Resistivity by Meter Bridge: Measuring the Unmeasurable

Resistivity, written ρ, is a material's intrinsic resistance to electric current: it tells how strongly a substance opposes current flow, independent of the shape or size of the sample. Its SI unit is the ohm-metre (Ω·m). Copper's tiny value (~1.7 × 10⁻⁸ Ω·m) is why it is used for wiring, while nichrome's much larger value makes it ideal for heaters.

A short thick wire's resistance is a fraction of an ohm, below what an ohmmeter reads reliably. The meter bridge compares the unknown resistance against a known one in a balanced Wheatstone bridge, and the resistivity follows from the wire's dimensions.

Key Concepts

• Resistance vs resistivity: resistance R depends on length and thickness; resistivity ρ depends only on the material and temperature.
• Wheatstone bridge: four resistors in a diamond; at balance no current flows through the middle branch and the resistance ratios are equal.
• Meter bridge: a 100 cm uniform wire with a sliding jockey; its two segments act as the ratio arms.
• Null point: the jockey position where the galvanometer reads zero; the balance condition is read from the two segment lengths.
• End corrections: small extra resistances at the wire's ends, handled by keeping the null point away from the ends.

Formulas

Balance condition: X / R = l₁ / (100 − l₁), where X is the unknown resistance, R the known resistance, and l₁ the balancing length in centimetres from the zero end. Resistivity: ρ = X A / L, with A the wire's cross-sectional area and L the length of the specimen whose resistance X was measured. For a round wire of diameter d, A = π d² / 4, so ρ = X π d² / (4 L). Assumptions: uniform bridge wire, null point between 30 and 70 cm, and small brief current so heating is negligible.

Worked Example

A wire specimen of length L = 0.50 m and diameter d = 0.40 mm is placed in the meter bridge against a known R = 2.0 Ω. The null point is found at l₁ = 40 cm.

1. Find the unknown resistance: X = R × l₁ / (100 − l₁) = 2.0 × 40 / 60 = 80 / 60 = 1.333 Ω.
2. Find the cross-sectional area: A = π d² / 4 = π × (4.0 × 10⁻⁴)² / 4 = π × 4.0 × 10⁻⁸ = 1.257 × 10⁻⁷ m².
3. Compute resistivity: ρ = X A / L = 1.333 × 1.257 × 10⁻⁷ / 0.50.
4. Multiply: 1.333 × 1.257 × 10⁻⁷ = 1.676 × 10⁻⁷, then divide by 0.50 to get ρ = 3.35 × 10⁻⁷ Ω·m.
5. This value suits a resistance-alloy wire of the kind used in school labs.

Common Misconceptions

• Resistivity is not resistance: doubling length doubles resistance but leaves resistivity, a material property, unchanged.
• A null point near the end is unreliable: end resistances distort the ratio, so choose R to put the balance near the middle.
• The bridge wire is not the specimen. The 100 cm wire is only the measuring scale; the specimen is the separate wire connected in the gap whose ρ is being found.

Real-World Applications

• Checking the purity and quality of copper and aluminium conductors before they are cabled.
• Characterising semiconductor wafers and thin films, where ρ reveals doping levels.
• Selecting heating-element alloys: high ρ materials give the needed resistance in a short coil.
• Earthing and soil-resistivity surveys use the same null-balance idea on a larger scale.

The meter bridge turns a tiny resistance into a length ratio on a metre of wire: balance the bridge, read the null point, combine with the specimen's dimensions, and the resistivity emerges.