Free Pearson Edexcel International A Level Physics lesson

Resistance, resistivity and a uniform wire

Separate the material property from the specimen geometry.

What you’ll learn

  • Use R = ρL/A with consistent area units.
  • Determine resistivity from a resistance–length gradient.
  • Explain the potential gradient along a uniform current-carrying wire.

One material, many different resistances

A longer uniform conductor offers a greater distance over which carriers experience scattering, while a larger cross-sectional area provides more parallel conducting paths. At fixed temperature the model R = ρL/A separates these geometric effects from the resistivity ρ of the material. The resistance of one wire is therefore not automatically a material constant. Two wires made of the same material can have very different resistances.

For a circular wire, A = πd²/4. Convert diameter to metres before squaring, or explicitly square the conversion factor. Doubling diameter makes area four times larger and resistance four times smaller if length and material remain unchanged. A radius/diameter confusion creates a factor-of-four error, often much larger than the experimental scatter that students carefully record elsewhere.

The model assumes reasonably uniform area, composition and temperature. A stretched wire may change both length and area; heating changes resistivity as well as potentially geometry. State which changes are neglected before applying simple ratios. The resistivity unit Ω m follows from ρ = RA/L because area contributes m² and length contributes m. This dimensional check helps reject a mistakenly calculated result in Ω m⁻¹.

R = ρL/A; A = πd²/4; ρ = RA/L

Symbols and units

  • R is resistance in Ω; ρ is material resistivity in Ω m; L is wire length in m; A is cross-sectional area in m²; d is diameter in m.

How the equation explains the idea

  1. For a uniform wire at fixed temperature, R = ρL/A. A circular cross section gives A = π(d/2)² = πd²/4.
  2. Doubling L doubles R, whereas doubling d quadruples A and reduces R to one quarter. Resistivity stays unchanged when only geometry changes.

Worked example

A thin alloy specimen

  1. A wire has length 1.20 m, diameter 0.40 mm and resistance 4.8 Ω.
  2. Area = π(0.40 × 10⁻³)²/4 = 1.257 × 10⁻⁷ m².
  3. Resistivity = 4.8 × 1.257 × 10⁻⁷/1.20 = 5.03 × 10⁻⁷ Ω m.

Geometry is needed before resistance can identify a material property.

Keep this idea. Use the full geometry and square the diameter conversion when calculating resistivity.

Core practical 7: measure a resistance–length gradient

Secure a uniform wire beside a ruler and connect it to a low-voltage supply with a current limiter and an ammeter in series. Connect a voltmeter across the selected length, placing contacts at measured positions. At several lengths, record voltage and current and calculate R = V/I. A resistance-versus-length graph should be approximately linear if area, material and temperature remain constant.

Measure diameter with a micrometer at several positions and orientations, checking and correcting its zero reading. This tests whether the specimen is uniform and reduces the influence of a single misleading diameter. Find area from the representative diameter and multiply the resistance–length gradient by that area to obtain resistivity. Use widely separated points on the best-fit line for the gradient rather than two adjacent raw observations.

Keep currents small and disconnect between readings to limit heating. Contacts must be stable, but excessive pressure can damage a thin wire or change the effective measurement points. Short lengths can make lead and contact contributions a large fraction of total resistance, so examine the intercept and choose a useful range. Repeat measurements and record raw values before processing; a straight graph alone does not prove that the temperature stayed fixed throughout the sequence.

Rmeasured = (ρ/A)L + Rextra; ρ = A × gradient of R against L

Symbols and units

  • Rmeasured and Rextra are resistances in Ω; L is selected wire length in m; A is wire area in m²; ρ is resistivity in Ω m.

How the equation explains the idea

  1. A fixed contact/lead contribution adds an intercept: Rmeasured = (ρ/A)L + Rextra. Subtract two readings to cancel that intercept.
  2. The R-versus-L gradient is ρ/A in Ω m⁻¹. Multiply by measured A to obtain ρ; do not multiply the intercept by area to estimate resistivity.
An extra contact resistance changes the intercept

Conditions. Uniform wire with ρ = 1.0 × 10⁻⁶ Ω m and cross-sectional area A = 2.5 × 10⁻⁷ m² at constant temperature. Compare zero extra resistance and a fixed 1 Ω contribution.

Wire length / m; Measured resistance / Ω.

Both lines have slope 4 Ω m⁻¹. Multiplying by A gives the same resistivity 1.0 × 10⁻⁶ Ω m. The 1 Ω intercept represents the added circuit contribution, not the wire’s resistivity.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Recover resistivity from a slope

  1. A resistance-versus-length fitted line has gradient 3.0 Ω m⁻¹.
  2. The wire area from corrected diameter measurements is 1.5 × 10⁻⁷ m².
  3. Resistivity = 3.0 × 1.5 × 10⁻⁷ = 4.5 × 10⁻⁷ Ω m.

A fixed intercept contribution does not belong in the gradient-derived material resistivity.

Keep this idea. Use a low-heating resistance–length series and independently measured cross-sectional area.

Potential changes steadily along a uniform wire

In a uniform wire carrying a steady current, equal lengths have equal resistance when material, cross-sectional area and temperature are unchanged. Each segment therefore has the same potential drop per unit length because Vsegment = IRsegment. A plot of potential against distance along the wire is a straight line, with sign depending on the direction chosen and the reference potential.

The reference point sets the numerical zero of potential but does not alter potential differences. Moving the reference adds a constant to every plotted potential, changing the intercept while leaving the gradient unchanged. The magnitude of the gradient is Iρ/A. This idea underlies using a uniform wire as a continuously adjustable potential divider, where a sliding contact selects a fraction of the total resistance and potential drop.

Uniform spacing does not guarantee uniform potential drop if a wire narrows, heats unevenly or contains different materials. Nor does the unloaded linear profile necessarily survive attaching a low-resistance device to a tapping point; the added path changes current distribution. State the uniform-wire and high-input-resistance measurement assumptions. A voltmeter’s nonzero loading can be analysed as an extra branch rather than being dismissed as an unexplained experimental error.

For a uniform unloaded wire: |ΔV/Δx| = Iρ/A; selected drop/total drop = selected length/total length

Symbols and units

  • I is steady current in A; ρ is resistivity in Ω m; A is area in m²; Δx is length in m; ΔV is potential change in V.

How the equation explains the idea

  1. A wire segment has resistance ρΔx/A, so its voltage-drop magnitude is IρΔx/A.
  2. For constant I, ρ and A, |ΔV/Δx| is constant. A quarter of the length has a quarter of the total drop only if loading does not change the current distribution.

Worked example

Select a fraction of the wire

  1. An unloaded uniform 1.20 m wire has 6.0 V across its ends.
  2. Its potential-drop magnitude per metre is 6.0/1.20 = 5.0 V m⁻¹.
  3. A 0.30 m segment therefore has 1.5 V across it.

The fractional voltage follows fractional length only under the uniform, negligibly loaded assumptions.

Keep this idea. Equal uniform-wire lengths give equal potential drops; reference shifts and loading have different effects.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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