Aim for FivePhysics C: E&M

Unit 11 · Topic 11.3

11.3 Resistance, Resistivity, and Ohm’s Law

Resistance measures how much an object opposes current. This topic covers how a wire's resistance depends on its material, length and thickness, how temperature changes resistivity, and Ohm's law, ΔV=IR\Delta V = IR, along with how to read current-versus-voltage graphs.

Key terms

  • resistance
  • resistivity
  • Ohm’s law
  • ohmic material
  • current–voltage graph

Resistance and resistivity

Resistance R is defined by the ratio of the potential difference across an object to the current through it: R=ΔVIR = \dfrac{\Delta V}{I}. It's measured in ohms (Ω), and 1 Ω = 1 V/A.

Resistivity ρ\rho is a property of the material itself, measured in Ω·m. Copper has a very low resistivity; rubber and glass have enormous ones. For a uniform wire, the resistance depends on the material and the shape: R=ρℓAR = \dfrac{\rho\ell}{A}, where ℓ\ell is the length and A is the cross-sectional area.

That formula matches intuition. A longer wire is like a longer hallway for charges to push through, so R grows with length. A thicker wire is like a wider hallway with more lanes, so R shrinks as area grows.

If a resistor's resistivity or thickness changes along its length, you can add up thin slices in series: R=∫ρ dxAR = \displaystyle\int \frac{\rho\,dx}{A}. On the microscopic level, resistivity also links the field in the material to the current density, E⃗=ρJ⃗\vec{E} = \rho\vec{J}.

Temperature and resistivity

For metals, resistivity rises as temperature rises. Hotter atoms vibrate more, and the drifting electrons collide with them more often. That's why a light bulb's filament has a much higher resistance when it's glowing than when it's cold.

Some materials behave differently. Semiconductors like silicon usually conduct better as they warm up, because heat frees more charge carriers. You only need the general idea that resistivity depends on temperature, and the metal case in particular.

Ohm's law and ohmic materials

Ohm's law says that for some materials the current is proportional to the potential difference: ΔV=IR\Delta V = IR with R constant. Those are ohmic materials, and most resistors at a steady temperature are close to ohmic.

ΔV=IR\Delta V = IR can always be used to define R at one instant. Ohm's law is the stronger claim that R doesn't change as you change ΔV\Delta V. Real bulbs, diodes and anything that heats up a lot are non-ohmic. On the exam, though, resistors and light bulbs are treated as ohmic, with constant resistance, unless a question says otherwise.

Reading current-versus-voltage graphs

Labs often graph current I on the vertical axis against potential difference ΔV\Delta V on the horizontal axis. For an ohmic resistor, the graph is a straight line through the origin, and its slope is IΔV=1R\dfrac{I}{\Delta V} = \dfrac{1}{R}. A steeper line means a smaller resistance.

Check which variable is on which axis. If the graph is drawn the other way, with ΔV vertical, the slope is R itself.

A filament bulb's graph curves and gets flatter at higher voltage, because its resistance grows as it heats. At any point on a curved graph, the resistance at that moment is ΔV divided by I at that point, not the slope of the curve there.

Change to a wireEffect on R
Double the lengthR doubles
Double the diameterarea × 4, so R ÷ 4
Use a material with twice the resistivityR doubles
Heat a metal wireR increases

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Resistance of a nichrome wire

    A heating wire is 2.0 m long with a diameter of 0.50 mm and a resistivity of 1.1 × 10⁻⁶ Ω·m. Find its resistance and the current through it when a 6.0 V battery is connected across it.

    Show the solution
    1. Step 1: The radius is 0.25 × 10⁻³ m, so A=π(0.25×10−3)2=1.96×10−7 m2A = \pi(0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7}\text{ m}^2.
    2. Step 2: R=ρℓA=(1.1×10−6)(2.0)1.96×10−7≈11 ΩR = \dfrac{\rho\ell}{A} = \dfrac{(1.1 \times 10^{-6})(2.0)}{1.96 \times 10^{-7}} \approx 11\ \Omega.
    3. Step 3: I=ΔVR=6.011.2≈0.54 AI = \dfrac{\Delta V}{R} = \dfrac{6.0}{11.2} \approx 0.54\text{ A}.

    Answer: R ≈ 11 Ω and I ≈ 0.54 A

  2. Example 2Calculator allowed

    Stretching a wire (classic trap)

    A metal wire has resistance R. It is drawn out to twice its original length while its volume stays the same. What is its new resistance?

    Show the solution
    1. Step 1: The volume is length × area. If the length doubles and the volume stays fixed, the area must be cut in half.
    2. Step 2: New resistance: R′=ρ(2ℓ)A/2=4ρℓA=4RR' = \dfrac{\rho(2\ell)}{A/2} = 4\dfrac{\rho\ell}{A} = 4R.
    3. Step 3: The trap is doubling only the length and answering 2R. Stretching a wire changes both length and area, and both changes raise R.

    Answer: 4R

Common mistakes

  • Using the diameter in place of the radius in A=πr2A = \pi r^2, or forgetting to convert millimeters to meters.
  • Mixing up resistance (a property of one object) with resistivity (a property of the material).
  • Reading the slope of an I-versus-ΔV graph as R. When I is on the vertical axis, the slope is 1/R.
  • Assuming every device obeys Ohm's law in a lab. A real bulb's resistance rises as it heats, so its I-versus-ΔV graph curves. (In exam problems, bulbs are ohmic unless the question says they aren't.)

On the exam

  • Lab-based questions often ask you to choose what to graph so the result is a straight line, such as R against ℓ\ell or R against 1/A, and then to find ρ\rho from the slope.
  • Proportional reasoning shows up a lot: one wire is twice as long and half as wide as another; how do their resistances compare? Write the ratio of the two formulas so the constants cancel.

Connected topics

Videos

Check yourself

4 questions on 11.3 Resistance, Resistivity, and Ohm’s Law. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A metal wire has resistance R. It is drawn out so that its length doubles while its volume and resistivity stay the same. What is its new resistance?

Question 2 of 4Calculator allowed

A 2.0 m length of nichrome wire (resistivity 1.1 × 10⁻⁶ Ω·m) has a diameter of 0.50 mm. What is its resistance?

Question 3 of 4Calculator allowed

A cylindrical rod of length L and uniform cross-sectional area A is made of a material whose resistivity increases along the rod as ρ(x)=ρ0(1+xL)\rho(x) = \rho_0\left(1 + \frac{x}{L}\right), where x is measured from one end. What is the rod’s resistance between its ends?

Question 4 of 4Calculator allowed

A student wants to find the resistivity of a metal. She measures the resistance R of several pieces of the same wire, with cross-sectional area A, cut to different lengths ℓ. Which graph gives a straight line, and how does she get the resistivity from it?

0 of 4 answered