Current and Resistance

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Electric Circuits· 14 min read

1. Electric Current, Drift Velocity, and Current Density★★☆☆☆⏱ 4 min

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.

📘 Definition

Electric Current

$I$

The rate of net charge flow through a given cross-sectional surface

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 $J$ , 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 $2.0 \times 1 0^{- 6} m^{2}$ carries a current of $10 A$ . Copper has $n = 8.5 \times 1 0^{28}$ 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 $e = 1.6 \times 1 0^{- 19} C$ 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.

2. Resistance, Resistivity, and Ohm's Law★★☆☆☆⏱ 4 min

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.

📘 Definition

Resistance and Resistivity

$R$ (resistance), $ρ$ (resistivity)

Resistivity $ρ$ : intrinsic material property with units $Ω \cdot m$ . Resistance $R$ : total opposition of a conductor to current flow, with units ohms ( $Ω$ ).

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 $2.0 cm$ and radius $0.5 cm$ . Carbon has resistivity $ρ = 3.5 \times 1 0^{- 5} Ω \cdot m$ . Find the resistance of the resistor. If connected across a $12 V$ 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.

3. Temperature Dependence of Resistance★★★☆☆⏱ 3 min

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 $2 0^{\circ} C$ ), and $α$ is the temperature coefficient of resistivity. For metals, $α$ is positive (resistance increases with temperature); for semiconductors, $α$ is negative (resistance decreases with increasing temperature).

4. Power Dissipation (Joule Heating) & Combined Applications★★★☆☆⏱ 3 min

Power dissipated as heat in a resistive material comes from the work done by the electric field on moving charge carriers. Starting from $P = d U / d t = V d Q / d t = V I$ , we can rewrite this into three equivalent forms using Ohm's law:

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

📐 Worked Example

A tungsten light bulb filament has a resistance of $R_{0} = 10 Ω$ at $2 0^{\circ} C$ , and $α = 4.5 \times 1 0^{- 3}^{\circ} C^{- 1}$ . When operating connected to a $120 V$ outlet, the filament reaches a temperature of $250 0^{\circ} C$ . Find the power dissipated by the bulb when it is first turned on (still at $2 0^{\circ} C$ ) and when it is at operating temperature.

1
At turn-on, resistance is $R_{0} = 10 Ω$ , 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:
$P_{\text{operating}} = \frac{(120)^2}{121.6} \approx 118\ \text{W}$

✓ Quick check

Test your understanding of resistance change for a stretched wire:

A wire of length $L$ and cross-sectional area $A$ has resistance $R$ . The wire is drawn uniformly to a new length of $3 L$ , with total volume remaining constant. What is the new resistance of the wire?
- $3 R$
- $R /3$
- $9 R$
- $R /9$
Reveal answer
2 —
Volume is constant: $L A = 3 L A^{'} ⟹ A^{'} = A /3$ . New resistance $R^{'} = ρ (3 L) / (A /3) = 9 ρ L / A = 9 R$ . Distractor A comes from only changing length, not area, D comes from incorrectly inverting the area change.

5. Common Pitfalls

Wrong move:

Using electron flow direction as current direction when answering direction questions

Why:

Students confuse the microscopic motion of negative electrons with the AP convention of conventional current

Correct move:

Always state current direction as the direction positive charge would flow, opposite to electron drift direction, unless explicitly asked for electron flow

Wrong move:

Claiming any device with $R = V / I$ obeys Ohm's law

Why:

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 move:

Only label a device as ohmic (obeying Ohm's law) if $V \propto I$ and $R$ is constant for all applied voltages/currents

Wrong move:

Using circumference or surface area instead of cross-sectional area for a cylindrical wire

Why:

Students mix up the surface area of the wire with the cross-sectional area perpendicular to current flow

Correct move:

Draw a diagram marking current direction along the wire length, then calculate area perpendicular to this direction as $A = π r^{2}$ for a cylinder

Wrong move:

Claiming resistivity changes when you cut a wire in half

Why:

Students mix up intrinsic resistivity and extrinsic resistance

Correct move:

Resistivity is a property of the material, so it never changes when you change the size/shape of the wire; only resistance changes

Wrong move:

Forgetting that volume is constant when stretching a wire, and only updating the length term in $R = ρ L / A$

Why:

Students focus on the obvious length change and miss that stretching reduces cross-sectional area to keep volume constant

Correct move:

Always apply $V = L A = constant$ to find the new area when a wire is stretched or compressed uniformly

6. Quick Reference Cheatsheet

Category	Formula	Notes
Electric Current	$I = \frac{d Q}{d t}$	Scalar, direction = conventional (positive charge flow), units: A = C/s
Current-Drift Velocity Relation	$I = n e v_{d} A$ , $J = n e v_{d}$	$n$ = free electron number density, $v_{d}$ = drift velocity, $J = I / A$ for uniform flow
Resistance and Resistivity	$R = ρ \frac{L}{A}$	$ρ$ = intrinsic material property ( $Ω \cdot$ m), $R$ = extrinsic (depends on size), units: $Ω$
Ohm's Law	$V = I R$ , $J = \frac{1}{ρ} E$	Only applies to ohmic materials; $R = V / I$ is always true for a given operating point
Temperature Dependence	$R (T) = R_{0} [1 + α (T - T_{0})]$	$α > 0$ for metals, $α < 0$ for semiconductors; $Δ T$ same in °C and K
Power Dissipation	$P = I V = I^{2} R = \frac{V ^{2}}{R}$	All forms equivalent for ohmic resistors, units: W = J/s

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Drift velocity calculation
2022 · FRQ
Temperature dependent resistance
2021 · MCQ
Resistance of stretched wire

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.