A resistor obeys Ohm's law with R = 20\,. Find the current at 10 V. [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do resistance and resistivity arise, and what does Ohm's law say about them?

Topic 11.3 Resistance, Resistivity, and Ohm's Law: relate resistance to resistivity and geometry, apply Ohm's law, and distinguish ohmic from non-ohmic behavior.

A calculus-based answer to AP Physics C E&M Topic 11.3, covering Ohm's law, resistance from resistivity and geometry, the microscopic form J = sigma E, temperature dependence, and ohmic versus non-ohmic devices.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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Quick answer

Resistance $R$ relates the voltage across a conductor to the current through it by Ohm's law, $V = IR$ (resistance in ohms, $\Omega$ ). For a uniform conductor, resistance comes from the material's resistivity $\rho$ and the geometry: $R = \dfrac{\rho L}{A}$ , larger for a longer or thinner wire. The microscopic statement of Ohm's law is $\vec{J} = \sigma\vec{E}$ , where $\sigma = \dfrac{1}{\rho}$ is the conductivity; this links the current density to the field driving it. Resistivity rises with temperature in metals (more thermal scattering), approximately $\rho = \rho_0[1 + \alpha(T - T_0)]$ . A device is ohmic if $R$ is constant, so its $I$ - $V$ graph is a straight line through the origin; non-ohmic devices (diodes, filament bulbs) have a curved $I$ - $V$ graph because $R$ changes with conditions.

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What this topic is asking
Ohm's law and resistance
Resistance from resistivity and geometry
The microscopic form and temperature
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What this topic is asking

The College Board (Topic 11.3) wants you to relate resistance to resistivity and geometry, apply Ohm's law $V = IR$ , understand its microscopic form $\vec{J} = \sigma\vec{E}$ , and distinguish ohmic from non-ohmic devices. Resistance is what opposes current and converts electrical energy to heat.

Ohm's law and resistance

Resistance measures how strongly a conductor opposes current: for a given voltage, a higher $R$ allows less current. Ohm's law is an empirical relation, not a fundamental law, and holds for metals over a wide range but not for every device.

Resistance from resistivity and geometry

The resistance of a uniform conductor depends on the material and its shape:

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

where $\rho$ is the resistivity (an intrinsic property of the material, in $\Omega\,$ m), $L$ the length, and $A$ the cross-sectional area. A longer wire has more resistance; a thicker wire has less. Resistivity ranges over many orders of magnitude, from good conductors (copper, $\sim10^{-8}\,\Omega\,$ m) to insulators ( $\sim10^{16}\,\Omega\,$ m).

The microscopic form and temperature

At the level of the current density and field, Ohm's law reads

\vec{J} = \sigma\vec{E}, \qquad \sigma = \frac{1}{\rho}

with $\sigma$ the conductivity. This says the local current density is proportional to the local field. In a metal, resistivity rises with temperature because hotter ions vibrate more and scatter the drifting electrons more often:

\rho = \rho_0\left[1 + \alpha(T - T_0)\right]

where $\alpha$ is the temperature coefficient.

Resistance and current of a copper wire

A copper wire ( $\rho = 1.7\times10^{-8}\,\Omega\,$ m) is $2.0$ m long with cross-sectional area $1.0\times10^{-6}$ m squared. A $0.50$ V potential difference is applied. Find the resistance and the current.

step 1 Find the resistance

$R = \dfrac{\rho L}{A} = \dfrac{(1.7\times10^{-8})(2.0)}{1.0\times10^{-6}} = 3.4\times10^{-2}\,\Omega$ .

step 2 Apply Ohm's law for the current

$I = \dfrac{V}{R} = \dfrac{0.50}{3.4\times10^{-2}} = 14.7$ A.

step 3 Check the size

The very low resistivity of copper gives a tiny resistance, so even a small voltage drives a large current. This is why copper is used for wiring.

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Q1. A resistor obeys Ohm's law with $R = 20\,\Omega$ . Find the current at $10$ V. [2 points]

Cue. $I = \dfrac{V}{R} = \dfrac{10}{20} = 0.50$ A.

Q2. State how the $I$ - $V$ graph of an ohmic resistor looks. [1 point]

Cue. A straight line through the origin (constant slope $1/R$ ).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). A wire of resistance

R

is stretched to twice its length while its volume stays constant. Its new resistance is (A)

R

(B)

2R

(C)

4R

(D)

R/2

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on resistance and geometry. The answer is (C).

$R = \dfrac{\rho L}{A}$ . Doubling $L$ at constant volume halves $A$ (since $V = LA$ is fixed). So $R \to \rho\dfrac{2L}{A/2} = 4\dfrac{\rho L}{A} = 4R$ . The trap is (B): only the length is considered, missing that the cross-section also shrinks.

AP 2024 (style)4 marksSection II (FRQ, quantitative and conceptual). A cylindrical resistor has length

0.50

m, cross-sectional area

1.0\times10^{-6}

m squared, and resistivity

5.0\times10^{-7}\,\Omega\,\text{m}

. (a) Calculate its resistance. (b) If

3.0

V is applied, calculate the current. (c) State what distinguishes an ohmic from a non-ohmic device on an

I

V

graph.

Show worked answer →

A 4-point FRQ on resistance and Ohm's law.

(a) Resistance (2 points): $R = \dfrac{\rho L}{A} = \dfrac{(5.0\times10^{-7})(0.50)}{1.0\times10^{-6}} = 0.25\,\Omega$ .
(b) Current (1 point): $I = \dfrac{V}{R} = \dfrac{3.0}{0.25} = 12$ A.
(c) Ohmic vs non-ohmic (1 point): an ohmic device gives a straight line through the origin on an $I$ - $V$ graph (constant $R$ ); a non-ohmic device (a diode, a filament bulb) gives a curved line, so $R$ changes with voltage.

Markers reward $R = \rho L/A$ , Ohm's law, and the straight-versus-curved distinction.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)