text{m}, \quad r = 0.005\ \text{m}$$
2. Calculate cross-sectional area perpendicular to current flow:

   $$A = \pi r^2 = \pi (0.005)^2 \approx 7.85 \times 10^{-5}\ \text{m}^2$$
3. Substitute into the resistance formula:

   $$R = \frac{\rho L}{A} = \frac{(3.5 \times 10^{-5})(0.02)}{7.85 \times 10^{-5}} \approx 8.9 \times 10^{-3}\ \Omega = 9\ \text{m}\Omega$$
4. Use Ohm's law to find current:

   $$I = \frac{V}{R} = \frac{12\ \text{V}}{0.0089\ \Omega} \approx 1350\ \text{A}$$

> **Exam tip:** AP MCQ distractors almost always include the answer you get from leaving length/area units in centimeters. Always convert to meters before calculating resistance, and do a quick unit check to catch this mistake.

## Temperature Dependence of Resistance

Resistivity depends on temperature because higher temperatures increase lattice vibrations in metals, leading to more scattering of charge carriers and higher resistivity. For small temperature changes, we use a linear empirical relation that applies to both resistivity and total resistance:

$$R(T) = R_0 \left[1 + \alpha (T - T_0)\right]$$

Where $R_0$ is resistance at reference temperature $T_0$ (usually $20^\circ\text{C}$), and $\alpha$ is the temperature coefficient of resistivity. For metals, $\alpha$ is positive (resistance increases with temperature); for semiconductors, $\alpha$ is negative (resistance decreases with increasing temperature).

## Power Dissipation (Joule Heating) & Combined Applications

Power dissipated as heat in a resistive material comes from the work done by the electric field on moving charge carriers. Starting from $P = dU/dt = V dQ/dt = VI$, we can rewrite this into three equivalent forms using Ohm's law:

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

> **tip**
>
> Choose the simplest form for your problem: use $P = V^2/R$ when voltage is constant (connected to a battery), use $P = I^2 R$ when current is constant (resistor in series).

**Worked example:** A tungsten light bulb filament has a resistance of $R_0 = 10\ \Omega$ at $20^\circ\text{C}$, and $\alpha = 4.5 \times 10^{-3}\ ^\circ\text{C}^{-1}$. When operating connected to a $120\ \text{V}$ outlet, the filament reaches a temperature of $2500^\circ\text{C}$. Find the power dissipated by the bulb when it is first turned on (still at $20^\circ\text{C}$) and when it is at operating temperature.

1. At turn-on, resistance is $R_0 = 10\ \Omega$, so use $P = V^2/R$ for constant voltage:

   $$P_{\text{turn-on}} = \frac{(120)^2}{10} = 1440\ \text{W}$$
2. Calculate resistance at operating temperature using the temperature dependence formula:

   $$R = R_0\left[1 + \alpha(T-T_0)\right] = 10\left[1 + 0.0045(2500 - 20)\right] = 121.6\ \Omega \approx 122\ \Omega$$
3. Calculate operating power, still at constant 120 V:

   $$P_{\text{operating}} = \frac{(120)^2}{121.6} \approx 118\ \text{W}$$

**Check your understanding**

Test your understanding of resistance change for a stretched wire:

1. A wire of length $L$ and cross-sectional area $A$ has resistance $R$. The wire is drawn uniformly to a new length of $3L$, with total volume remaining constant. What is the new resistance of the wire?

   - $3R$
   - $R/3$
   - $9R$
   - $R/9$

   *Answer:* $9R$

   *Why:* Volume is constant: $LA = 3L A' \implies A' = A/3$. New resistance $R' = \rho (3L)/(A/3) = 9 \rho L/A = 9R$. Distractor A comes from only changing length, not area, D comes from incorrectly inverting the area change.

## Common pitfalls

- **Wrong:** Using electron flow direction as current direction when answering direction questions
  - Why it fails: Students confuse the microscopic motion of negative electrons with the AP convention of conventional current
  - Correct: Always state current direction as the direction positive charge would flow, opposite to electron drift direction, unless explicitly asked for electron flow
- **Wrong:** Claiming any device with $R = V/I$ obeys Ohm's law
  - Why it fails: Students confuse the definition of resistance at an operating point with Ohm's law, which requires proportionality between $V$ and $I$ across all operating points
  - Correct: Only label a device as ohmic (obeying Ohm's law) if $V \propto I$ and $R$ is constant for all applied voltages/currents
- **Wrong:** Using circumference or surface area instead of cross-sectional area for a cylindrical wire
  - Why it fails: Students mix up the surface area of the wire with the cross-sectional area perpendicular to current flow
  - Correct: Draw a diagram marking current direction along the wire length, then calculate area perpendicular to this direction as $A = \pi r^2$ for a cylinder
- **Wrong:** Claiming resistivity changes when you cut a wire in half
  - Why it fails: Students mix up intrinsic resistivity and extrinsic resistance
  - Correct: Resistivity is a property of the material, so it never changes when you change the size/shape of the wire; only resistance changes
- **Wrong:** Forgetting that volume is constant when stretching a wire, and only updating the length term in $R = \rho L/A$
  - Why it fails: Students focus on the obvious length change and miss that stretching reduces cross-sectional area to keep volume constant
  - Correct: Always apply $V = LA = \text{constant}$ to find the new area when a wire is stretched or compressed uniformly

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Electric Current | $I = \frac{dQ}{dt}$ | Scalar, direction = conventional (positive charge flow), units: A = C/s |
| Current-Drift Velocity Relation | $I = n e v_d A$, $\vec{J} = n e \vec{v}_d$ | $n$ = free electron number density, $v_d$ = drift velocity, $J = I/A$ for uniform flow |
| Resistance and Resistivity | $R = \rho \frac{L}{A}$ | $\rho$ = intrinsic material property ($\Omega \cdot$m), $R$ = extrinsic (depends on size), units: $\Omega$ |
| Ohm's Law | $V = IR$, $\vec{J} = \frac{1}{\rho} \vec{E}$ | Only applies to ohmic materials; $R = V/I$ is always true for a given operating point |
| Temperature Dependence | $R(T) = R_0\left[1 + \alpha (T-T_0)\right]$ | $\alpha > 0$ for metals, $\alpha < 0$ for semiconductors; $\Delta T$ same in °C and K |
| Power Dissipation | $P = IV = I^2 R = \frac{V^2}{R}$ | All forms equivalent for ohmic resistors, units: W = J/s |

## What's next

Current and Resistance is the foundational topic for all electric circuit analysis, which makes up the entire rest of Unit 3 for AP Physics C: E&M. Next you will apply the resistance rules you learned here to analyze series and parallel combinations of resistors, then move on to Kirchhoff's rules for multi-loop circuits, and finally RC circuits with time-varying current. Without a solid understanding of how to calculate resistance from material properties, how power is dissipated, and the definition of current, you cannot correctly set up Kirchhoff's current or voltage laws, the core skill for all circuit free-response questions on the AP exam. This topic also connects back to electric fields (the field drives current through the drift velocity relation) and feeds into energy concepts in circuits that are frequently tested across the exam.

- [Kirchhoff's Rules](https://www.owlsprep.com/study/ap-physics-c-em-u3-kirchhoff-s-rules/)
- [Steady-State Direct Current Circuits](https://www.owlsprep.com/study/ap-physics-c-em-u3-steady-state-direct-current-circuits/)
- [Capacitors in Circuits](https://www.owlsprep.com/study/ap-physics-c-em-u3-capacitors-in-circuits/)

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