Current and Resistance — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Definition of electric current, common direction conventions, drift velocity, current density, resistance, and resistivity; derivation of Ohm's law from microscopic principles; temperature dependence of resistance; and power dissipation in resistive materials.

You should already know: Electric potential and potential difference across conductors. Conservation of charge for isolated systems. Properties of electric fields in conductive materials.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Current and Resistance?

Current and resistance are the two foundational quantities for all electric circuit analysis, and make up approximately 7-8% of the total AP Physics C: E&M exam score, as part of Unit 3: Electric Circuits (which is 20-25% of the exam total). This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as a foundation for more complex circuit problems, but also as standalone MCQ questions testing conceptual understanding. Electric current is defined as the net flow of electric charge through a cross-sectional area over time. Resistance is the opposition of a material object to the flow of electric current through it. AP Physics C exclusively uses the convention of conventional current, where current is defined as the direction positive charge would flow, even though in most metallic conductors the actual moving charge carriers are negatively charged electrons (which flow opposite to conventional current). This topic connects microscopic properties of materials (number density of free charges, scattering of charges by lattice vibrations) to macroscopic behavior (current, voltage, resistance) that we measure in circuits, a connection frequently tested on the exam.

2. Electric Current, Drift Velocity, and Current Density

Electric current is defined as the rate of net charge flow through a surface, with the formula: $$I=\frac{dQ}{dt}$$ Current is a scalar quantity, with units of amperes ($1\text{A}=1\text{C/s}$). For uniform current through a material of constant cross-sectional area $A$, we define current density $\vec{J}$, a vector pointing in the direction of conventional current, with magnitude $J=I/A$. More generally, current is the flux of current density through a surface: $I=\int \vec{J}\cdot d\vec{A}$.

To connect current to microscopic behavior, consider that when an electric field is applied to a conductor, free charge carriers accelerate briefly before scattering off lattice defects, resulting in a net average velocity called drift velocity $v_d$. For a material with $n$ charge carriers per unit volume, each with charge $q$, we derive that $J=nq{v}_d$, so rearranged: $$I=nq{v}_{d}A$$ This relation shows that current depends on the density of charge carriers, their charge, drift speed, and the cross-sectional area of the conductor. For typical metals, drift velocity is very slow (on the order of $10^{-4}\text{m/s}$), much slower than the speed of the electric signal that propagates along the wire.

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. First calculate current density, which is just current divided by cross-sectional area for uniform flow: $J=I/A=10\text{A}/(2.0\times 10^{-6}{\text{m}}^2)=5.0\times 10^6{\text{A/m}}^2$.
2. Use the microscopic relation $J=ne{v}_d$, where $e=1.6\times 10^{-19}\text{C}$ is the charge of an electron. Rearrange to solve for $v_d$: $v_d=J/(ne)$.
3. Substitute values: $v_d=(5.0\times 10^6)/(8.5\times 10^{28}\times 1.6\times 10^{-19})\approx 3.7\times 10^{-4}\text{m/s}$.
4. Both values are consistent with expected behavior for copper wire at 10 A: slow drift velocity, high current density.

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.

3. Resistance, Resistivity, and Ohm's Law

Resistivity $\rho$ is an intrinsic material property that describes how strongly a material opposes current flow, with units of $\Omega \cdot \text{m}$. For a uniform conductor of length $L$ (along the direction of current flow) and cross-sectional area $A$, the total resistance $R$ (units: ohms, $\Omega$) of the object is: $$R=\rho \frac{L}{A}$$ Resistance is an extrinsic property: it depends on both the material (via $\rho$) and the size/shape of the object, while resistivity depends only on the material and temperature.

Ohm's law is an empirical law that states for ohmic materials, the potential difference across the material is proportional to the current through it: $$V=IR$$ Microscopically, Ohm's law comes from $\vec{J}=\sigma \vec{E}$, where $\sigma =1/\rho$ is conductivity. For non-ohmic materials (e.g., diodes, incandescent bulbs at varying temperatures), $V$ is not proportional to $I$, so Ohm's law does not hold. Note that $R=V/I$ is always true for any device at a specific operating point, even non-ohmic ones — this is just the definition of resistance at that point, not a statement of Ohm's law.

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 meters: $L=0.02\text{m}$, $r=0.005\text{m}$.
2. Calculate cross-sectional area: $A=\pi r^2=\pi (0.005{)}^2\approx 7.85\times 10^{-5}{\text{m}}^2$.
3. Substitute into the resistance formula: $R=\rho L/A=(3.5\times 10^{-5}\times 0.02)/(7.85\times 10^{-5})\approx 8.9\times 10^{-3}\Omega \approx 9\text{m}\Omega$.
4. Use Ohm's law to find current: $I=V/R=12\text{V}/0.0089\Omega \approx 1350\text{A}$.

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 on your final answer to catch this mistake.

4. Temperature Dependence and Power Dissipation

Resistivity depends on temperature because higher temperatures increase lattice vibration in metals, which scatters charge carriers more effectively, increasing resistivity. For small temperature changes, the empirical relation is: $$\rho (T)={\rho }_0\left[1+\alpha (T-T_0)\right]$$ where ${\rho }_0$ is resistivity at reference temperature $T_0$ (usually $20^{\circ }\text{C}$), and $\alpha$ is the temperature coefficient of resistivity (units: ${}^{\circ }{\text{C}}^{-1}$). Since resistance is proportional to resistivity, this relation also holds for resistance: $$R(T)=R_0\left[1+\alpha (T-T_0)\right]$$ For metals, $\alpha$ is positive; for semiconductors, $\alpha$ is negative.

Power dissipated as heat (Joule heating) in a resistor is given by three equivalent forms, derived from $P=dU/dt=VdQ/dt=VI$: $$P=VI=I^{2}R=\frac{V^2}{R}$$

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 $P=V^2/R$: $P_{\text{turn-on}}=(120{)}^2/10=1440\text{W}$.
2. Calculate resistance at operating temperature: $R=R_0\left[1+\alpha (T-T_0)\right]=10\left[1+0.0045(2500-20)\right]=10(1+11.16)=121.6\Omega \approx 122\Omega$.
3. Calculate operating power: $P_{\text{operating}}=(120{)}^2/121.6\approx 118\text{W}$.

Exam tip: If a problem states voltage is constant (e.g., connected to a battery), use $P=V^2/R$ to avoid extra calculation. If current is constant (e.g., a resistor in series), use $P=I^{2}R$.

5. Common Pitfalls (and how to avoid them)

• 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 $A=\pi r^2$ 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 $R=\rho L/A$. Why: Students focus on the obvious length change and miss that stretching reduces cross-sectional area to keep volume constant. Correct move: Always apply $V=LA=\text{constant}$ to find the new area when a wire is stretched or compressed uniformly.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

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? A) $3R$ B) $R/3$ C) $9R$ D) $R/9$

Worked Solution: Volume of the wire is $V=LA$, which is constant. When length increases to $3L$, the new cross-sectional area $A^{\prime }$ satisfies $LA=(3L)A^{\prime }$, so $A^{\prime }=A/3$. Substitute into the resistance formula: $R^{\prime }=\rho (3L)/(A/3)=9(\rho L/A)=9R$. The other options correspond to common mistakes: A comes from only changing length, not area, D comes from incorrectly inverting the area change. The correct answer is C.

Question 2 (Free Response)

A cylindrical aluminum power transmission line has resistivity ${\rho }_0=2.82\times 10^{-8}\Omega \cdot \text{m}$ at $20^{\circ }\text{C}$, and temperature coefficient $\alpha =3.9\times 10^{-3}{}^{\circ }{\text{C}}^{-1}$. The wire is $10\text{km}$ long and has a radius of $1.5\text{cm}$. (a) Calculate the resistance of the wire at $20^{\circ }\text{C}$. (b) On a hot day, the wire temperature rises to $45^{\circ }\text{C}$. Calculate the new resistance of the wire. (c) The wire carries a constant current of $200\text{A}$ from the power plant to a city. By what percentage does power dissipated as heat increase between $20^{\circ }\text{C}$ and $45^{\circ }\text{C}$?

Worked Solution: (a) Convert units: $L=10\text{km}=10^4\text{m}$, $r=1.5\text{cm}=0.015\text{m}$. Cross-sectional area: $A=\pi r^2=\pi (0.015{)}^2\approx 7.07\times 10^{-4}{\text{m}}^2$. Resistance: $R_0={\rho }_{0}L/A=(2.82\times 10^{-8}\cdot 10^4)/7.07\times 10^{-4}\approx 0.40\Omega$. (b) Use the temperature dependence formula: $R=R_0\left[1+\alpha (T-T_0)\right]=0.40\left[1+0.0039(45-20)\right]=0.40(1.0975)\approx 0.44\Omega$. (c) Power dissipated is $P=I^{2}R$, and $I$ is constant, so $P/P_0=R/R_0=1.0975$. Percentage increase: $\frac{P-P_0}{P_0}\times 100\%=9.8\%$.

Question 3 (Application / Real-World Style)

An external defibrillator passes a total charge of $0.5\text{C}$ through a patient's chest in $2.0\text{ms}$ to restart a stopped heart. The current path can be modeled as a cylindrical region of length $30\text{cm}$ and diameter $20\text{cm}$, with average resistivity of human tissue equal to $5.0\Omega \cdot \text{m}$. Estimate the average current, effective resistance of the path, and total energy dissipated during the shock.

Worked Solution: Average current is $I_{\text{avg}}=\Delta Q/\Delta t=0.5\text{C}/0.002\text{s}=250\text{A}$. Convert units for resistance calculation: $L=0.30\text{m}$, radius $r=0.10\text{m}$, so $A=\pi r^2\approx 0.0314{\text{m}}^2$. Resistance is $R=\rho L/A=(5.0\cdot 0.30)/0.0314\approx 48\Omega$. Total energy is $U=P\Delta t=I^{2}R\Delta t=(250{)}^2(48)(0.002)=6000\text{J}=6\text{kJ}$. This energy is consistent with the output of standard clinical defibrillators, enough to depolarize the heart and restart normal rhythm.

7. Quick Reference Cheatsheet

8. What's Next

Current and Resistance is the foundational topic for all electric circuit analysis, which makes up the entire rest of Unit 3. 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 FRQ 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.

• Resistors in Series and Parallel
• Kirchhoff's Circuit Rules
• RC Circuits